AC-Circuits and Power in AC-Circuits — JEE Main Physics MCQs with Solutions
Free JEE Main Physics AC-Circuits and Power in AC-Circuits MCQs with step-by-step solutions (31 questions). Part of Alternating Current. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — AC-Circuits and Power in AC-Circuits · medium · numerical
A $40\,\mu\text{F}$ capacitor is connected to a $200\,\text{V}$, $50\,\text{Hz}$ AC supply. The rms value of the current in the circuit is, nearly
A. $2.05\,\text{A}$
B. $2.5\,\text{A}$ ✓ Correct
C. $25.1\,\text{A}$
D. $1.7\,\text{A}$
Solution: $X_C=\dfrac{1}{2\pi fC}=\dfrac{1}{2\pi(50)(40\times10^{-6})}\approx79.6\,\Omega$. $I_{rms}=\dfrac{V_{rms}}{X_C}=\dfrac{200}{79.6}\approx2.5\,\text{A}$.
Q2 — AC-Circuits and Power in AC-Circuits · medium · theory
A small signal voltage $V(t)=V_0\sin\omega t$ is applied across an ideal capacitor $C$.
A. over a full cycle the capacitor $C$ does not consume any energy from the voltage source ✓ Correct
B. current $I(t)$ is in phase with voltage $V(t)$
C. current $I(t)$ leads voltage $V(t)$ by $180^\circ$
D. current $I(t)$ lags voltage $V(t)$ by $90^\circ$
Solution: For a purely capacitive AC circuit, the current leads the voltage by $\dfrac{\pi}{2}$ (i.e., $90^\circ$), so the phase difference $\phi=\dfrac{\pi}{2}$. Average power $P=V_{rms}I_{rms}\cos\phi=V_{rms}I_{rms}\cos90^\circ=0$. Hence over a full cycle the capacitor consumes no net energy from the source.
Q3 — AC-Circuits and Power in AC-Circuits · medium · theory
A coil of self-inductance $L$ is connected in series with a bulb $B$ and an AC source. The brightness of the bulb decreases when
A. frequency of the AC source is decreased
B. number of turns in the coil is reduced
C. a capacitance of reactance $X_C=X_L$ is included in the same circuit
D. an iron rod is inserted in the coil ✓ Correct
Solution: Current in the circuit $I=\dfrac{E}{\sqrt{\omega^2L^2+R^2}}$. Since $L=\dfrac{\mu_0\mu_rN^2A}{l}$, inserting an iron rod increases the relative permeability $\mu_r$, which increases $L$ (and hence $X_L=\omega L$), increasing the impedance and decreasing the current, so the bulb's brightness decreases.
Q4 — AC-Circuits and Power in AC-Circuits · medium · numerical
An AC voltage is applied to a resistance $R$ and an inductor $L$ in series. If $R$ and the inductive reactance are both equal to $3\,\Omega$, the phase difference between the applied voltage and the current in the circuit is
A. $\dfrac{\pi}{4}$ ✓ Correct
B. $\dfrac{\pi}{2}$
C. zero
D. $\dfrac{\pi}{6}$
Solution: $\tan\phi = \dfrac{X_L}{R} = \dfrac{3}{3} = 1 \Rightarrow \phi = 45^\circ = \dfrac{\pi}{4}\text{ rad}$.
Q5 — AC-Circuits and Power in AC-Circuits · medium · theory
For a series $LCR$ circuit, the power loss at resonance is
A. $\dfrac{V^2}{\omega L-\dfrac{1}{\omega C}}$
B. $i^2C\omega$
C. $i^2R$ ✓ Correct
D. $\dfrac{V^2}{\omega C}$
Solution: At resonance, $X_L = X_C$, so total impedance $Z = R$ and the circuit behaves as purely resistive. Power loss $P = V_{rms}i_{rms}\cos\phi = V_{rms}i_{rms}\;(\because \cos\phi=1) = (i_{rms}R)i_{rms} = i_{rms}^2R$.
Q6 — AC-Circuits and Power in AC-Circuits · medium · theory
The reactance of a capacitor of capacitance $C$ is $X$. If both the frequency and the capacitance be doubled, then the new reactance will be
A. $X$
B. $2X$
C. $4X$
D. $\dfrac{X}{4}$ ✓ Correct
Solution: Capacitive reactance $X = \dfrac{1}{\omega C} = \dfrac{1}{2\pi fC}$, i.e. $X \propto \dfrac{1}{fC}$. Doubling both $f$ and $C$: $\dfrac{X'}{X} = \dfrac{fC}{(2f)(2C)} = \dfrac{1}{4} \Rightarrow X' = \dfrac{X}{4}$.
