Electricity · Class 10 Science: 217 MCQs with answers

Showing 200 of 217 multiple-choice questions from this kit.

Warm-up · 24 questions

Q1. No current flows in a circuit when the plug key is open because:

Beginner
A the cell drains B resistance becomes zero C the conducting path is broken D electrons vanish
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the conducting path is broken

Current needs a continuous closed loop.

Q2. Define electric current. State and define its SI unit.

Beginner
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Q3. Define potential difference between two points. State and define its SI unit.

Beginner
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Q4. Distinguish between conventional current and electron flow.

Beginner
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Q5. A 6 V battery moves charge through itself. How much energy does every coulomb pick up?

Beginner
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Q6. Name the components needed to make the simplest working circuit that lights a bulb, and state the function of each.

Beginner
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Q7. 1 ohm equals:

Beginner
A 1 A/V B 1 J/C C 1 C/s D 1 V/A
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1 V/A

One volt per ampere.

Q8. The resistance of a wire is directly proportional to its:

Beginner
A area B temperature squared C length D colour
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length

R ∝ l — longer wire, more collisions.

Q9. If the cross-section area of a wire increases, its resistance:

Beginner
A increases B stays the same C becomes zero D decreases
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decreases

R ∝ 1/A — a wider road for electrons.

Q10. A device used to change the current in a circuit gradually is a:

Beginner
A rheostat B fuse C voltmeter D cell
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rheostat

A rheostat is a variable resistance with a slider.

Q11. State Ohm's law. Draw (describe) the shape of the V–I graph for an ohmic conductor and state what its slope gives.

Beginner
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Q12. Define resistance of a conductor and define 1 ohm.

Beginner
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Q13. On what factors does the resistance of a uniform metallic wire depend? Write the formula.

Beginner
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Q14. The equivalent of 2 Ω, 3 Ω and 5 Ω in series is:

Beginner
A 30 Ω B 1 Ω C 2.5 Ω D 10 Ω
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10 Ω

Rs = 2 + 3 + 5 = 10 Ω.

Q15. The equivalent of two 10 Ω resistors in parallel is:

Beginner
A 20 Ω B 10 Ω C 5 Ω D 100 Ω
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5 Ω

Equal pair: R/n = 10/2 = 5 Ω.

Q16. In parallel, the branch with the smallest resistance carries the:

Beginner
A smallest current B same current as others C no current D largest current
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largest current

I = V/R with common V — small R, large I.

Q17. Five 1 Ω resistors in series give:

Beginner
A 0.2 Ω B 1 Ω C 25 Ω D 5 Ω
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5 Ω

nR = 5 × 1 = 5 Ω.

Q18. Five 1 Ω resistors in parallel give:

Beginner
A 5 Ω B 1 Ω C 0.2 Ω D 0.5 Ω
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0.2 Ω

R/n = 1/5 = 0.2 Ω.

Q19. Why is the series arrangement not used for connecting domestic electrical appliances?

Beginner
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Q20. What happens to the total resistance and the current drawn when more resistors are added (a) in series, (b) in parallel?

Beginner
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Q21. Two lamps rated for 220 V are connected in parallel to a 220 V supply. State two advantages of this connection over series.

Beginner
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Q22. 1 kWh equals:

Beginner
A 3.6 × 10³ J B 6.3 × 10⁶ J C 10⁶ J D 3.6 × 10⁶ J
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3.6 × 10⁶ J

1000 W × 3600 s = 3.6 × 10⁶ J.

Q23. A fuse is connected in:

Beginner
A series with the live wire B parallel with the appliance C series with the earth wire D parallel with the switch
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series with the live wire

It must carry (and cut) the full circuit current.

Q24. The formula P = V²/R is most convenient when you know:

Beginner
A current and resistance B charge and time C current and time D voltage and resistance
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voltage and resistance

V and R plug straight in — ideal for mains appliances.

Building up · 83 questions

Q25. A charge of one coulomb passes a point in one second. The current is:

Intermediate
A 1 V B 1 A C 1 C D 1 W
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1 A

One ampere is defined as one coulomb per second, so I = Q/t = 1/1 = 1 A.

Q26. A torch bulb draws 300 mA. Expressed in amperes this is:

Intermediate
A 0.3 A B 3 A C 0.03 A D 300 A
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0.3 A

1 mA = 10⁻³ A, so 300 mA = 300 × 10⁻³ = 0.3 A.

Q27. A current of 2 A flows through a lamp for half a minute. The charge that passes through it is:

Intermediate
A 1 C B 30 C C 120 C D 60 C
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60 C

Q = It with the time in seconds: Q = 2 × 30 = 60 C. Leaving the time in minutes is the usual slip.

Q28. A battery does 24 J of work moving 3 C of charge through a circuit. The potential difference is:

Intermediate
A 3 V B 8 V C 24 V D 72 V
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8 V

V = W/Q = 24/3 = 8 V. One volt is one joule per coulomb.

Q29. In a circuit diagram, conventional current is shown leaving the cell from the terminal drawn as:

Intermediate
A the long thin line, the positive terminal B the short thick line, the negative terminal C either line, since the symbol is symmetric D the line nearest the switch
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the long thin line, the positive terminal

The long thin line is the positive terminal, and conventional current flows out of it into the external circuit.

Q30. A current of 0.2 A flows for 5 minutes. The charge that passes is:

Intermediate
A 1 C B 60 C C 12 C D 0.04 C
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60 C

Q = It with the time in seconds: Q = 0.2 × 300 = 60 C.

Q31. 10 J of work moves 2 C of charge between two points. The p.d. is:

Intermediate
A 20 V B 0.2 V C 5 V D 12 V
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5 V

V = W/Q = 10/2 = 5 V.

Q32. A 6 V battery gives every coulomb passing through it:

Intermediate
A 6 A of current B 6 C of charge C 6 W of power D 6 J of energy
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6 J of energy

Volts are joules per coulomb — each coulomb receives 6 J.

Q33. How much energy is given to 4 C of charge by a 12 V battery?

Intermediate
A 3 J B 16 J C 0.33 J D 48 J
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48 J

W = VQ = 12 × 4 = 48 J.

Q34. The number of electrons in 1 coulomb of charge is about:

Intermediate
A 6.25 × 10¹⁸ B 1.6 × 10⁻¹⁹ C 9 × 10⁹ D 3 × 10⁸
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6.25 × 10¹⁸

n = 1/(1.6 × 10⁻¹⁹) ≈ 6.25 × 10¹⁸.

Q35. The same current flows through every part of a simple series loop because:

Intermediate
A voltage is constant B charge is conserved and has only one path C resistance is zero D electrons speed up
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charge is conserved and has only one path

Charges cannot pile up or vanish — one path forces one common current.

Q36. The bulb lights the instant the switch closes because:

Intermediate
A electrons travel at light speed B the electric field is set up through the whole circuit almost instantly C the filament stores light D current jumps the gap
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the electric field is set up through the whole circuit almost instantly

All free electrons everywhere start drifting at once, though each drifts slowly.

Q37. In a torch circuit, the correct order along the loop is:

Intermediate
A cell → switch → bulb → back to cell B bulb must come before the switch C switch must touch the cell D any order works, as long as the loop is closed
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any order works, as long as the loop is closed

In a single series loop, component order doesn't matter — the same current flows regardless.

