The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 31 — Alternating Current

Problems with worked solutions · 13 points · 3 extra credit

Chapter 31 — Alternating Current

Core: 13 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

1. 1. (1 pt •) A household outlet supplies a sinusoidal voltage with amplitude $V_0=170\,\text{V}$ at $f=60\,\text{Hz}$.

   (a) Find the rms voltage.

   (b) A resistive heater draws a current of amplitude $I_0=2.5\,\text{A}$. Find the rms current.

   Solution

   1.

   (a) $V_\text{rms}=\dfrac{V_0}{\sqrt2}=\dfrac{170}{\sqrt2}\approx 120\,\text{V}$.

   (b) $I_\text{rms}=\dfrac{I_0}{\sqrt2}=\dfrac{2.5}{\sqrt2}\approx 1.8\,\text{A}$.

2. 2. (2 pts ••) A $0.25\,\text{H}$ inductor and, separately, an $8.0\,\mu\text{F}$ capacitor are each driven at $f=60\,\text{Hz}$.

   (a) Find the inductive reactance $X_L$ and the capacitive reactance $X_C$.

   (b) If each element alone is connected across a $120\,\text{V}$ (rms) source at this frequency, find the rms current through each.

   (c) Without recomputing, state how each reactance changes if the frequency is doubled.

   Solution

   2.

   (a) $\omega=2\pi f=2\pi(60)\approx 377\,\text{rad/s}$.

   $X_L=\omega L=(377)(0.25)\approx 94\,\Omega$.

   $X_C=\dfrac{1}{\omega C}=\dfrac{1}{(377)(8.0\times10^{-6})}\approx 3.3\times10^{2}\,\Omega$.

   (b) Inductor: $I_\text{rms}=\dfrac{120}{94.2}\approx 1.3\,\text{A}$.

   Capacitor: $I_\text{rms}=\dfrac{120}{331.6}\approx 0.36\,\text{A}$.

   (c) Doubling $f$ doubles $\omega$, so $X_L=\omega L$ doubles (to $\approx 188\,\Omega$) while $X_C=1/\omega C$ halves (to $\approx 166\,\Omega$).

3. 3. (2 pts ••) In a series circuit driven at a single frequency, the voltage amplitudes measured across the resistor, inductor, and capacitor are $V_R=45\,\text{V}$, $V_L=80\,\text{V}$, and $V_C=40\,\text{V}$.

   (a) Using a phasor diagram, find the amplitude of the source voltage.

   (b) Find the phase angle of the source voltage relative to the current, and state whether the current leads or lags the source.

   Solution

   3.

   (a) $V_R$ is in phase with the current; $V_L$ leads by $90^\circ$ and $V_C$ lags by $90^\circ$, so they subtract along the perpendicular axis. The source voltage amplitude is

   $$V=\sqrt{V_R^2+(V_L-V_C)^2}=\sqrt{45^2+(80-40)^2}=\sqrt{2025+1600}\approx 60\,\text{V}.$$

   (b) $\phi=\tan^{-1}\!\dfrac{V_L-V_C}{V_R}=\tan^{-1}\!\dfrac{40}{45}\approx 42^\circ$. Since $V_L>V_C$ the net reactance is inductive, so the current lags the source voltage by $\approx 42^\circ$.

4. 4. (3 pts •••) A series R–L–C circuit has $R=200\,\Omega$, $L=0.40\,\text{H}$, and $C=5.0\,\mu\text{F}$, driven by a source of rms voltage $120\,\text{V}$ at $f=60\,\text{Hz}$.

   (a) Find $X_L$, $X_C$, and the impedance $Z$.

   (b) Find the rms current.

   (c) Find the phase angle $\phi$, and state whether the circuit is inductive or capacitive.

   (d) Find the average power delivered by the source, and check it against $I_\text{rms}^2 R$.

   Solution

   4.

   (a) $\omega=2\pi(60)\approx 377\,\text{rad/s}$.

   $X_L=\omega L=(377)(0.40)\approx 151\,\Omega$.

   $X_C=\dfrac{1}{\omega C}=\dfrac{1}{(377)(5.0\times10^{-6})}\approx 531\,\Omega$.

   $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{200^2+(151-531)^2}=\sqrt{200^2+(-380)^2}\approx 429\,\Omega$.

