A series LCR circuit is connected to an a.c. source of \( 220\ \mathrm{V} \) at \( 60\ \mathrm{Hz} \). The circuit contains a resistor of \( 200\ \Omega \). When only the capacitor is disconnected, the supply voltage leads the current by \( 60^{\circ} \). When only inductor is disconnected, the current leads the supply voltage by \( 60^{\circ} \). The peak current flowing in the complete LCR circuit is approximately
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.55\ \mathrm{A} \)
Each partial circuit gives one reactance from the phase angle, \( \tan\phi = X/R \).
Capacitor removed (R and L only; voltage leads by \( 60^{\circ} \)):
\( X_{L} = R\tan60^{\circ} = 200\sqrt{3}\,\Omega \)
Inductor removed (R and C only; current leads by \( 60^{\circ} \)):
\( X_{C} = R\tan60^{\circ} = 200\sqrt{3}\,\Omega \)
In the complete circuit \( X_{L} = X_{C} \), so the circuit is at resonance and the impedance equals the resistance:
\( \displaystyle Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} = R = 200\,\Omega \)
The source voltage 220 V is an rms value:
\( \displaystyle I_{\mathrm{rms}} = \frac{220\,\mathrm{V}}{200\,\Omega} = 1.1\,\mathrm{A} \)
\( I_{0} = \sqrt{2}\,I_{\mathrm{rms}} = 1.414\times1.1\,\mathrm{A} \approx 1.55\,\mathrm{A} \)