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:
BeginnerShow answer & explanation
the conducting path is broken
Current needs a continuous closed loop.
Q2. Define electric current. State and define its SI unit.
BeginnerShow answer & explanation
—
Q3. Define potential difference between two points. State and define its SI unit.
BeginnerShow answer & explanation
—
Q4. Distinguish between conventional current and electron flow.
BeginnerShow answer & explanation
—
Q5. A 6 V battery moves charge through itself. How much energy does every coulomb pick up?
BeginnerShow answer & explanation
—
Q6. Name the components needed to make the simplest working circuit that lights a bulb, and state the function of each.
BeginnerShow answer & explanation
—
Q7. 1 ohm equals:
BeginnerShow answer & explanation
1 V/A
One volt per ampere.
Q8. The resistance of a wire is directly proportional to its:
BeginnerShow answer & explanation
length
R ∝ l — longer wire, more collisions.
Q9. If the cross-section area of a wire increases, its resistance:
BeginnerShow answer & explanation
decreases
R ∝ 1/A — a wider road for electrons.
Q10. A device used to change the current in a circuit gradually is a:
BeginnerShow answer & explanation
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.
BeginnerShow answer & explanation
—
Q12. Define resistance of a conductor and define 1 ohm.
BeginnerShow answer & explanation
—
Q13. On what factors does the resistance of a uniform metallic wire depend? Write the formula.
BeginnerShow answer & explanation
—
Q14. The equivalent of 2 Ω, 3 Ω and 5 Ω in series is:
BeginnerShow answer & explanation
10 Ω
Rs = 2 + 3 + 5 = 10 Ω.
Q15. The equivalent of two 10 Ω resistors in parallel is:
BeginnerShow answer & explanation
5 Ω
Equal pair: R/n = 10/2 = 5 Ω.
Q16. In parallel, the branch with the smallest resistance carries the:
BeginnerShow answer & explanation
largest current
I = V/R with common V — small R, large I.
Q17. Five 1 Ω resistors in series give:
BeginnerShow answer & explanation
5 Ω
nR = 5 × 1 = 5 Ω.
Q18. Five 1 Ω resistors in parallel give:
BeginnerShow answer & explanation
0.2 Ω
R/n = 1/5 = 0.2 Ω.
Q19. Why is the series arrangement not used for connecting domestic electrical appliances?
BeginnerShow answer & explanation
—
Q20. What happens to the total resistance and the current drawn when more resistors are added (a) in series, (b) in parallel?
BeginnerShow answer & explanation
—
Q21. Two lamps rated for 220 V are connected in parallel to a 220 V supply. State two advantages of this connection over series.
BeginnerShow answer & explanation
—
Q22. 1 kWh equals:
BeginnerShow answer & explanation
3.6 × 10⁶ J
1000 W × 3600 s = 3.6 × 10⁶ J.
Q23. A fuse is connected in:
BeginnerShow answer & explanation
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:
BeginnerShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
5 V
V = W/Q = 10/2 = 5 V.
Q32. A 6 V battery gives every coulomb passing through it:
IntermediateShow answer & explanation
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?
IntermediateShow answer & explanation
48 J
W = VQ = 12 × 4 = 48 J.
Q34. The number of electrons in 1 coulomb of charge is about:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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?
IntermediateShow answer & explanation
current
Series = one path = one current through both.
Q40. In the water analogy of circuits, the pump corresponds to the:
IntermediateShow answer & explanation
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?
IntermediateShow answer & explanation
—
Q42. Why is a voltmeter connected in parallel and made with very high resistance?
IntermediateShow answer & explanation
—
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.
IntermediateShow answer & explanation
—
Q44. A charge of 3 C is carried between two points separated by 15 V. What work does that take?
IntermediateShow answer & explanation
—
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.
IntermediateShow answer & explanation
—
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.
IntermediateShow answer & explanation
—
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
5 Ω
R = V/I = 10/2 = 5 Ω.
Q50. The current through a 30 Ω resistor connected to a 6 V battery is:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
20 V
V = IR = 0.5 × 40 = 20 V.
Q52. If the p.d. across a fixed resistor is halved, the current:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
P has more resistance
Steeper V–I slope = larger V/I = larger R.
Q58. The resistance of a metal wire increases when heated because:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
4 Ω
R' = ρ(2l)/(2A) = ρl/A = 4 Ω — the changes cancel.
Q61. Which has the HIGHEST resistivity?
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
ρ
Resistivity is a material property — unchanged by length.
Q63. Define resistivity. Why is it called a characteristic property of the material?
IntermediateShow answer & explanation
—
Q64. Why are the coils of electric heating devices made of an alloy rather than a pure metal?
