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AC through Resistor, Inductor & Capacitor — MH-CET Physics MCQs with Solutions
Free MH-CET Physics AC through Resistor, Inductor & Capacitor MCQs with step-by-step solutions (21 questions). Part of AC Circuits. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — AC through Resistor, Inductor & Capacitor · easy · theory
In a purely resistive AC circuit, the current and the applied voltage are:
A. Out of phase by $\pi$
B. Out of phase by $\dfrac{\pi}{2}$
C. Out of phase by $\dfrac{\pi}{4}$
D. In phase with each other ✓ Correct
Solution: A resistor offers no reactance, so the current follows the voltage instantaneously.
Q2 — AC through Resistor, Inductor & Capacitor · easy · theory
In a purely inductive AC circuit, the current:
A. Leads the applied voltage by $\dfrac{\pi}{2}$
B. Lags behind the applied voltage by $\dfrac{\pi}{2}$ ✓ Correct
C. Is in phase with the voltage
D. Lags behind the voltage by $\pi$
Solution: The back EMF of the inductor opposes the rise of current, delaying it by a quarter cycle.
Q3 — AC through Resistor, Inductor & Capacitor · easy · theory
In a purely capacitive AC circuit, the current:
A. Leads the applied voltage by $\dfrac{\pi}{2}$ ✓ Correct
B. Is in phase with the voltage
C. Leads the voltage by $\pi$
D. Lags behind the applied voltage by $\dfrac{\pi}{2}$
Solution: Charge must flow before the voltage across the plates can build up, so current leads by a quarter cycle.
Q4 — AC through Resistor, Inductor & Capacitor · easy · theory
The inductive reactance of a coil of inductance $L$ at angular frequency $\omega$ is:
A. $\omega^2L$
B. $\omega L$ ✓ Correct
C. $\dfrac{\omega}{L}$
D. $\dfrac{1}{\omega L}$
Solution: It is measured in ohm and grows in proportion to the frequency.
Q5 — AC through Resistor, Inductor & Capacitor · easy · theory
The capacitive reactance of a capacitor $C$ at angular frequency $\omega$ is:
A. $\dfrac{\omega}{C}$
B. $\dfrac{1}{\omega^2C}$
C. $\dfrac{1}{\omega C}$ ✓ Correct
D. $\omega C$
Solution: Capacitive reactance falls as the frequency rises, which is why a capacitor passes high frequencies readily.
Q6 — AC through Resistor, Inductor & Capacitor · medium · theory
With increasing frequency, the inductive and capacitive reactances respectively:
A. Both increase
B. Increase and decrease ✓ Correct
C. Both decrease
D. Decrease and increase
Solution: $X_L = \omega L$ rises with $\omega$, while $X_C = \dfrac{1}{\omega C}$ falls.
Q7 — AC through Resistor, Inductor & Capacitor · medium · theory
An inductor connected to a steady DC supply offers:
A. Zero reactance, behaving as an ordinary conductor ✓ Correct
B. A reactance that grows with time
C. The same reactance as at $50\text{ Hz}$
D. Infinite reactance, blocking the current
Solution: For DC the frequency is zero, so $X_L = \omega L = 0$ once the steady state is reached.
Q8 — AC through Resistor, Inductor & Capacitor · medium · theory
A capacitor connected to a steady DC supply:
A. Offers zero reactance
B. Blocks the current once fully charged ✓ Correct
C. Passes the current freely
D. Behaves as a pure resistor
Solution: At $f = 0$, $X_C = \dfrac{1}{\omega C} \to \infty$, so no steady current flows.
Q9 — AC through Resistor, Inductor & Capacitor · medium · theory
The average power consumed over a full cycle in a pure inductor or a pure capacitor is:
A. Maximum
B. Equal to $V_{rms}I_{rms}$
C. Zero ✓ Correct
D. Half of $V_{rms}I_{rms}$
Solution: With a phase difference of $\dfrac{\pi}{2}$, $\cos\phi = 0$; energy is stored and returned each quarter cycle.
Q10 — AC through Resistor, Inductor & Capacitor · medium · numerical
An alternating voltage $e = 100\sin(100\pi t)\text{ V}$ is applied across an inductor of $L = \dfrac{1}{\pi}\text{ H}$. The inductive reactance is:
A. $50\,\Omega$
B. $314\,\Omega$
C. $200\,\Omega$
D. $100\,\Omega$ ✓ Correct
Solution: $X_L = \omega L = 100\pi \times \dfrac{1}{\pi} = 100\,\Omega$.
