Series LCR Circuit & Impedance — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Series LCR Circuit & Impedance 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 — Series LCR Circuit & Impedance · easy · theory
The impedance of a series LCR circuit is given by:
A. $Z = R + X_L + X_C$
B. $Z = \sqrt{R^2 - (X_L - X_C)^2}$
C. $Z = \sqrt{R^2 + (X_L - X_C)^2}$ ✓ Correct
D. $Z = \sqrt{R^2 + X_L^2 + X_C^2}$
Solution: Resistance and net reactance add as perpendicular sides of the impedance triangle.
Q2 — Series LCR Circuit & Impedance · medium · theory
The phase angle between voltage and current in a series LCR circuit is given by:
A. $\sin\phi = \dfrac{X_L - X_C}{R}$
B. $\tan\phi = \dfrac{X_L - X_C}{R}$ ✓ Correct
C. $\cos\phi = \dfrac{X_L - X_C}{Z}$
D. $\tan\phi = \dfrac{R}{X_L - X_C}$
Solution: The net reactance is the opposite side of the impedance triangle and $R$ the adjacent side.
Q3 — Series LCR Circuit & Impedance · medium · theory
In a series LCR circuit with $X_L > X_C$, the circuit is:
A. Capacitive, with the current leading the voltage
B. At resonance
C. Purely resistive
D. Inductive, with the current lagging behind the voltage ✓ Correct
Solution: The larger inductive reactance dominates, so the circuit behaves like an inductor with some resistance.
Q4 — Series LCR Circuit & Impedance · medium · theory
In a series LCR circuit with $X_C > X_L$, the circuit is:
A. Purely resistive
B. Non-conducting
C. Capacitive, with the current leading the voltage ✓ Correct
D. Inductive, with the current lagging the voltage
Solution: The dominant capacitive reactance makes the current lead the applied voltage.
Q5 — Series LCR Circuit & Impedance · easy · theory
In a series LCR circuit, the phase angle between the applied voltage and the current is zero when:
A. $X_L = X_C$ ✓ Correct
B. $X_L < X_C$
C. $X_L > X_C$
D. $R = 0$
Solution: The reactances cancel, the impedance reduces to $R$, and the circuit is purely resistive.
Q6 — Series LCR Circuit & Impedance · medium · theory
In the impedance triangle of a series LCR circuit, the hypotenuse represents:
A. The impedance $Z$ ✓ Correct
B. The net reactance
C. The phase angle
D. The resistance $R$
Solution: The two perpendicular sides are $R$ and $(X_L - X_C)$, and the angle between $Z$ and $R$ is the phase angle.
Q7 — Series LCR Circuit & Impedance · easy · theory
The SI unit of impedance is the:
A. Volt
B. Henry
C. Farad
D. Ohm ✓ Correct
Solution: Impedance is the AC generalisation of resistance and carries the same unit.
Q8 — Series LCR Circuit & Impedance · medium · theory
In a series LCR circuit, the quantity that is the same for all three elements is the:
A. Voltage
B. Current ✓ Correct
C. Power dissipated
D. Phase angle
Solution: Series elements carry a common current, while the voltages across them differ in both magnitude and phase.
Q9 — Series LCR Circuit & Impedance · hard · numerical
A series LCR circuit has $R = 30\,\Omega$, $X_L = 80\,\Omega$ and $X_C = 40\,\Omega$. Its impedance and power factor are:
A. $50\,\Omega$ and $0.8$
B. $70\,\Omega$ and $0.43$
C. $30\,\Omega$ and $1.0$
D. $50\,\Omega$ and $0.6$ ✓ Correct
Solution: Net reactance $= 40\,\Omega$, so $Z = \sqrt{30^2 + 40^2} = 50\,\Omega$ and $\cos\phi = \dfrac{30}{50} = 0.6$.
Q10 — Series LCR Circuit & Impedance · easy · numerical
A series circuit has $R = 3\,\Omega$ and $X_L = 4\,\Omega$ with no capacitor. Its impedance is:
A. $5\,\Omega$ ✓ Correct
B. $1\,\Omega$
C. $12\,\Omega$
D. $7\,\Omega$
Solution: $Z = \sqrt{3^2 + 4^2} = 5\,\Omega$.