Q38. An ammeter mistakenly connected in parallel across a bulb causes:

Intermediate
A a very large current through the ammeter (short circuit) B no change C the bulb to brighten safely D current to stop
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a very large current through the ammeter (short circuit)

Its near-zero resistance bypasses the bulb — dangerous overload.

Q39. Which quantity is the SAME for the two bulbs when they are connected one after another (series) to a cell?

Intermediate
A current B potential difference C power D brightness always
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current

Series = one path = one current through both.

Q40. In the water analogy of circuits, the pump corresponds to the:

Intermediate
A wire B switch C bulb D cell
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cell

The pump maintains pressure difference as the cell maintains potential difference.

Q41. Why is an ammeter connected in series and made with very low resistance?

Intermediate
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Q42. Why is a voltmeter connected in parallel and made with very high resistance?

Intermediate
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Q43. A current of 0.5 A is drawn by a filament bulb for 10 minutes. Find the amount of charge that flows through the circuit.

Intermediate
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Q44. A charge of 3 C is carried between two points separated by 15 V. What work does that take?

Intermediate
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Q45. Explain, using the water analogy, the roles of the cell, potential difference and current in an electric circuit. Then state the exact definitions and units of current and potential difference.

Intermediate
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Analogy mapped correctly: pump/cell, pressure/p.d., flow/current (2), current definition + ampere (1.5), p.d. definition + volt (1.5).

Q46. Draw (describe) a circuit containing a battery of two cells, a plug key, an ammeter, a resistor, and a voltmeter measuring the p.d. across the resistor. Explain the placement of both meters, the consequences of interchanging them, and the direction of the current.

Intermediate
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Correct loop with all five elements (1.5), series/parallel placements with reasons (1.5), both interchange consequences (1.5), conventional direction stated (0.5).

Q47. A conductor carries 0.5 A when 6 V is applied across it. Its resistance is:

Intermediate
A 3 Ω B 6 Ω C 12 Ω D 0.08 Ω
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12 Ω

R = V/I = 6/0.5 = 12 Ω. One ohm is one volt per ampere.

Q48. Silver conducts better than copper, yet household wiring is copper. The deciding factor is that silver is:

Intermediate
A a poor conductor at high temperature B far more expensive C too brittle to draw into wire D magnetic
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far more expensive

Silver has the lowest resistivity of all metals, but cost rules it out for kilometres of house wiring. Copper is the practical compromise.

Q49. A p.d. of 10 V drives 2 A through a conductor. Its resistance is:

Intermediate
A 20 Ω B 0.2 Ω C 12 Ω D 5 Ω
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5 Ω

R = V/I = 10/2 = 5 Ω.

Q50. The current through a 30 Ω resistor connected to a 6 V battery is:

Intermediate
A 5 A B 180 A C 0.5 A D 0.2 A
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0.2 A

I = V/R = 6/30 = 0.2 A.

Q51. To drive 0.5 A through a 40 Ω element, the required p.d. is:

Intermediate
A 80 V B 8 V C 20 V D 0.0125 V
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20 V

V = IR = 0.5 × 40 = 20 V.

Q52. If the p.d. across a fixed resistor is halved, the current:

Intermediate
A halves B doubles C stays the same D becomes zero
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halves

I ∝ V for fixed R.

Q53. Two wires of the same material and length; wire B is twice as thick (twice the diameter). Then:

Intermediate
A R_B = 2R_A B R_B = R_A/2 C R_B = 4R_A D R_B = R_A/4
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R_B = R_A/4

Double diameter → 4× area → quarter resistance.

Q54. A wire of resistance 8 Ω is cut into four equal pieces. Each piece has resistance:

Intermediate
A 32 Ω B 8 Ω C 2 Ω D 0.5 Ω
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2 Ω

Quarter length → quarter resistance: 8/4 = 2 Ω.

Q55. The filament of a glowing bulb does not obey Ohm's law across its range because:

Intermediate
A tungsten is an insulator B its temperature — and hence resistance — keeps rising C current flows backwards D the glass interferes
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its temperature — and hence resistance — keeps rising

The constant-temperature condition is violated as it heats.

Q56. On an I–V graph (I on the y-axis), the slope equals:

Intermediate
A R B 1/R C V² D power
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1/R

I = V/R, so slope = I/V = 1/R here.

Q57. Of two V–I lines drawn for resistors P and Q (V on y-axis), P's line is steeper. Then:

Intermediate
A P has less resistance B both are equal C P has more resistance D cannot say
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P has more resistance

Steeper V–I slope = larger V/I = larger R.

Q58. The resistance of a metal wire increases when heated because:

Intermediate
A atomic vibrations increase, causing more collisions B electrons slow down permanently C the wire gets longer only D electrons leave the wire
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atomic vibrations increase, causing more collisions

More thermal agitation = more electron-atom collisions.

Q59. Manganin and constantan are used in standard resistance coils because their resistivity:

Intermediate
A is the lowest known B hardly changes with temperature C is infinite D changes rapidly with temperature
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hardly changes with temperature

Stable resistance = reliable standard.

Q60. A wire has R = 4 Ω. Another wire of the SAME material with twice the length and twice the area has resistance:

Intermediate
A 16 Ω B 8 Ω C 1 Ω D 4 Ω
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4 Ω

R' = ρ(2l)/(2A) = ρl/A = 4 Ω — the changes cancel.

Q61. Which has the HIGHEST resistivity?

Intermediate
A rubber B copper C nichrome D aluminium
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rubber

Insulators like rubber sit 18+ orders of magnitude above metals.

Q62. The resistivity of a copper wire 10 m long is ρ. For a copper wire 20 m long of the same thickness, the resistivity is:

Intermediate
A 2ρ B ρ/2 C 4ρ D ρ
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ρ

Resistivity is a material property — unchanged by length.

Q63. Define resistivity. Why is it called a characteristic property of the material?

Intermediate
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Q64. Why are the coils of electric heating devices made of an alloy rather than a pure metal?

Intermediate
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Q65. How much current flows through an electric bulb of resistance 1100 Ω when connected to a 220 V supply?

Intermediate
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Q66. The V–I graphs of two conductors A and B are drawn with V on the y-axis; A's line is steeper. Which conductor has more resistance, and why?

Intermediate
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Q67. Why does the connecting cord of an electric heater not glow while its heating element does?

Intermediate
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Q68. Calculate the resistance of a 2 m long nichrome wire of cross-section 10⁻⁶ m², given nichrome's resistivity is 1.1 × 10⁻⁶ Ω·m.

Intermediate
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Q69. Describe an experiment to verify Ohm's law. Include the circuit, procedure, observations, the graph, and the conclusion.

Intermediate
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Correct circuit incl. meter placements (1.5), stepwise procedure (1), constant V/I observation (1), straight-line graph through origin (1), conclusion + temperature precaution (0.5).

Q70. (a) A 9 V battery drives a current through a series circuit containing a rheostat set to 30 Ω. Find the current. (b) The rheostat is now adjusted to 15 Ω. What is the new current? (c) What must the rheostat be set to for a current of 0.9 A? (d) State what this experiment demonstrates about the role of a rheostat.