Q7 — AC-Circuits and Power in AC-Circuits · medium · theory
A wire of resistance $R$ is connected in series with an inductor of reactance $\omega L$. Then the quality factor of the $RL$ circuit is
A. $\dfrac{R}{\omega L}$
B. $\dfrac{\omega L}{R}$ ✓ Correct
C. $\dfrac{R}{\sqrt{R^2+\omega^2L^2}}$
D. $\dfrac{\omega L}{\sqrt{R^2+\omega^2L^2}}$
Solution: Quality factor $Q = 2\pi\dfrac{\text{total energy stored in the circuit}}{\text{loss in energy in each cycle}}$. Energy stored $\propto Li_{rms}^2$, energy lost per cycle $\propto \dfrac{i_{rms}^2R}{f}$, giving $Q = \dfrac{2\pi fL}{R} = \dfrac{\omega L}{R}$.
Q8 — AC-Circuits and Power in AC-Circuits · medium · theory
In an AC circuit with voltage $V$ and current $i$, the power dissipated is
A. Depends on the phase between $V$ and $i$ ✓ Correct
B. $\dfrac{1}{\sqrt{2}}Vi$
C. $\dfrac{1}{2}Vi$
D. $Vi$
Solution: In an AC circuit with voltage $V$ and current $i$, the power dissipated is $P = Vi\cos\phi$, where $\phi$ is the phase difference and $\cos\phi$ is the power factor. Thus, the power dissipated depends on the phase between voltage $V$ and current $i$.
Q9 — AC-Circuits and Power in AC-Circuits · medium · numerical
In an experiment, $200\text{ V}$ AC is applied at the ends of an $LCR$ circuit. The circuit consists of an inductive reactance $(X_L) = 50\,\Omega$, capacitive reactance $(X_C) = 50\,\Omega$ and ohmic resistance $(R) = 10\,\Omega$. The impedance of the circuit is
A. $10\,\Omega$ ✓ Correct
B. $20\,\Omega$
C. $30\,\Omega$
D. $40\,\Omega$
Solution: Impedance $Z = \sqrt{R^2+(X_L-X_C)^2} = \sqrt{(10)^2+(50-50)^2} = \sqrt{100} = 10\,\Omega$.
Q10 — AC-Circuits and Power in AC-Circuits · medium · theory
An $LCR$ series circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of
A. $\pi$
B. $\dfrac{\pi}{2}$
C. $\dfrac{\pi}{4}$
D. zero ✓ Correct
Solution: At resonance, $X_L = X_C$, so the impedance $Z = \sqrt{R^2+(X_L-X_C)^2} = R$; the circuit behaves as purely resistive, so the power factor $\cos\phi = R/Z = 1$, meaning voltage and current are in phase (phase difference is zero).
Q11 — AC-Circuits and Power in AC-Circuits · medium · theory
In an AC circuit, the rms value of current, $i_{rms}$, is related to the peak current, $i_0$, by the relation
A. $i_{rms} = \sqrt{2}\,i_0$
B. $i_{rms} = \pi i_0$
C. $i_{rms} = \dfrac{i_0}{\pi}$
D. $i_{rms} = \dfrac{1}{\sqrt{2}}i_0$ ✓ Correct
Solution: The rms current is defined as $i_{rms} = \sqrt{\overline{i^2}}$, taken over a complete cycle, with $i = i_0\sin\omega t$. Averaging $i_0^2\sin^2\omega t$ over a cycle gives $\overline{i^2} = \dfrac{i_0^2}{2}$, so $i_{rms} = \dfrac{i_0}{\sqrt{2}}$.
Q12 — AC-Circuits and Power in AC-Circuits · medium · theory
The time constant of a $C$-$R$ circuit is
A. $\dfrac{1}{CR}$
B. $\dfrac{C}{R}$
C. $CR$ ✓ Correct
D. $\dfrac{R}{C}$
Solution: The quantity $CR$ is called the time constant (capacitive time constant) of a $CR$ circuit, since it has the dimensions of time: as $R = V/i$ and $C = q/V$, $RC = \dfrac{V}{i}\cdot\dfrac{q}{V} = \dfrac{q}{i} = \dfrac{i\,t}{i} = t$, so $[RC]=[t]=[T]$.