   (b) $I_\text{rms}=\dfrac{V_\text{rms}}{Z}=\dfrac{120}{429}\approx 0.28\,\text{A}$.

   (c) $\phi=\tan^{-1}\!\dfrac{X_L-X_C}{R}=\tan^{-1}\!\dfrac{-380}{200}\approx -62^\circ$. Because $X_C>X_L$ the net reactance is negative: the circuit is capacitive, and the current leads the source voltage by $\approx 62^\circ$.

   (d) $\cos\phi=\cos(-62.2^\circ)\approx 0.47$.

   $P_\text{av}=V_\text{rms}I_\text{rms}\cos\phi=(120)(0.2796)(0.466)\approx 16\,\text{W}$.

   Check: $I_\text{rms}^2 R=(0.2796)^2(200)\approx 16\,\text{W}$. ✓

5. 5. (2 pts ••) For the same components as Problem 4 ($L=0.40\,\text{H}$, $C=5.0\,\mu\text{F}$, $R=200\,\Omega$, source rms voltage $120\,\text{V}$):

   (a) Find the resonance angular frequency $\omega_0$ and the corresponding frequency $f_0$.

   (b) Find the rms current and the average power delivered at resonance, and explain why the power is larger than in Problem 4.

   Solution

   5.

   (a) $\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{(0.40)(5.0\times10^{-6})}}\approx 707\,\text{rad/s}$.

   $f_0=\dfrac{\omega_0}{2\pi}\approx 113\,\text{Hz}$.

   (b) At resonance $X_L=X_C$, so $Z=R=200\,\Omega$.

   $I_\text{rms}=\dfrac{120}{200}=0.60\,\text{A}$; $\phi=0$, so $P_\text{av}=V_\text{rms}I_\text{rms}=(120)(0.60)=72\,\text{W}$.

   The power is much larger than in Problem 4 because at $60\,\text{Hz}$ the large net reactance ($\approx 380\,\Omega$) chokes the current; at resonance the reactances cancel, $Z$ drops to just $R$, the current rises, and the power factor becomes $1$.

6. 6. (2 pts ••) An AC motor operating from a $240\,\text{V}$ (rms) line draws an rms current of $5.0\,\text{A}$ with a power factor of $0.80$ (current lagging).

   (a) Find the average power consumed by the motor.

   (b) Find the phase angle between the line voltage and the current.

   (c) The apparent power is $V_\text{rms}I_\text{rms}$. Find it, and explain in one sentence why it exceeds the average power.

   Solution

   6.

   (a) $P_\text{av}=V_\text{rms}I_\text{rms}\cos\phi=(240)(5.0)(0.80)=960\,\text{W}$.

   (b) $\phi=\cos^{-1}(0.80)\approx 37^\circ$ (current lagging).

   (c) Apparent power $=V_\text{rms}I_\text{rms}=(240)(5.0)=1200\,\text{V}\cdot\text{A}$. It exceeds $P_\text{av}$ because the reactive part of the current carries energy that flows back and forth to the source each cycle without being dissipated; only the fraction $\cos\phi$ contributes to average power.

7. 7. (1 pt •) An ideal transformer has a primary of $800$ turns and a secondary of $100$ turns, with $120\,\text{V}$ (rms) applied to the primary.

   (a) Find the secondary voltage.

   (b) The secondary drives a $3.0\,\Omega$ resistive load. Find the secondary current and the primary current.

   Solution

   7.

   (a) $V_2=V_1\dfrac{N_2}{N_1}=120\cdot\dfrac{100}{800}=15\,\text{V}$.

   (b) Secondary current: $I_2=\dfrac{V_2}{R}=\dfrac{15}{3.0}=5.0\,\text{A}$.

   Primary current (ideal, power conserved): $I_1=I_2\dfrac{N_2}{N_1}=5.0\cdot\dfrac{100}{800}=0.63\,\text{A}$.

   (Check: $P_1=V_1 I_1=(120)(0.625)=75\,\text{W}=V_2 I_2=(15)(5.0)=75\,\text{W}$. ✓)

8. Extra Credit (3 pts •••)

   A series R–L–C circuit is driven by a source $v(t)=V_0\cos\omega t$, and carries current $i(t)=I_0\cos(\omega t-\phi)$, where $\phi$ is the phase by which the current lags the voltage.