Intermediate
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0.3 A (1), 0.6 A with inverse observation (1.5), 10 Ω (1.5), rheostat role stated via Ohm's law (1).

Q71. Three resistors are joined in parallel across a 12 V battery. The potential difference across the middle one is:

Intermediate
A 4 V B 36 V C 12 V D it depends on its resistance
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12 V

Every branch of a parallel combination is connected across the same two points, so each gets the full 12 V regardless of its resistance.

Q72. An old string of fairy lights goes completely dark when a single bulb fails. This tells you the bulbs were connected:

Intermediate
A in parallel B in series C half in series and half in parallel D directly to separate cells
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in series

Only a series circuit has a single path, so one break stops the current everywhere. Modern strings are wired so that one failure does not do this.

Q73. A 6 Ω and a 12 Ω resistor are joined in parallel. Their equivalent resistance is:

Intermediate
A 18 Ω B 9 Ω C 2 Ω D 4 Ω
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4 Ω

Product over sum: (6 × 12)/(6 + 12) = 72/18 = 4 Ω — smaller than 6 Ω, as a parallel combination must be.

Q74. The equivalent of 4 Ω and 12 Ω in parallel is:

Intermediate
A 16 Ω B 3 Ω C 8 Ω D 48 Ω
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3 Ω

48/16 = 3 Ω — and correctly below the smaller (4 Ω).

Q75. A 6 Ω and a 3 Ω resistor in parallel are joined in series with a 4 Ω resistor. The total resistance is:

Intermediate
A 6 Ω B 13 Ω C 2 Ω D 9 Ω
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6 Ω

3∥6 = 2 Ω; 2 + 4 = 6 Ω.

Q76. Through the 3 Ω branch of that pair flows:

Intermediate
A 2/3 A B 2 A C 1 A D 4/3 A
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4/3 A

I = V/R = 4/3 ≈ 1.33 A; the 6 Ω takes 0.67 A.

Q77. Two bulbs of resistance 10 Ω and 20 Ω in SERIES on a battery: which glows brighter?

Intermediate
A the 20 Ω B the 10 Ω C equal brightness D neither glows
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the 20 Ω

Same I; power = I²R is larger for the larger R in series.

Q78. Adding one more resistor in parallel to an existing circuit makes the total resistance:

Intermediate
A increase B decrease C stay the same D double
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decrease

An extra path always eases the total flow.

Q79. Adding one more resistor in series makes the current drawn from the battery:

Intermediate
A increase B stay the same C reverse D decrease
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decrease

Total R rises, so I = V/R falls.

Q80. To get 5 Ω from two 10 Ω resistors, connect them in:

Intermediate
A series B parallel C either way D impossible
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parallel

10∥10 = 5 Ω.

Q81. To get 15 Ω from 10 Ω and 5 Ω resistors, connect them in:

Intermediate
A series B parallel C either D impossible
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series

10 + 5 = 15 Ω in series.

Q82. Which single value CANNOT be made from two 6 Ω resistors?

Intermediate
A 6 Ω (using both) B 12 Ω C 3 Ω D any of these can be made
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6 Ω (using both)

Series = 12, parallel = 3; there is no two-resistor arrangement of both that returns 6 Ω.

Q83. Three 3 Ω resistors: two in parallel, that pair in series with the third. Total:

Intermediate
A 9 Ω B 1 Ω C 6 Ω D 4.5 Ω
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4.5 Ω

3∥3 = 1.5; 1.5 + 3 = 4.5 Ω.

Q84. An ammeter reads 3 A entering a junction of two branches carrying I₁ and 1.2 A. I₁ is:

Intermediate
A 1.8 A B 4.2 A C 3 A D 1.2 A
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1.8 A

Currents at a junction add: I₁ = 3 − 1.2 = 1.8 A.

Q85. Derive the expression for the equivalent resistance of three resistors connected in series.

Intermediate
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Q86. Derive the expression for the equivalent resistance of three resistors connected in parallel.

Intermediate
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Q87. Find the equivalent resistance of 5 Ω and 20 Ω connected in parallel, and the current drawn from a 12 V battery connected across them.

Intermediate
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Q88. Three resistors of 10 Ω, 20 Ω and 30 Ω are in series across a 12 V battery. Find the current and the p.d. across each resistor.

Intermediate
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Q89. A current of 6 A enters a parallel combination of 2 Ω and 4 Ω. Find the current in each branch.

Intermediate
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Q90. An appliance converts 500 joules of electrical energy every second. Its power is:

Intermediate
A 500 J B 500 V C 500 A D 500 W
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500 W

Power is energy per unit time, and one watt is one joule per second, so 500 J/s = 500 W.

Q91. An electric iron, an electric fan and a tube light all run from the same supply. Which group names the MAIN energy conversion in each correctly?

Intermediate
A iron: light; fan: heat; tube light: kinetic B iron: heat; fan: kinetic; tube light: light C iron: kinetic; fan: light; tube light: heat D iron: heat; fan: light; tube light: kinetic
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iron: heat; fan: kinetic; tube light: light

The iron uses the heating effect, the fan the magnetic effect to turn a motor, and the tube light a gas discharge that produces light.

Q92. The heat produced by 2 A through 10 Ω in 5 s is:

Intermediate
A 100 J B 50 J C 200 J D 400 J
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200 J

H = I²Rt = 4 × 10 × 5 = 200 J.

Q93. The power of an appliance drawing 3 A at 220 V is:

Intermediate
A 73 W B 220 W C 660 W D 990 W
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660 W

P = VI = 220 × 3 = 660 W.

Q94. The resistance of a 220 V, 550 W appliance is:

Intermediate
A 44 Ω B 2.5 Ω C 121 Ω D 88 Ω
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88 Ω

R = V²/P = 48400/550 = 88 Ω.

Q95. The current drawn by a 220 V, 440 W appliance is:

Intermediate
A 0.5 A B 2 A C 4 A D 20 A
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2 A

I = P/V = 440/220 = 2 A.

Q96. A suitable fuse for a 220 V, 1000 W heater (I ≈ 4.5 A) is:

Intermediate
A 1 A B 5 A C 3 A D 15 A
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5 A

Just above the normal current — 5 A. (15 A would defeat the protection.)

Q97. A 100 W bulb runs 10 hours. Energy consumed:

Intermediate
A 1 kWh B 10 kWh C 0.1 kWh D 100 kWh
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1 kWh

0.1 kW × 10 h = 1 kWh — one unit.

Q98. A 1.5 kW geyser runs 40 minutes. Units consumed:

Intermediate
A 1 B 60 C 1.5 D 0.5
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1

1.5 kW × (2/3) h = 1 kWh = 1 unit.

Q99. At ₹5 per unit, a 2 kW heater running 3 hours daily for 10 days costs:

Intermediate
A ₹30 B ₹150 C ₹300 D ₹60
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₹300

2 × 3 × 10 = 60 units; 60 × 5 = ₹300.

Q100. Two bulbs, 40 W and 60 W, both rated 220 V. Which has more resistance?

Intermediate
A 60 W B equal C 40 W D cannot say
Show answer & explanation

40 W

R = V²/P — the lower-power bulb has the higher resistance.