Q13 — AC-Circuits and Power in AC-Circuits · medium · numerical
A series L-C-R circuit containing a $5.0\,\text{H}$ inductor, $80\,\mu\text{F}$ capacitor and $40\,\Omega$ resistor is connected to a $230\,\text{V}$ variable frequency AC source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be
A. $25\,\text{rad/s}$ and $75\,\text{rad/s}$
B. $50\,\text{rad/s}$ and $25\,\text{rad/s}$
C. $46\,\text{rad/s}$ and $54\,\text{rad/s}$ ✓ Correct
D. $42\,\text{rad/s}$ and $58\,\text{rad/s}$
Solution: Resonant angular frequency $\omega_r=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{5\times80\times10^{-6}}}=50\,\text{rad/s}$. Half-power angular frequencies are $\omega_r\pm\Delta\omega$ where $\Delta\omega=\dfrac{R}{2L}=\dfrac{40}{2(5)}=4\,\text{rad/s}$. So $\omega=(50\pm4)\,\text{rad/s}=46\,\text{rad/s}$ and $54\,\text{rad/s}$.
Q14 — AC-Circuits and Power in AC-Circuits · medium · numerical
A series L-C-R circuit is connected to an AC voltage source. When $L$ is removed from the circuit, the phase difference between current and voltage is $\dfrac{\pi}{3}$. If instead $C$ is removed from the circuit, the phase difference is again $\dfrac{\pi}{3}$ between current and voltage. The power factor of the circuit is
A. $0.5$
B. $1.0$ ✓ Correct
C. $-1.0$
D. $\text{zero}$
Solution: With $L$ removed, $\tan\dfrac{\pi}{3}=\dfrac{X_C}{R}\Rightarrow X_C=\sqrt{3}R$. With $C$ removed, $\tan\dfrac{\pi}{3}=\dfrac{X_L}{R}\Rightarrow X_L=\sqrt{3}R$. Since $X_L=X_C$, the circuit is at resonance, so impedance $Z=R$ and power factor $\cos\phi=\dfrac{R}{Z}=1.0$.
Q15 — AC-Circuits and Power in AC-Circuits · medium · theory
A series R-C circuit is connected to an alternating voltage source. Consider two situations: (1) when the capacitor is air filled, and (2) when the capacitor is mica filled. The current through the resistor is $i$ and the voltage across the capacitor is $V$. Taking subscript $a$ for situation 1 and $b$ for situation 2, then
A. $V_a<V_b$
B. $V_a>V_b$ ✓ Correct
C. $i_a<i_b$
D. $V_a=V_b$
Solution: Filling the capacitor with mica increases its capacitance $C$, so the capacitive reactance $X_C=\dfrac{1}{2\pi fC}$ decreases, and the circuit current $i=\dfrac{V_{source}}{\sqrt{R^2+X_C^2}}$ increases (i.e. $i_a<i_b$). However, the voltage across the capacitor $V=iX_C=\dfrac{V_{source}}{\sqrt{4\pi^2f^2C^2R^2+1}}$ decreases as $C$ increases, so the voltage across the capacitor is larger for the air-filled case: $V_a>V_b$.
Q16 — AC-Circuits and Power in AC-Circuits · medium · numerical
In an electrical circuit, $R$, $L$, $C$ and an AC voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\dfrac{\pi}{3}$. If instead $C$ is removed from the circuit, the phase difference is again $\dfrac{\pi}{3}$. The power factor of the circuit is
A. $1/2$
B. $1/\sqrt{2}$
C. $1$ ✓ Correct
D. $\sqrt{3}/2$
Solution: The phase difference for the L-C-R series circuit is $\tan\phi=\dfrac{X_L-X_C}{R}$. When $L$ is removed, $\tan\dfrac{\pi}{3}=\dfrac{X_C}{R}\Rightarrow X_C=\sqrt{3}R$. When $C$ is removed, $\tan\dfrac{\pi}{3}=\dfrac{X_L}{R}\Rightarrow X_L=\sqrt{3}R$. So $X_L=X_C$, which is the condition of resonance: $\tan\phi=\dfrac{\sqrt{3}R-\sqrt{3}R}{R}=0$, so $\phi=0$ and the power factor $\cos\phi=1$.