Q11 — AC through Resistor, Inductor & Capacitor · hard · numerical
An alternating voltage $V = 200\sqrt{2}\sin(100t)\text{ V}$ is applied across a $50\,\mu\text{F}$ capacitor. The RMS current is:
A. $1.414\text{ A}$
B. $0.707\text{ A}$
C. $2.0\text{ A}$
D. $1.0\text{ A}$ ✓ Correct
Solution: $X_C = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 50 \times 10^{-6}} = 200\,\Omega$, so $I_{rms} = \dfrac{200}{200} = 1.0\text{ A}$.
Q12 — AC through Resistor, Inductor & Capacitor · medium · numerical
The inductive reactance of a $0.1\text{ H}$ coil at $50\text{ Hz}$ is approximately:
A. $314\,\Omega$
B. $3.14\,\Omega$
C. $31.4\,\Omega$ ✓ Correct
D. $5\,\Omega$
Solution: $X_L = 2\pi fL = 2\pi \times 50 \times 0.1 \approx 31.4\,\Omega$.
Q13 — AC through Resistor, Inductor & Capacitor · hard · numerical
The capacitive reactance of a $100\,\mu\text{F}$ capacitor at $50\text{ Hz}$ is approximately:
A. $100\,\Omega$
B. $3.18\,\Omega$
C. $318\,\Omega$
D. $31.8\,\Omega$ ✓ Correct
Solution: $X_C = \dfrac{1}{2\pi fC} = \dfrac{1}{314 \times 10^{-4}} \approx 31.8\,\Omega$.
Q14 — AC through Resistor, Inductor & Capacitor · easy · numerical
If the frequency of the supply is doubled, the inductive reactance of a coil:
A. Remains unchanged
B. Halves
C. Doubles ✓ Correct
D. Becomes four times
Solution: $X_L = 2\pi fL \propto f$.
Q15 — AC through Resistor, Inductor & Capacitor · easy · numerical
If the frequency of the supply is doubled, the capacitive reactance of a capacitor:
A. Halves ✓ Correct
B. Remains unchanged
C. Becomes four times
D. Doubles
Solution: $X_C = \dfrac{1}{2\pi fC} \propto \dfrac{1}{f}$.
Q16 — AC through Resistor, Inductor & Capacitor · easy · numerical
An inductor of $0.5\text{ H}$ is connected to a supply of angular frequency $100\text{ rad/s}$. Its reactance is:
A. $50\,\Omega$ ✓ Correct
B. $200\,\Omega$
C. $0.005\,\Omega$
D. $100\,\Omega$
Solution: $X_L = \omega L = 100 \times 0.5 = 50\,\Omega$.
Q17 — AC through Resistor, Inductor & Capacitor · medium · numerical
A $10\,\mu\text{F}$ capacitor is connected to a supply of angular frequency $1000\text{ rad/s}$. Its reactance is:
A. $100\,\Omega$ ✓ Correct
B. $0.01\,\Omega$
C. $10\,\Omega$
D. $1000\,\Omega$
Solution: $X_C = \dfrac{1}{\omega C} = \dfrac{1}{1000 \times 10^{-5}} = 100\,\Omega$.
Q18 — AC through Resistor, Inductor & Capacitor · easy · numerical
An RMS voltage of $100\text{ V}$ is applied across a $50\,\Omega$ resistor. The RMS current is:
A. $0.5\text{ A}$
B. $5000\text{ A}$
C. $1.41\text{ A}$
D. $2\text{ A}$ ✓ Correct
Solution: $I_{rms} = \dfrac{V_{rms}}{R} = \dfrac{100}{50} = 2\text{ A}$.
Q19 — AC through Resistor, Inductor & Capacitor · medium · numerical
A pure inductor of reactance $20\,\Omega$ carries an RMS voltage of $100\text{ V}$. The RMS current is:
A. $7.07\text{ A}$
B. $5\text{ A}$ ✓ Correct
C. $0.2\text{ A}$
D. $2000\text{ A}$
Solution: $I_{rms} = \dfrac{V_{rms}}{X_L} = \dfrac{100}{20} = 5\text{ A}$.
Q20 — AC through Resistor, Inductor & Capacitor · medium · numerical
In an AC circuit, the inductive and capacitive reactances are equal. The circuit then behaves as:
A. An open circuit
B. Purely inductive
C. Purely capacitive
D. Purely resistive ✓ Correct
Solution: The two reactances cancel, leaving only the resistance, which is the condition for resonance.
Q21 — AC through Resistor, Inductor & Capacitor · medium · numerical
A coil of inductance $L$ is connected to a DC source in the steady state. The reactance it offers is:
A. $\omega L$ with $\omega = 50$
B. Equal to its resistance
C. Infinite
D. Zero ✓ Correct
Solution: For DC, $f = 0$ so $X_L = 2\pi fL = 0$; only the ohmic resistance of the winding remains.