IntermediateShow answer & explanation
—
Q65. How much current flows through an electric bulb of resistance 1100 Ω when connected to a 220 V supply?
IntermediateShow answer & explanation
—
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?
IntermediateShow answer & explanation
—
Q67. Why does the connecting cord of an electric heater not glow while its heating element does?
IntermediateShow answer & explanation
—
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.
IntermediateShow answer & explanation
—
Q69. Describe an experiment to verify Ohm's law. Include the circuit, procedure, observations, the graph, and the conclusion.
IntermediateShow answer & explanation
—
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.
IntermediateShow answer & explanation
—
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
6 Ω
3∥6 = 2 Ω; 2 + 4 = 6 Ω.
Q76. Through the 3 Ω branch of that pair flows:
IntermediateShow answer & explanation
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?
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
decrease
An extra path always eases the total flow.
Q79. Adding one more resistor in series makes the current drawn from the battery:
IntermediateShow answer & explanation
decrease
Total R rises, so I = V/R falls.
Q80. To get 5 Ω from two 10 Ω resistors, connect them in:
IntermediateShow answer & explanation
parallel
10∥10 = 5 Ω.
Q81. To get 15 Ω from 10 Ω and 5 Ω resistors, connect them in:
IntermediateShow answer & explanation
series
10 + 5 = 15 Ω in series.
Q82. Which single value CANNOT be made from two 6 Ω resistors?
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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.
IntermediateShow answer & explanation
—
Q86. Derive the expression for the equivalent resistance of three resistors connected in parallel.
IntermediateShow answer & explanation
—
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.
IntermediateShow answer & explanation
—
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.
IntermediateShow answer & explanation
—
Q89. A current of 6 A enters a parallel combination of 2 Ω and 4 Ω. Find the current in each branch.
IntermediateShow answer & explanation
—
Q90. An appliance converts 500 joules of electrical energy every second. Its power is:
IntermediateShow answer & explanation
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?
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
200 J
H = I²Rt = 4 × 10 × 5 = 200 J.
Q93. The power of an appliance drawing 3 A at 220 V is:
IntermediateShow answer & explanation
660 W
P = VI = 220 × 3 = 660 W.
Q94. The resistance of a 220 V, 550 W appliance is:
IntermediateShow answer & explanation
88 Ω
R = V²/P = 48400/550 = 88 Ω.
Q95. The current drawn by a 220 V, 440 W appliance is:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
1 kWh
0.1 kW × 10 h = 1 kWh — one unit.
Q98. A 1.5 kW geyser runs 40 minutes. Units consumed:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
₹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?
IntermediateShow 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:
IntermediateShow answer & explanation
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:
IntermediateShow 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:
IntermediateShow answer & explanation
10 A
I = P/V = 2200/220 = 10 A.
Q104. The energy consumed by a 40 W bulb in 1 minute is:
IntermediateShow answer & explanation
2400 J
E = Pt = 40 × 60 = 2400 J.
Q105. Long-distance power lines use high voltage mainly to:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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:
IntermediateShow answer & explanation
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?
AdvancedShow answer & explanation
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)
AdvancedShow 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:
AdvancedShow answer & explanation
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:
AdvancedShow answer & explanation
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:
AdvancedShow answer & explanation
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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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?
AdvancedShow 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?
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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.
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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.
AdvancedShow 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?
AdvancedShow 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.
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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 =
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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.
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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.
AdvancedShow answer & explanation
—
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.
AdvancedShow 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?
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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.
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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]:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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.
AdvancedShow answer & explanation
—
Q180. Why do the headlights of a car become slightly dimmer when the starter motor runs?
AdvancedShow answer & explanation
—
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.
AdvancedShow 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.
AdvancedShow 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.
AdvancedShow 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).
AdvancedShow 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?
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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:
AdvancedShow 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?
AdvancedShow 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?
AdvancedShow answer & explanation
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?
AdvancedShow answer & explanation
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?
AdvancedShow answer & explanation
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 =
AdvancedShow answer & explanation
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:
AdvancedShow answer & explanation
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.
AdvancedShow answer & explanation
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:
AdvancedShow answer & explanation
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:
AdvancedShow answer & explanation
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:
AdvancedShow answer & explanation
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:
AdvancedShow answer & explanation
halved
Same V: P = V²/R — halving R doubles P, halving the required time.
Your notes can be a kit like these
Every kit here was made from real study material. Point Veda at a chapter, a PDF, a photo or a link and get notes, flashcards, practice MCQs, games and a scored test — free, in 42 languages.
- Chapter notes
- Flashcards
- Practice MCQs
- Scored tests
- Exam prep
Get Veda free See what students get
Free to start · No ads · Works in 42 languages