Q17 — AC-Circuits and Power in AC-Circuits · medium · theory
In a circuit, $L$, $C$ and $R$ are connected in series with an alternating voltage source of frequency $f$. The current leads the voltage by $45^\circ$. The value of $C$ is
A. $\dfrac{1}{2\pi f(2\pi f L-R)}$
B. $\dfrac{1}{\pi f(2\pi f L-R)}$
C. $\dfrac{1}{2\pi f(2\pi f L+R)}$ ✓ Correct
D. $\dfrac{1}{\pi f(2\pi f L+R)}$
Solution: Since the current leads the voltage, the circuit is net capacitive: $\tan\phi = \dfrac{\dfrac{1}{2\pi f C}-2\pi f L}{R}$. With $\phi=45^\circ$: $1 = \dfrac{\dfrac{1}{2\pi f C}-2\pi f L}{R} \Rightarrow \dfrac{1}{2\pi f C} = 2\pi f L+R \Rightarrow C = \dfrac{1}{2\pi f(2\pi f L+R)}$.
Q18 — AC-Circuits and Power in AC-Circuits · medium · numerical
An inductor of inductance $L$, a capacitor of capacitance $C$ and a resistor of resistance $R$ are connected in series to an AC source of potential difference $V$ volts. The potential difference across $L$, $C$ and $R$ is $40\,\text{V}$, $10\,\text{V}$ and $40\,\text{V}$, respectively. The amplitude of current flowing through the L-C-R series circuit is $10\sqrt{2}\,\text{A}$. The impedance of the circuit is
A. $4\sqrt{2}\,\Omega$
B. $5\sqrt{2}\,\Omega$
C. $4\,\Omega$
D. $5\,\Omega$ ✓ Correct
Solution: $V_L=40\,\text{V}$, $V_C=10\,\text{V}$, $V_R=40\,\text{V}$, and $I_0=10\sqrt{2}\,\text{A}$, so $I_{rms}=\dfrac{I_0}{\sqrt{2}}=10\,\text{A}$. $V_{rms}=\sqrt{V_R^2+(V_L-V_C)^2}=\sqrt{40^2+(40-10)^2}=\sqrt{1600+900}=50\,\text{V}$. Impedance $Z=\dfrac{V_{rms}}{I_{rms}}=\dfrac{50}{10}=5\,\Omega$.
Q19 — AC-Circuits and Power in AC-Circuits · medium · numerical
A circuit when connected to an AC source of $12\,\text{V}$ gives a current of $0.2\,\text{A}$. The same circuit when connected to a DC source of $12\,\text{V}$, gives a current of $0.4\,\text{A}$. The circuit is
A. series LR ✓ Correct
B. series RC
C. series LC
D. series LCR
Solution: With AC, impedance $Z=\dfrac{V}{I}=\dfrac{12}{0.2}=60\,\Omega$. With DC (steady state), only resistance matters: $R=\dfrac{V}{I}=\dfrac{12}{0.4}=30\,\Omega$. Since a steady current flows with DC, the circuit cannot contain a capacitor in series (which would block DC in steady state). So it must be a series L-R circuit, where $Z=\sqrt{R^2+X_L^2}=60\,\Omega$ is consistent with $R=30\,\Omega$.
Q20 — AC-Circuits and Power in AC-Circuits · medium · numerical
In an electromagnetic wave in free space the root mean square value of the electric field is $E_{rms}=6\,\text{V/m}$. The peak value of the magnetic field is
A. $1.41\times10^{-8}\,\text{T}$
B. $2.83\times10^{-8}\,\text{T}$ ✓ Correct
C. $0.70\times10^{-8}\,\text{T}$
D. $4.23\times10^{-8}\,\text{T}$
Solution: Peak electric field $E_0=\sqrt{2}\,E_{rms}=\sqrt{2}(6)\approx8.49\,\text{V/m}$. Since $c=\dfrac{E_0}{B_0}$, $B_0=\dfrac{E_0}{c}=\dfrac{8.49}{3\times10^8}\approx2.83\times10^{-8}\,\text{T}$.
