Series RLC Circuit Analysis

C · $Z = \sqrt{R^2 + (X_L - X_C)^2}$

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

D · The resistor only

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

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

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

B · $P_{avg} = I_{rms}^2 R$

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.

D · $V_{rms} = \frac{V_m}{\sqrt{2}}$

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

A · True

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

C · $126.06\,\Omega$

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

C · 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.

A · 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.

D · $0.7933\,\text{A}$

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

B · $X_C > X_L$, 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.

D · $Z = R$, 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.

C · 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.

A · $-37.51^\circ$

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

A · $31.46\,\text{W}$

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

D · 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.

B · -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.

B · 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.