Q101. When the voltage across a fixed resistor doubles, the power dissipated becomes:

Intermediate
A double B four times C half D unchanged
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four times

P = V²/R — square the doubling.

Q102. The element of a room heater stays red-hot but its copper leads stay cool because the leads have:

Intermediate
A thicker insulation B much smaller resistance C much larger resistance D no current
Show answer & explanation

much smaller resistance

Same I; H ∝ R puts the heat in the nichrome.

Q103. An electric kettle rated 220 V, 2.2 kW draws a current of:

Intermediate
A 10 A B 1 A C 22 A D 0.1 A
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10 A

I = P/V = 2200/220 = 10 A.

Q104. The energy consumed by a 40 W bulb in 1 minute is:

Intermediate
A 40 J B 240 J C 2400 J D 4 J
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2400 J

E = Pt = 40 × 60 = 2400 J.

Q105. Long-distance power lines use high voltage mainly to:

Intermediate
A reduce I²R losses by lowering the current B make current flow faster C brighten bulbs D avoid lightning
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reduce I²R losses by lowering the current

Same power at higher V means lower I — losses fall as I².

Q106. An MCB differs from a fuse in that it:

Intermediate
A allows higher currents forever B can be reset instead of replaced C has no rating D is connected in parallel
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can be reset instead of replaced

It trips a switch rather than melting a wire.

Q107. If the current through a heater coil is halved (same coil), the heat produced in the same time becomes:

Intermediate
A half B double C one-fourth D unchanged
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one-fourth

H ∝ I² — halve the current, quarter the heat.

Topper level · 93 questions

Q108. Consider these statements about electric current. I. It is the rate of flow of charge. II. Its SI unit is the ampere. III. In a metal the moving charges are electrons. IV. Conventional current flows in the same direction as the electrons. Which statements are correct?

Advanced
A I, II and III only B I and IV only C II and III only D All four
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I, II and III only

I, II and III are correct. IV is the reverse of the truth — conventional current runs from + to − outside the cell, while electrons drift from − to +.

Q109. About how many electrons must pass a point each second to make a current of 1 A? (e = 1.6 × 10⁻¹⁹ C)

Advanced
A 1.6 × 10⁻¹⁹ B 6.25 × 10⁻¹⁹ C 6.25 × 10¹⁸ D 1.6 × 10¹⁹
Show answer & explanation

6.25 × 10¹⁸

n = Q/e = 1/(1.6 × 10⁻¹⁹) ≈ 6.25 × 10¹⁸ electrons every second — which is why the coulomb is such a large unit.

Q110. A student says "current and electrons both flow from the negative terminal to the positive one outside the cell". The statement is:

Advanced
A completely correct B wrong about the electrons only C wrong about both D wrong about the conventional current only
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wrong about the conventional current only

The electrons do drift from − to +, so that half is right. Conventional current is defined the other way round, from + to − outside the cell.

Q111. A lamp lights the instant a switch is closed, even though electrons drift through the wire slower than a walking ant. The explanation is that:

Advanced
A the electrons speed up enormously once the switch closes B the light is produced before the electrons arrive C the current travels through the air rather than the wire D the electric field is set up through the whole circuit almost instantly, so electrons everywhere start moving at once
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the electric field is set up through the whole circuit almost instantly, so electrons everywhere start moving at once

It is the field, not the electrons, that propagates at nearly the speed of light. Like a pipe already full of water, a push at one end produces flow at the other immediately.

Q112. Saying that a cell is marked "1.5 V" means that it gives every coulomb passing through it:

Advanced
A 1.5 coulombs of charge B 1.5 amperes of current C 1.5 joules of energy D 1.5 watts of power
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1.5 joules of energy

A volt is a joule per coulomb, so a 1.5 V cell does 1.5 J of work on each coulomb it drives round the circuit.

Q113. A cell goes flat after long use. What has actually stopped is:

Advanced
A the chemical reaction that keeps charge separated at the two terminals B the supply of electrons inside it C the flow of current through the external circuit only D the resistance of the connecting wires
Show answer & explanation

the chemical reaction that keeps charge separated at the two terminals

A cell does not store electrons and run out of them. It maintains a potential difference by a chemical reaction, and when the reactants are used up the separation, and so the push, stops.

Q114. An ammeter must carry the whole current it is measuring. It follows that it is connected in:

Advanced
A series, and should have very high resistance B parallel, and should have very low resistance C parallel, and should have very high resistance D series, and should have very low resistance
Show answer & explanation

series, and should have very low resistance

Only a series connection passes the full current through the meter, and low resistance keeps the meter from changing the very circuit it is measuring.

Q115. A voltmeter must read the potential difference ACROSS a component. It follows that it is connected in:

Advanced
A series, and should have very low resistance B series, and should have very high resistance C parallel, and should have very high resistance D parallel, and should have very low resistance
Show answer & explanation

parallel, and should have very high resistance

Across means parallel, one terminal each side. High resistance means almost no current is diverted through the meter, so the reading reflects the circuit as it was.

Q116. Why would an ammeter of high resistance give a misleading reading?

Advanced
A it would burn out immediately B it would read the voltage instead C it would reverse the direction of the current D it would add resistance to the circuit and reduce the very current it is meant to measure
Show answer & explanation

it would add resistance to the circuit and reduce the very current it is meant to measure

A measuring instrument should disturb what it measures as little as possible. An ammeter sits in the path of the current, so its own resistance must be negligible.

Q117. Which group correctly pairs each meter with how it is connected and the resistance it should have?

Advanced
A ammeter: parallel, low resistance; voltmeter: series, high resistance B ammeter: series, low resistance; voltmeter: parallel, high resistance C ammeter: series, high resistance; voltmeter: parallel, low resistance D ammeter: parallel, high resistance; voltmeter: series, low resistance
Show answer & explanation

ammeter: series, low resistance; voltmeter: parallel, high resistance

Series and low for the ammeter, parallel and high for the voltmeter. Reversing either pairing both spoils the reading and, for an ammeter in parallel, creates a deliberate short circuit.

Q118. A current flows because a potential difference pushes charge round a circuit. Which statement best describes what a cell actually supplies?

Advanced
A the energy that does work on each coulomb of charge passing through it B a reservoir of electrons that is gradually used up by the circuit C the resistance that limits how much current can flow D the magnetic field that drives the electrons forward
Show answer & explanation

the energy that does work on each coulomb of charge passing through it

A cell does not supply electrons — the circuit already has them. It supplies ENERGY, doing work on every coulomb, which is exactly what a potential difference measures.

Q119. Which group lists three quantities that are ALL measured in different units from one another?

Advanced
A current, ampere, coulomb B charge, coulomb, volt C current, charge, potential difference D potential difference, volt, joule
Show answer & explanation

current, charge, potential difference

Current, charge and potential difference are three distinct quantities, measured in amperes, coulombs and volts. Each of the other groups mixes a quantity with the name of its own unit.

Q120. A student connects a voltmeter in series with a bulb instead of across it. The result is that:

Advanced
A the bulb glows more brightly than before B the voltmeter's very high resistance almost stops the current, so the bulb does not light C the voltmeter is destroyed by a large current D nothing changes, since the voltmeter reads the same either way
Show answer & explanation

the voltmeter's very high resistance almost stops the current, so the bulb does not light

A voltmeter is built with a very high resistance so it draws almost no current. Put in series, that resistance becomes part of the circuit and chokes it.