Q21 — AC-Circuits and Power in AC-Circuits · medium · numerical
An inductor $20\,\text{mH}$, a capacitor $50\,\mu\text{F}$ and a resistor $40\,\Omega$ are connected in series across a source of emf $V=10\sin340t$. The power loss in the AC circuit is
A. $0.67\,\text{W}$
B. $0.76\,\text{W}$
C. $0.89\,\text{W}$
D. $0.51\,\text{W}$ ✓ Correct
Solution: $L=20\times10^{-3}\,\text{H}$, $C=50\times10^{-6}\,\text{F}$, $R=40\,\Omega$, $\omega=340\,\text{rad/s}$. $X_L=\omega L=340\times0.02=6.8\,\Omega$. $X_C=\dfrac{1}{\omega C}=\dfrac{1}{340\times50\times10^{-6}}\approx58.8\,\Omega$. $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{40^2+(6.8-58.8)^2}=\sqrt{1600+2704}\approx65.6\,\Omega$. $E_{rms}=\dfrac{10}{\sqrt{2}}$. Power loss $P_{av}=\left(\dfrac{E_{rms}}{Z}\right)^2R\approx0.51\,\text{W}$.
Q22 — AC-Circuits and Power in AC-Circuits · medium · numerical
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
A. $R=20\,\Omega,\ L=1.5\,\text{H},\ C=35\,\mu\text{F}$
B. $R=25\,\Omega,\ L=2.5\,\text{H},\ C=45\,\mu\text{F}$
C. $R=15\,\Omega,\ L=3.5\,\text{H},\ C=30\,\mu\text{F}$ ✓ Correct
D. $R=25\,\Omega,\ L=1.5\,\text{H},\ C=45\,\mu\text{F}$
Solution: For better tuning, the peak of current growth at resonance must be sharp, which requires a high quality factor $Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$. Among the given options, $R=15\,\Omega,\ L=3.5\,\text{H},\ C=30\,\mu\text{F}$ gives the highest value of $Q$.
Q23 — AC-Circuits and Power in AC-Circuits · medium · numerical
The potential differences across the resistance, capacitance and inductance are $80\,\text{V}$, $40\,\text{V}$ and $100\,\text{V}$ respectively in an L-C-R circuit. The power factor of this circuit is
A. $0.4$
B. $0.5$
C. $0.8$ ✓ Correct
D. $1.0$
Solution: Power factor $\cos\phi=\dfrac{V_R}{\sqrt{V_R^2+(V_L-V_C)^2}}=\dfrac{80}{\sqrt{80^2+(100-40)^2}}=\dfrac{80}{\sqrt{6400+3600}}=\dfrac{80}{100}=0.8$.
Q24 — AC-Circuits and Power in AC-Circuits · medium · numerical
A $100\,\Omega$ resistance and a capacitor of $100\,\Omega$ reactance are connected in series across a $220\,\text{V}$ source. When the capacitor is 50% charged, the peak value of the displacement current is
A. $2.2\,\text{A}$ ✓ Correct
B. $11\,\text{A}$
C. $4.4\,\text{A}$
D. $11\sqrt{2}\,\text{A}$
Solution: Impedance $Z=\sqrt{R^2+X_C^2}=\sqrt{100^2+100^2}=100\sqrt{2}\,\Omega$. Peak current $I_{max}=\dfrac{V_{max}}{Z}=\dfrac{220\sqrt{2}}{100\sqrt{2}}=2.2\,\text{A}$. The displacement current inside the capacitor equals the conduction current in the circuit at every instant, regardless of how much the capacitor is charged, so the peak displacement current is also $2.2\,\text{A}$.
Q25 — AC-Circuits and Power in AC-Circuits · medium · numerical
A resistance $R$ draws power $P$ when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes $Z$, the power drawn will be
A. $P\left(\dfrac{R}{Z}\right)^2$ ✓ Correct
B. $P\sqrt{\dfrac{R}{Z}}$
C. $P\left(\dfrac{R}{Z}\right)$
D. $P$
Solution: With only $R$: $P=V_{rms}I_{rms}=\dfrac{V_{rms}^2}{R}$. With $L$ added in series, power drawn $P'=V_{rms}I_{rms}\cos\phi=\dfrac{V_{rms}^2}{Z}\cdot\dfrac{R}{Z}=\dfrac{V_{rms}^2R}{Z^2}=P\dfrac{R^2}{Z^2}=P\left(\dfrac{R}{Z}\right)^2$.