   (a) Starting from the instantaneous power $p(t)=v(t)\,i(t)$, integrate over one full period $T=2\pi/\omega$ to derive the average power

   $$P_\text{av}=\tfrac{1}{2}V_0 I_0\cos\phi = V_\text{rms}I_\text{rms}\cos\phi.$$

   (b) Write $Z(\omega)=\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}$. By setting $dZ/d\omega=0$, prove that the impedance is a minimum (and therefore the current amplitude a maximum) exactly at $\omega_0=1/\sqrt{LC}$.

   (c) Using part (a), show that the average power dissipated is maximized at $\omega_0$, and give its value there in terms of $V_0$ and $R$.

   Solution

   Extra Credit

   (a) The instantaneous power is

   $$p(t)=v(t)\,i(t)=V_0 I_0\cos\omega t\,\cos(\omega t-\phi).$$

   Expand with $\cos(\omega t-\phi)=\cos\omega t\cos\phi+\sin\omega t\sin\phi$:

   $$p(t)=V_0 I_0\left[\cos^2\omega t\,\cos\phi+\cos\omega t\sin\omega t\,\sin\phi\right].$$

   Average over one period $T=2\pi/\omega$ using

   $$\frac{1}{T}\int_0^T \cos^2\omega t\,dt=\tfrac12,\qquad \frac{1}{T}\int_0^T \cos\omega t\sin\omega t\,dt=0.$$

   (The first follows from $\cos^2\theta=\tfrac12(1+\cos2\theta)$; the second because $\cos\omega t\sin\omega t=\tfrac12\sin2\omega t$ integrates to zero over a whole number of cycles.) Therefore

   $$P_\text{av}=\frac{1}{T}\int_0^T p(t)\,dt=V_0 I_0\cos\phi\cdot\tfrac12=\tfrac12 V_0 I_0\cos\phi.$$

   Writing $V_0=\sqrt2\,V_\text{rms}$ and $I_0=\sqrt2\,I_\text{rms}$ gives $P_\text{av}=V_\text{rms}I_\text{rms}\cos\phi$. $\blacksquare$

   (b) With $Z(\omega)=\sqrt{R^2+g(\omega)^2}$ where $g(\omega)=\omega L-\dfrac{1}{\omega C}$, we have

   $$\frac{dZ}{d\omega}=\frac{g(\omega)\,g'(\omega)}{\sqrt{R^2+g(\omega)^2}},\qquad g'(\omega)=L+\frac{1}{\omega^2 C}.$$

   Since $g'(\omega)>0$ for all $\omega>0$, and $\sqrt{R^2+g^2}>0$, the only way $dZ/d\omega=0$ is $g(\omega)=0$:

   $$\omega L-\frac{1}{\omega C}=0\ \Longrightarrow\ \omega^2=\frac{1}{LC}\ \Longrightarrow\ \omega_0=\frac{1}{\sqrt{LC}}.$$

   This is a minimum: for $\omega<\omega_0$, $g<0$ so $dZ/d\omega<0$ (falling); for $\omega>\omega_0$, $g>0$ so $dZ/d\omega>0$ (rising). At $\omega_0$, $Z=\sqrt{R^2+0}=R$, the smallest possible value, so the current amplitude $I_0=V_0/Z$ is maximal there. $\blacksquare$

   (c) From part (a), $P_\text{av}=\tfrac12 V_0 I_0\cos\phi$. With $I_0=V_0/Z$ and $\cos\phi=R/Z$,

   $$P_\text{av}=\frac12\,V_0\cdot\frac{V_0}{Z}\cdot\frac{R}{Z}=\frac{V_0^2 R}{2Z^2}=\frac{V_0^2 R}{2\left[R^2+g(\omega)^2\right]}.$$

   This is largest when the denominator is smallest, i.e. when $g(\omega)=0$, which is exactly $\omega=\omega_0$ from part (b). At resonance $Z=R$, so

   $$P_\text{av,max}=\frac{V_0^2 R}{2R^2}=\frac{V_0^2}{2R}=\frac{V_\text{rms}^2}{R}.$$