Q121. A steady current of 1 A flows for 1 hour. The total charge that passes is:

Advanced
A 1 C B 60 C C 3600 C D 216000 C
Show answer & explanation

3600 C

Q = It = 1 × 3600 = 3600 C, since an hour is 3600 seconds. This is why a battery's capacity is quoted in ampere-hours.

Q122. If 1.2 × 10²⁰ electrons pass a point in 4 s, the current is about:

Advanced
A 1.2 A B 4.8 A C 19.2 A D 0.48 A
Show answer & explanation

4.8 A

Q = ne = 1.2×10²⁰ × 1.6×10⁻¹⁹ = 19.2 C; I = 19.2/4 = 4.8 A.

Q123. A charge of 2 C moves between two points, gaining 8 J at one and losing all of it in a bulb. The p.d. across the bulb is:

Advanced
A 16 V B 4 V C 0.25 V D 2 V
Show answer & explanation

4 V

V = W/Q = 8/2 = 4 V.

Q124. Assertion: The direction of conventional current is opposite to electron flow. Reason: The convention was fixed before the electron was discovered, assuming positive carriers.

Advanced
A Both true, reason explains assertion B Both true, reason irrelevant C Assertion true, reason false D Both false
Show answer & explanation

Both true, reason explains assertion

The + → − convention predates the electron's discovery (1897) — and stuck.

Q125. Two identical cells (1.5 V each) are joined in series (+ of one to − of the next). A voltmeter across the pair reads:

Advanced
A 1.5 V B 0 V C 0.75 V D 3 V
Show answer & explanation

3 V

Series cells add their voltages: 1.5 + 1.5 = 3 V — that is a battery.

Q126. Two identical cells of 1.5 V each are joined + terminal to + terminal, so that they oppose each other. A voltmeter across the pair reads:

Advanced
A 3 V B 0 V C 1.5 V D 6 V
Show answer & explanation

0 V

Equal and opposite emfs cancel exactly. Joined the usual way, + to −, the same two cells would give 3 V — which is why the order of connection matters.

Q127. A wire carries 0.8 A. In 100 s, the number of electrons crossing any section is about:

Advanced
A 8 × 10¹⁹ B 1.6 × 10¹⁹ C 5 × 10²⁰ D 6.25 × 10¹⁸
Show answer & explanation

5 × 10²⁰

Q = 80 C; n = 80/1.6×10⁻¹⁹ = 5 × 10²⁰ electrons.

Q128. A car battery is rated 40 Ah. It can deliver 2 A for:

Advanced
A 80 hours B 2 hours C 20 hours D 40 hours
Show answer & explanation

20 hours

t = 40 Ah ÷ 2 A = 20 h — ampere-hours are stored charge.

Q129. Case: In a lab circuit, the ammeter reads 0.25 A before a resistor and a second ammeter reads after it. In the same series loop, the second reading is:

Advanced
A less than 0.25 A B more than 0.25 A C 0.25 A exactly D zero
Show answer & explanation

0.25 A exactly

Series current is identical everywhere — components use energy, not current.

Q130. A student wants to measure both the current through a bulb and the p.d. across it. The correct setup is:

Advanced
A ammeter in series with the bulb, voltmeter in parallel across it B both meters in series with the bulb C both in parallel with the bulb D voltmeter in series, ammeter in parallel
Show answer & explanation

ammeter in series with the bulb, voltmeter in parallel across it

The classic measuring arrangement — SAil and PoVer.

Q131. Which statement about a 9 V battery made of six 1.5 V cells is TRUE?

Advanced
A cells are in parallel B each cell supplies 9 V C cells are in series, voltages adding D the battery halves each cell's voltage
Show answer & explanation

cells are in series, voltages adding

6 × 1.5 = 9 V requires series stacking.

Q132. Calculate the number of electrons flowing per second through a conductor carrying a current of 2 A.

Advanced
Show answer & explanation

—

Q133. (a) How many electrons make one coulomb? Show the calculation. (b) A device passes 6 × 10²⁰ electrons in 40 s. Find the current. (c) If the device works on a 10 V supply, how much energy does it use in that time?

Advanced
Show answer & explanation

—

6.25 × 10¹⁸ shown (1), Q = 96 C (1), I = 2.4 A (1.5), W = 960 J (1.5).

Q134. A student sets up a circuit and reports: 'The bulb glows dimly. When I open the switch the glow disappears. When I add a second cell in series the glow brightens. When I accidentally connected the second cell the opposite way, the bulb went off completely.' Explain each of the four observations using the concepts of closed circuits and potential difference.

Advanced
Show answer & explanation

—

Each observation correctly explained (1 × 4), explicit closing statement linking closed loop + net p.d. (1).

Q135. Consider these statements about Ohm's law. I. It states that V is directly proportional to I. II. It holds only at constant temperature. III. Its V–I graph is a straight line through the origin. IV. Every electrical device obeys it. Which statements are correct?

Advanced
A I, II and III only B I and IV only C II and III only D All four
Show answer & explanation

I, II and III only

I, II and III are correct. IV is false — a filament bulb, a diode and a thermistor are all non-ohmic, and their V–I graphs are curved.

Q136. A filament bulb's V–I graph curves instead of being straight. The reason is that as the current rises:

Advanced
A the voltage across the bulb falls B the bulb begins to store charge C the ammeter becomes inaccurate D the filament heats up, and its resistance increases with temperature
Show answer & explanation

the filament heats up, and its resistance increases with temperature

Ohm's law requires constant temperature. A filament runs at about 2500 °C when bright, so its resistance changes greatly over the range of the graph — it is non-ohmic.

Q137. Why does the V–I graph of an ohmic conductor pass through the origin as well as being straight?

Advanced
A because the resistance of the conductor is always zero at the start B because the ammeter always reads zero before the experiment begins C because no current can flow when there is no potential difference across it D because the graph is conventionally drawn from the origin
Show answer & explanation

because no current can flow when there is no potential difference across it

With V = 0 there is nothing to drive the charge, so I = 0 — the origin is a real measurement, not a drawing convention. Straightness then comes from V and I rising in proportion.

Q138. Resistivity is found from ρ = RA/l. Its SI unit must therefore be the:

Advanced
A ohm B ohm-metre C ohm per metre D metre per ohm
Show answer & explanation

ohm-metre

Ohm × metre² ÷ metre leaves ohm × metre. Working the unit out from the formula is safer than trying to recall it.

Q139. A copper wire is cut into two equal halves. Which quantity is the SAME for each half as for the original wire?

Advanced
A resistance B resistivity C length D mass
Show answer & explanation

resistivity

Resistance, length and mass all halve. Resistivity is a property of the MATERIAL, not of the piece, so cutting the wire leaves it unchanged.

Q140. Nichrome, not copper, is used for a heating element because nichrome has:

Advanced
A a much lower melting point B a much lower cost per metre C a much higher resistivity, so it produces far more heat for the same current D almost no resistance at all
Show answer & explanation

a much higher resistivity, so it produces far more heat for the same current

High resistivity means a large H = I²Rt for a modest current. Nichrome also melts high and does not oxidise when red-hot, which copper would.