Q26 — AC-Circuits and Power in AC-Circuits · medium · numerical
An AC voltage $e = 200\sqrt{2}\sin(100t)$ volt is connected to a capacitor of capacity $1\,\mu\text{F}$. The rms value of the current in the circuit is
A. 100 mA
B. 200 mA
C. 20 mA ✓ Correct
D. 10 mA
Solution: Comparing with $e = e_m\sin\omega t$, $e_{rms} = 200\text{ V}$, $\omega = 100\text{ rad/s}$, $C = 1\times10^{-6}\text{ F}$. So $X_C = \dfrac{1}{\omega C} = \dfrac{1}{100\times10^{-6}} = 10^4\,\Omega$. Hence $i_{rms} = \dfrac{e_{rms}}{X_C} = \dfrac{200}{10^4} = 2\times10^{-2}\text{ A} = 20\text{ mA}$.
Q27 — AC-Circuits and Power in AC-Circuits · medium · numerical
In a series circuit consisting of an inductor $L$, a capacitor $C$ and a resistor $R = 100\,\Omega$ connected across a $220\text{ V}$, $50\text{ Hz}$ AC source (with an ammeter $A$ in the main line and voltmeters $V_1$, $V_2$, $V_3$ measuring the voltages across $L$, $C$ and $R$ respectively), the readings of voltmeter $V_1$ and $V_2$ are $300\text{ V}$ each. The readings of the voltmeter $V_3$ and ammeter $A$ are respectively
A. 150 V, 2.2 A
B. 220 V, 2.2 A ✓ Correct
C. 220 V, 2.0 A
D. 100 V, 2.0 A
Solution: For a series $LCR$ circuit, $V = \sqrt{V_R^2+(V_L-V_C)^2}$. Since $V_L = V_C$ (both read $300\text{ V}$), $V = V_R = 220\text{ V}$ (the applied voltage). Also $i = \dfrac{V}{R} = \dfrac{220}{100} = 2.2\text{ A}$.
Q28 — AC-Circuits and Power in AC-Circuits · medium · theory
Power dissipated in an $L$-$C$-$R$ series circuit connected to an AC source of emf $\varepsilon$ is
A. $\dfrac{\varepsilon^2 R}{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}$ ✓ Correct
B. $\dfrac{\varepsilon^2\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}}{R}$
C. $\dfrac{\varepsilon^2\left[R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2\right]}{R}$
D. $\dfrac{\varepsilon^2 R}{\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}}$
Solution: $P = i_{rms}^2R = \dfrac{\varepsilon_{rms}^2}{|Z|^2}R = \dfrac{\varepsilon^2R}{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}$, since $I_{rms}=\dfrac{V_{rms}}{Z}$.
Q29 — AC-Circuits and Power in AC-Circuits · medium · numerical
What is the value of inductance $L$ for which the current is maximum in a series $LCR$ circuit with $C = 10\,\mu\text{F}$ and $\omega = 1000\,\text{s}^{-1}$?
A. 100 mH ✓ Correct
B. 1 mH
C. Cannot be calculated unless $R$ is known
D. 10 mH
Solution: Current is maximum at resonance, when $\omega L = \dfrac{1}{\omega C}$, i.e. $L = \dfrac{1}{\omega^2 C}$. Substituting $\omega = 1000\,\text{s}^{-1}$, $C = 10\times10^{-6}\text{ F}$: $L = \dfrac{1}{(1000)^2\times10\times10^{-6}} = 0.1\text{ H} = 100\text{ mH}$.
Q30 — AC-Circuits and Power in AC-Circuits · medium · numerical
A coil of inductive reactance $31\,\Omega$ has a resistance of $8\,\Omega$. It is placed in series with a condenser of capacitive reactance $25\,\Omega$. The combination is connected to an AC source of $110\text{ V}$. The power factor of the circuit is
A. 0.56
B. 0.64
C. 0.80 ✓ Correct
D. 0.33
Solution: Power factor $\cos\phi = \dfrac{R}{Z} = \dfrac{R}{\sqrt{R^2+(X_L-X_C)^2}}$. With $R=8\,\Omega$, $X_L=31\,\Omega$, $X_C=25\,\Omega$: $\cos\phi = \dfrac{8}{\sqrt{8^2+(31-25)^2}} = \dfrac{8}{\sqrt{64+36}} = \dfrac{8}{10} = 0.80$.