Q11 — Series LCR Circuit & Impedance · easy · numerical
A series circuit has $R = 6\,\Omega$ and net reactance $8\,\Omega$. Its impedance is:
A. $2\,\Omega$
B. $48\,\Omega$
C. $14\,\Omega$
D. $10\,\Omega$ ✓ Correct
Solution: $Z = \sqrt{36 + 64} = 10\,\Omega$.
Q12 — Series LCR Circuit & Impedance · medium · numerical
A series circuit has $R = 40\,\Omega$ and $X_L = 30\,\Omega$. Its impedance is:
A. $35\,\Omega$
B. $50\,\Omega$ ✓ Correct
C. $70\,\Omega$
D. $10\,\Omega$
Solution: $Z = \sqrt{1600 + 900} = 50\,\Omega$.
Q13 — Series LCR Circuit & Impedance · medium · numerical
A series circuit has $R = 8\,\Omega$ and $X_C = 6\,\Omega$ with no inductor. Its impedance is:
A. $2\,\Omega$
B. $14\,\Omega$
C. $10\,\Omega$ ✓ Correct
D. $48\,\Omega$
Solution: $Z = \sqrt{64 + 36} = 10\,\Omega$.
Q14 — Series LCR Circuit & Impedance · easy · numerical
An RMS voltage of $100\text{ V}$ is applied to a series circuit of impedance $50\,\Omega$. The RMS current is:
A. $5000\text{ A}$
B. $0.5\text{ A}$
C. $2\text{ A}$ ✓ Correct
D. $1.41\text{ A}$
Solution: $I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{100}{50} = 2\text{ A}$.
Q15 — Series LCR Circuit & Impedance · hard · numerical
A series LCR circuit has $R = 30\,\Omega$ and net reactance $40\,\Omega$. The phase angle is approximately:
A. $45^\circ$
B. $53.1^\circ$ ✓ Correct
C. $30^\circ$
D. $36.9^\circ$
Solution: $\tan\phi = \dfrac{40}{30} = 1.333$, so $\phi \approx 53.1^\circ$.
Q16 — Series LCR Circuit & Impedance · medium · numerical
A series LCR circuit has $R = 100\,\Omega$, $X_L = 100\,\Omega$ and $X_C = 100\,\Omega$. Its impedance is:
A. Zero
B. $141\,\Omega$
C. $300\,\Omega$
D. $100\,\Omega$ ✓ Correct
Solution: The reactances cancel, leaving $Z = R = 100\,\Omega$.
Q17 — Series LCR Circuit & Impedance · medium · numerical
A series circuit has $R = 5\,\Omega$ and $X_L = 12\,\Omega$. Its impedance is:
A. $13\,\Omega$ ✓ Correct
B. $7\,\Omega$
C. $60\,\Omega$
D. $17\,\Omega$
Solution: $Z = \sqrt{25 + 144} = \sqrt{169} = 13\,\Omega$.
Q18 — Series LCR Circuit & Impedance · easy · numerical
In a series LCR circuit a current of $2\text{ A}$ flows through a resistance of $30\,\Omega$. The voltage across the resistor is:
A. $15\text{ V}$
B. $120\text{ V}$
C. $30\text{ V}$
D. $60\text{ V}$ ✓ Correct
Solution: $V_R = IR = 2 \times 30 = 60\text{ V}$.
Q19 — Series LCR Circuit & Impedance · medium · numerical
In a series LCR circuit a current of $2\text{ A}$ flows through an inductive reactance of $80\,\Omega$. The voltage across the inductor is:
A. $80\text{ V}$
B. $320\text{ V}$
C. $40\text{ V}$
D. $160\text{ V}$ ✓ Correct
Solution: $V_L = IX_L = 2 \times 80 = 160\text{ V}$.
Q20 — Series LCR Circuit & Impedance · medium · numerical
A series LCR circuit has $R = 6\,\Omega$ and impedance $10\,\Omega$. Its power factor is:
A. $0.4$
B. $0.8$
C. $0.6$ ✓ Correct
D. $1.67$
Solution: $\cos\phi = \dfrac{R}{Z} = \dfrac{6}{10} = 0.6$.
Q21 — Series LCR Circuit & Impedance · medium · numerical
A series LCR circuit has $R = 9\,\Omega$ and net reactance $12\,\Omega$. Its impedance is:
A. $15\,\Omega$ ✓ Correct
B. $3\,\Omega$
C. $21\,\Omega$
D. $108\,\Omega$
Solution: $Z = \sqrt{81 + 144} = \sqrt{225} = 15\,\Omega$.