Q141. The same current flows through an electric iron's copper cord and its nichrome element, yet only the element gets hot. The reason is that:

Advanced
A H = I²Rt, and with I and t equal the heat depends on R — which is far larger in the element B the cord is thicker than the element C copper cannot conduct heat D the current in the cord is alternating and in the element is direct
Show answer & explanation

H = I²Rt, and with I and t equal the heat depends on R — which is far larger in the element

Copper's very low resistivity gives the cord a tiny resistance, so almost all the heat appears in the high-resistance nichrome — exactly where it is wanted.

Q142. A nichrome wire is stretched uniformly to 3 times its length. Its resistance becomes:

Advanced
A 3 times B 6 times C 9 times D 1/3
Show answer & explanation

9 times

n² rule: length ×3, area ÷3 → R × 9.

Q143. A wire of resistance 12 Ω is folded exactly in half (the halves side by side, joined at both ends). The new resistance is:

Advanced
A 3 Ω B 24 Ω C 6 Ω D 12 Ω
Show answer & explanation

3 Ω

Half length (6 Ω each) in parallel: 6/2 = 3 Ω — or n² rule in reverse.

Q144. The dimensions of wire X: length l, area A. Wire Y (same material): length 2l, area A/2. Then R_Y/R_X =

Advanced
A 1 B 2 C 1/4 D 4
Show answer & explanation

4

R_Y = ρ(2l)/(A/2) = 4ρl/A = 4R_X.

Q145. A copper wire (ρ = 1.6 × 10⁻⁸ Ω·m), l = 5 m, A = 0.8 × 10⁻⁶ m². Its resistance is:

Advanced
A 0.01 Ω B 0.1 Ω C 1 Ω D 10 Ω
Show answer & explanation

0.1 Ω

R = (1.6×10⁻⁸ × 5)/(0.8×10⁻⁶) = 0.1 Ω.

Q146. A student's V–I data: (2 V, 0.2 A), (4 V, 0.4 A), (6 V, 0.6 A). The resistance is:

Advanced
A 5 Ω B 10 Ω C 20 Ω D 0.1 Ω
Show answer & explanation

10 Ω

V/I = 2/0.2 = 10 Ω, constant across all readings — ohmic.

Q147. Assertion: A thick copper wire stays cooler than a thin one carrying the same current. Reason: The thick wire's larger area gives it lower resistance, so less heat (I²R) is produced.

Advanced
A Both true, reason explains assertion B Both true, reason irrelevant C Assertion true, reason false D Both false
Show answer & explanation

Both true, reason explains assertion

Same I, smaller R → smaller I²R heating.

Q148. The element of a room heater glows red-hot but its connecting copper wires stay cool because:

Advanced
A copper cannot heat up B current skips the wires C the element's much higher resistance concentrates the I²R heating there D the wires are longer
Show answer & explanation

the element's much higher resistance concentrates the I²R heating there

Same current everywhere; heat is produced where R is large.

Q149. Doubling BOTH the voltage across and the length of a given wire (same thickness, same material) changes the current by a factor of:

Advanced
A 4 B 2 C 1 (unchanged) D 1/2
Show answer & explanation

1 (unchanged)

V doubles and R doubles (R ∝ l): I = V/R is unchanged.

Q150. A 0.5 mm² copper wire and a 0.5 mm² nichrome wire of equal length carry equal currents. Which develops the greater p.d. across it?

Advanced
A copper B nichrome C equal D cannot say
Show answer & explanation

nichrome

Same I and dimensions; nichrome's ρ (and hence R) is ~60× larger → larger V = IR.

Q151. Case: A lab has two mystery wires of the same metal. Wire 1: l, area A, resistance 6 Ω. Wire 2 has length 3l and area 2A. Its resistance is:

Advanced
A 4 Ω B 9 Ω C 36 Ω D 1 Ω
Show answer & explanation

9 Ω

R₂ = ρ(3l)/(2A) = 1.5 × R₁ = 9 Ω.

Q152. Why do bulb filaments usually blow at the instant of switching ON?

Advanced
A the cold filament's low resistance lets a surge of current through B voltage is highest then C electrons pile up D the glass expands
Show answer & explanation

the cold filament's low resistance lets a surge of current through

Cold tungsten's R is ~1/10th of its glowing value — the initial current spike can melt weak spots.

Q153. The resistance of an ideal superconductor is:

Advanced
A 1 Ω B very high C negative D exactly zero
Show answer & explanation

exactly zero

Below its critical temperature, a superconductor has zero resistance.

Q154. For a wire at constant temperature, which graph is a straight line through the origin?

Advanced
A R vs l B R vs A C ρ vs l D R vs 1/l
Show answer & explanation

R vs l

R = (ρ/A)·l — R is directly proportional to length. (R vs A is a curve; ρ doesn't vary.)

Q155. A wire of resistance 10 Ω is stretched uniformly to twice its original length. Calculate its new resistance.

Advanced
Show answer & explanation

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Q156. Derive the factors governing the resistance of a wire in the form R = ρl/A, explaining physically why R grows with length and shrinks with thickness. Then define resistivity, give its unit, and compare the resistivity classes of conductors, alloys and insulators with one use of each.

Advanced
Show answer & explanation

—

Collision reasoning for l and A dependence (1.5), R = ρl/A assembled (0.5), resistivity definition + unit (1), three classes with ranges (1.5), one use each (0.5).

Q157. A copper wire has diameter 0.5 mm and resistivity 1.6 × 10⁻⁸ Ω·m. (a) Find the length of this wire needed to make a 10 Ω resistance. (b) If the diameter is doubled, how does the required length change for the same 10 Ω? What does this tell you about practical resistor design?

Advanced
Show answer & explanation

—

Correct area (1), l ≈ 122.7 m (1.5), 4× length reasoning (1.5), design insight (1).

Q158. Consider these statements about resistors in series. I. The same current flows through every resistor. II. The voltages across them add to the supply voltage. III. The equivalent resistance is larger than the largest resistor. IV. Each resistor has the full supply voltage across it. Which statements are correct?

Advanced
A I, II and III only B I and IV only C II and IV only D All four
Show answer & explanation

I, II and III only

I, II and III describe a series circuit. IV is what happens in PARALLEL — in series the supply voltage is shared out, not repeated.

Q159. Resistors of 3 Ω, 5 Ω and 12 Ω are joined in series. Without calculating, you can say the equivalent resistance must be:

Advanced
A less than 3 Ω B between 3 Ω and 12 Ω C greater than 12 Ω D exactly 12 Ω
Show answer & explanation

greater than 12 Ω

Series resistances add, so the total is always larger than the largest member — here 20 Ω. That check catches arithmetic slips immediately.

Q160. Resistors of 3 Ω, 5 Ω and 12 Ω are joined in parallel. Without calculating, the equivalent resistance must be:

Advanced
A between 3 Ω and 12 Ω B greater than 12 Ω C less than 3 Ω D exactly 5 Ω
Show answer & explanation

less than 3 Ω

A parallel combination is always easier to drive current through than any single branch, so the total is smaller than the smallest — less than 3 Ω.

Q161. Why does one failed bulb in a house not switch off every other appliance?

