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Alternating Current
Series LCR Circuit
3 questions from 3 papers, each with its answer.
- questions
- 3
- papers
- 3
- years asked, 2017–2026
- 1 of 10
- of Alternating Current
- 5%
Appears in
Concepts tested
- Impedance and Phase Angle
- Resonance
- Resonance and power factor
All Questions
2025 1 question
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To an AC power supply of $220\text{ V}$ at $50\text{ Hz}$, a resistor of $20\,\Omega$, a capacitor of reactance $25\,\Omega$ and an inductor of reactance $45\,\Omega$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively:
Impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}\,\Omega$. Current $I = 220 / (20\sqrt{2}) \approx 7.8\text{ A}$ and $\tan\phi = 20/20 = 1 \implies \phi = 45^\circ$.
Given $R = 20\,\Omega$, $X_L = 45\,\Omega$, $X_C = 25\,\Omega$. Net reactance $X = X_L - X_C = 45 - 25 = 20\,\Omega$. Total circuit impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2} \approx 28.28\,\Omega$. The RMS current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{28.28} \approx 7.78\text{ A} \approx 7.8\text{ A}$. The phase angle $\phi$ is $\tan\phi = \frac{X_L - X_C}{R} = \frac{20}{20} = 1 \implies \phi = 45^\circ$.
2012 AIPMT-PRE 1 question
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In an electrical circuit R, L, C and an a.c. 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 $\frac{\pi}{3}$. If instead, C is removed from the circuit, the phase difference is again $\frac{\pi}{3}$. The power factor of the circuit isWith L removed (RC circuit): $\tan\frac{\pi}{3} = \frac{X_C}{R}$
With C removed (RL circuit): $\tan\frac{\pi}{3} = \frac{X_L}{R}$
So $X_L = X_C$: the full LCR circuit is at resonance.
$\tan\phi = \frac{X_L - X_C}{R} = 0 \Rightarrow \phi = 0$
Power factor $= \cos\phi = 1$
2010 AIPMT-PRE 1 question
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In the given circuit the readings of voltmeters $V_1$ and $V_2$ are 300 volts each. The readings of the voltmeter $V_3$ and ammeter A are respectively
The circuit is a series LCR circuit across a 220 V, 50 Hz source, with $V_1$ across L, $V_2$ across C and $V_3$ across R = 100 $\Omega$.
$V_L = V_C = 300$ V. The voltages across L and C are $180^\circ$ out of phase, so they cancel: $V_L - V_C = 0$. The circuit is at resonance.
Source voltage: $V = \sqrt{V_R^2 + (V_L - V_C)^2} = V_R$
So $V_3 = V_R = 220$ V
At resonance the impedance is Z = R, so the current is $I = \frac{V}{R} = \frac{220}{100} = 2.2$ A
The readings are 220 V and 2.2 A.