Series RLC Circuit Analysis: 18 MCQs with answers

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

Impedance combines resistance with the NET reactance $(X_L - X_C)$ as a phasor triangle, so $Z = \sqrt{R^2 + (X_L - X_C)^2}$.

The resistor only

Inductors and capacitors store and release energy but dissipate none over a full cycle; only the resistor converts energy to heat.

$\phi = \arctan\left(\frac{X_L - X_C}{R}\right)$

The tangent of the phase angle is the net reactance divided by the resistance: $\phi = \arctan\left(\frac{X_L - X_C}{R}\right)$.

$P_{avg} = I_{rms}^2 R$

Only the resistor dissipates average power, so $P_{avg} = I_{rms}^2 R$. Using $Z$, $X_L$, or $X_C$ would overstate or misplace the dissipation.

$V_{rms} = \frac{V_m}{\sqrt{2}}$

For a sinusoid the RMS value is the amplitude divided by $\sqrt{2}$, so $V_{rms} = \frac{V_m}{\sqrt{2}}$.

True

In a series RLC circuit, the impedance Z is the vector sum of resistance R and the net reactance ($X_L - X_C$).

$126.06\,\Omega$

$Z = \sqrt{100^2 + (188.5 - 265.25)^2} = \sqrt{10000 + 5890.6} \approx 126.06\,\Omega$.

The current leads the voltage and the circuit is capacitive

A negative $\phi$ means $X_L - X_C < 0$, so $X_C > X_L$ and the current leads the voltage — a capacitive, leading circuit.

Because $X_L$ and $X_C$ oppose each other and are 90° out of phase with R, so they combine as vectors

All three quantities share the ohm as unit, but $X_L$ and $X_C$ act in opposite directions and are 90° out of phase with R, so they must be combined as phasors.

$0.7933\,\text{A}$

$I_m = \frac{V_m}{Z} = \frac{100}{126.06} \approx 0.7933\,\text{A}$. The value $0.5609\,\text{A}$ is the RMS current, not the amplitude.

$X_C > X_L$, so the current leads the voltage

A leading power factor means the current leads the voltage, which requires the capacitive reactance to exceed the inductive reactance.

$Z = R$, because the reactances cancel

When $X_L = X_C$ the net reactance is zero, so $Z = \sqrt{R^2 + 0} = R$. The circuit is at resonance and the power factor is 1.

Most of the supplied power is converted into useful work

Power factor is the ratio of real power to apparent power, so a value near 1 means little reactive circulation and efficient power use.

$-37.51^\circ$

$\phi = \arctan\left(\frac{188.5 - 265.25}{100}\right) = \arctan(-0.7675) \approx -37.51^\circ$. The sign is negative because $X_C > X_L$.

$31.46\,\text{W}$

$I_{rms} = \frac{0.7933}{\sqrt{2}} \approx 0.5609\,\text{A}$, so $P_{avg} = (0.5609)^2 \times 100 \approx 31.46\,\text{W}$. Using the peak current directly would give the wrong $62.93\,\text{W}$.

It has the same angular frequency $\omega$ as the applied voltage but is shifted by $\phi$

In a linear AC circuit the current keeps the source frequency $\omega$; only its amplitude $I_m$ and phase shift $\phi$ are set by R, L, and C.

-37.51^\circ

The phase angle is calculated as $\phi = \arctan\left(\frac{188.5 - 265.25}{100}\right) = \arctan(-0.7675) \approx -37.51^\circ$. A negative phase angle indicates that the current leads the voltage.

0.7933 A

The amplitude of the current is calculated as $I_m = \frac{100}{126.06} \approx 0.7933 A$. This calculation uses the relationship between voltage and impedance in an AC circuit.