Advanced
A the appliances are wired in parallel, so a break in one branch leaves the others complete B each appliance has its own fuse C the mains voltage rises to compensate D the failed bulb is bypassed automatically
Show answer & explanation

the appliances are wired in parallel, so a break in one branch leaves the others complete

Parallel wiring gives each appliance its own path. In series a single break anywhere would open the whole circuit and everything would stop.

Q162. Two unequal resistors are joined in series across a battery. Why does the larger one have the larger voltage across it?

Advanced
A because a larger resistor always draws a larger current through itself B because the same current flows through both, so V = IR makes the voltage follow the resistance C because the battery deliberately sends more energy to the larger resistor D because most of the current bypasses the smaller resistor
Show answer & explanation

because the same current flows through both, so V = IR makes the voltage follow the resistance

In series the current is common to both, so the voltage each takes is proportional to its own resistance. Nothing is bypassed, and the battery does not choose.

Q163. A combination of resistors with a total resistance of 6 Ω is connected to a 12 V battery. The main current drawn from the battery is:

Advanced
A 0.5 A B 2 A C 1 A D 6 A
Show answer & explanation

2 A

Ohm's law applied to the whole circuit: I = V/R = 12/6 = 2 A. The total resistance is what the battery sees, whatever the arrangement inside.

Q164. A 4 Ω resistor is in series with a parallel pair whose combined resistance is 2 Ω, the whole across a 12 V battery. The potential difference across the parallel pair is:

Advanced
A 12 V B 4 V C 8 V D 2 V
Show answer & explanation

4 V

Total resistance 6 Ω gives a current of 2 A. Across the parallel pair, V = IR = 2 × 2 = 4 V, and the remaining 8 V is dropped across the 4 Ω resistor.

Q165. Two bulbs of resistance 10 Ω and 20 Ω are connected in PARALLEL across a battery. Which glows brighter?

Advanced
A the 20 Ω B equal brightness C neither glows D the 10 Ω
Show answer & explanation

the 10 Ω

In parallel both have the same voltage, so P = V²/R makes the SMALLER resistance brighter. In series the current is shared and the larger resistance is brighter — the opposite answer.

Q166. A wire of resistance R is cut into n equal parts, and all parts are connected in parallel. The equivalent resistance is:

Advanced
A R/n B nR C R/n² D n²R
Show answer & explanation

R/n²

Each part is R/n; n of them in parallel give (R/n)/n = R/n².

Q167. A 12 Ω wire is cut into three equal parts connected in parallel. The result is:

Advanced
A 4/3 Ω B 4 Ω C 36 Ω D 1.5 Ω
Show answer & explanation

4/3 Ω

Each part 4 Ω; 4/3 ≈ 1.33 Ω by the R/n² rule.

Q168. Four 2 Ω resistors can produce which of the following equivalents?

Advanced
A only 8 Ω and 0.5 Ω B 8, 0.5, 2, 5, 1.2 Ω (several arrangements) C only 2 Ω D only 8 Ω
Show answer & explanation

8, 0.5, 2, 5, 1.2 Ω (several arrangements)

All four in series give 8 Ω and all four in parallel give 0.5 Ω. Two pairs in series, joined in parallel, give 2 Ω. Two in series plus a parallel pair gives 2+2+1 = 5 Ω. And ((2∥2)+2) in parallel with 2 gives 3∥2 = 1.2 Ω. Every listed value is reachable.

Q169. Assertion: In domestic wiring, a short circuit in the toaster does not stop the fans. Reason: Each appliance sits on its own parallel branch across the mains.

Advanced
A Both true, reason explains assertion B Both true, reason irrelevant C Assertion true, reason false D Both false
Show answer & explanation

Both true, reason explains assertion

Parallel isolation is exactly why one faulty branch (once its fuse cuts) leaves the rest live.

Q170. Two resistors give 9 Ω in series and 2 Ω in parallel. The resistors are:

Advanced
A 6 Ω and 3 Ω B 4.5 Ω and 4.5 Ω C 5 Ω and 4 Ω D 8 Ω and 1 Ω
Show answer & explanation

6 Ω and 3 Ω

Sum 9, product = Rp × sum = 18 → the pair 6 and 3.

Q171. A battery drives 1 A through two series resistors, dropping 4 V and 8 V across them. Their values are:

Advanced
A 2 Ω and 4 Ω B 8 Ω and 16 Ω C 4 Ω and 8 Ω D 1 Ω and 2 Ω
Show answer & explanation

4 Ω and 8 Ω

R = V/I with I = 1 A → 4 Ω and 8 Ω (battery = 12 V).

Q172. Case: A student needs exactly 1 Ω but has only 2 Ω resistors. The minimum number needed is:

Advanced
A 1 B 3 C 2 D 4
Show answer & explanation

2

Two 2 Ω in parallel = 1 Ω.

Q173. Case: A hostel corridor has 10 identical lamps in parallel on 220 V, each of resistance 484 Ω. The total current drawn is about:

Advanced
A 0.45 A B 4.5 A C 45 A D 2.2 A
Show answer & explanation

4.5 A

Each lamp: I = 220/484 ≈ 0.45 A; ten branches: ≈ 4.5 A.

Q174. If one of those 10 lamps burns out, the current through EACH remaining lamp:

Advanced
A increases B decreases C stays the same D doubles
Show answer & explanation

stays the same

Each branch still sees the full 220 V — parallel branches are independent.

Q175. The total resistance between points A and B: a 2 Ω resistor in series with [4 Ω parallel 4 Ω]:

Advanced
A 10 Ω B 4 Ω C 6 Ω D 2.67 Ω
Show answer & explanation

4 Ω

4∥4 = 2; 2 + 2 = 4 Ω.

Q176. Two heaters of resistance R each boil a pot of water in time t when used one at a time. Used TOGETHER in parallel on the same supply, they boil it in about:

Advanced
A 2t B t C t/2 D t/4
Show answer & explanation

t/2

Parallel halves the resistance → doubles the power → halves the time.

Q177. Two heaters of resistance R each boil a pot of water in time t when used one at a time. Connected in SERIES on the same supply, they boil the same pot in about:

Advanced
A t/2 B t C 4t D 2t
Show answer & explanation

2t

In series the resistance doubles to 2R, so the power P = V²/R halves — and half the power takes twice the time. In parallel the resistance halves instead and the time comes down to t/2.

Q178. Which arrangement of a 1 Ω, 2 Ω and 3 Ω resistor gives an equivalent CLOSEST to 2 Ω?

Advanced
A all in series B all in parallel C 3 Ω in series with (1∥2) D 1 Ω in series with (2∥3)
Show answer & explanation

1 Ω in series with (2∥3)

1 + (2×3/5) = 1 + 1.2 = 2.2 Ω — closest to 2. (Series = 6, parallel ≈ 0.55, option c = 3.67.)

Q179. Show how you would connect three resistors, each of 6 Ω, to obtain (i) 9 Ω, (ii) 4 Ω.

Advanced
Show answer & explanation

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Q180. Why do the headlights of a car become slightly dimmer when the starter motor runs?

Advanced
Show answer & explanation

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Q181. With derivations, compare the series and parallel combinations of resistors under the headings: current, potential difference, equivalent resistance, effect of one component failing, and one practical application of each.

Advanced
Show answer & explanation

—

Series derivation (1), parallel derivation (1), correct same-quantity identification both sides (1), failure behaviour both (1), one application each (1).

Q182. A circuit has a 6 V battery, a 1 Ω resistor in series, and then a parallel pair of 4 Ω and 12 Ω. Calculate (a) the total resistance, (b) the main current, (c) the p.d. across the parallel pair, (d) the current in each branch, and (e) verify that the branch currents add up correctly.

Advanced
Show answer & explanation

—

Rp = 3 and total 4 Ω (1), I = 1.5 A (1), V = 4.5 V with check (1), both branch currents (1.5), verification statement (0.5).

Q183. You are given a box of identical 4 Ω resistors. Explain, with working, how to build: (a) 12 Ω, (b) 1 Ω, (c) 6 Ω, (d) 10 Ω — using as few resistors as possible each time. State the general principles you used.

Advanced
Show answer & explanation

—

Each construction correct with count (0.75 × 4), principles summarised (2).

Q184. Explain fully why parallel wiring is used in homes, and what would actually happen, appliance by appliance, if a house with a 220 V supply were wired in series with a bulb (480 Ω), a fan (240 Ω) and a heater (48 Ω).

Advanced
Show answer & explanation

—

Series total R and current (1), voltage/power consequences per appliance (1.5), interdependence + single-failure faults (1), parallel advantages tied to numbers (1.5).

Q185. Consider these statements about the heating effect of current. I. The heat produced is proportional to the square of the current. II. Doubling the current gives four times the heat. III. The heat is proportional to the resistance and to the time. IV. Halving the current halves the heat. Which statements are correct?

Advanced
A I, II and III only B I and IV only C II and IV only D All four
Show answer & explanation

I, II and III only

H = I²Rt makes I, II and III true. IV is wrong — halving the current gives a QUARTER of the heat, not a half, because of the square.

Q186. A heater carries 4 A through a 20 Ω element for 1 minute. The heat produced is:

Advanced
A 4800 J B 19200 J C 1200 J D 320 J
Show answer & explanation

19200 J

H = I²Rt = 4² × 20 × 60 = 16 × 20 × 60 = 19 200 J. The time must be in seconds.

Q187. A bulb marked "40 W, 220 V" is run at 220 V. The current it draws is about:

Advanced
A 5.5 A B 40 A C 8800 A D 0.18 A
Show answer & explanation

0.18 A

I = P/V = 40/220 ≈ 0.18 A. A quick check: a small bulb drawing 5 A would trip a domestic fuse instantly.

Q188. A 2 kW geyser runs for 45 minutes. The number of units of electricity consumed is:

Advanced
A 2.0 B 90 C 0.67 D 1.5
Show answer & explanation

1.5

Put the power in kilowatts and the time in hours: E = 2 × 0.75 = 1.5 kWh, and one unit is one kilowatt-hour.

Q189. An incandescent bulb is filled with an inert gas rather than air because:

Advanced
A the gas makes the filament glow brighter B the gas conducts the current to the filament C tungsten at 2500 °C would react with oxygen and burn away in moments D air would make the glass crack
Show answer & explanation

tungsten at 2500 °C would react with oxygen and burn away in moments

Argon or nitrogen will not react with the white-hot tungsten, and the gas also slows the filament's evaporation, so it lasts far longer.

Q190. Why must a fuse wire have a LOW melting point and a HIGHER resistance than the circuit wiring?

Advanced
A so that it can safely carry a larger current than the rest of the circuit B so that it can be reset by hand once a fault has been cleared C so that it reduces the voltage reaching the appliance D so that it heats faster than the wiring and melts first, breaking the circuit before damage is done
Show answer & explanation

so that it heats faster than the wiring and melts first, breaking the circuit before damage is done

The fuse is designed to be the weakest link. A higher resistance means more heating in it than anywhere else, and a low melting point means it gives way first.

Q191. Why is an electric fan not said to work by the heating effect, even though its motor does become warm?

Advanced
A because its purpose is rotation, produced by the magnetic effect — the warmth is wasted energy rather than the point B because no current actually flows through the motor of a fan C because the resistance of a fan motor is exactly zero D because alternating current produces no heating at all
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because its purpose is rotation, produced by the magnetic effect — the warmth is wasted energy rather than the point

Any resistance produces some heat, but in a fan that heat is a loss. The effect a fan is built around is magnetic: a force on a current-carrying coil in a field.

Q192. A 60 W and a 100 W bulb (both 220 V-rated) are connected in SERIES to the mains. Which glows brighter?

Advanced
A 100 W B equally C 60 W D neither glows
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60 W

Series → same I; P = I²R and the 60 W bulb's R is larger — it takes more power.

Q193. A 60 W and a 100 W bulb, both rated 220 V, are connected in PARALLEL to the 220 V mains. Which glows brighter?

Advanced
A 60 W B 100 W C equally D neither
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100 W

In parallel each bulb gets its rated 220 V and so draws its rated power — the 100 W bulb is brighter, as the ratings promise. In SERIES the answer reverses, because the 60 W bulb has the higher resistance and takes the larger share of the voltage.

Q194. Two identical heating coils (R each) on a fixed supply: connected in parallel they produce heat at rate H₁; in series at rate H₂. H₁/H₂ =

Advanced
A 1 B 2 C 1/4 D 4
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4

Parallel: R/2 → P = 2V²/R. Series: 2R → P = V²/2R. Ratio = 4.

Q195. An electric iron's rating plate says 750 W at 220 V. Heat produced in 10 minutes:

Advanced
A 450 kJ B 7500 J C 75 kJ D 4.5 kJ
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450 kJ

E = Pt = 750 × 600 = 450,000 J = 450 kJ.

Q196. Assertion: A fuse must be placed in the live wire, not the neutral. Reason: Breaking the live wire disconnects the appliance from the high potential, making it safe to touch.

Advanced
A Both true, reason explains assertion B Both true, reason irrelevant C Assertion true, reason false D Both false
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Both true, reason explains assertion

A fuse in the neutral would stop the current but leave the appliance connected to the live 220 V.

Q197. A 220 V line has a 5 A fuse. The maximum total power of appliances that can safely run is:

Advanced
A 44 W B 220 W C 1100 W D 4400 W
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1100 W

P_max = VI = 220 × 5 = 1100 W.

Q198. A bulb rated 220 V, 100 W is run on 110 V. Its power consumption becomes about:

Advanced
A 100 W B 25 W C 50 W D 200 W
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25 W

P = V²/R — half the voltage, one-quarter the power (ignoring temperature effects).

Q199. Case: Riya's home runs a fridge (average 100 W, all 24 h), a TV (100 W for 4 h) and a washing machine (500 W for 1 h) daily. The daily consumption is about:

Advanced
A 3.3 units B 1.2 units C 5.4 units D 8 units
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3.3 units

Fridge 2.4 + TV 0.4 + washer 0.5 = 3.3 kWh = 3.3 units.

Q200. An immersion rod raises the temperature of water using H = I²Rt. To heat the same water in HALF the time on the same supply, the rod's resistance must be:

Advanced
A doubled B quadrupled C unchanged D halved
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halved

Same V: P = V²/R — halving R doubles P, halving the required time.

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