An L-C-R series circuit with 100 Ω resistance is connected to an ac source of 220 V, ω = 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 30°. When only the inductance is removed, the current leads the voltage by 30°. The current in the L-C-R circuit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2.2 A
In a series circuit the phase angle between current and voltage is given by the net reactance divided by the resistance. Removing the capacitor leaves an L-R circuit, and removing the inductor leaves a C-R circuit.
\( \displaystyle \text{L-R circuit: } \tan 30^\circ = \frac{X_L}{R} \)
\( \displaystyle \text{C-R circuit: } \tan 30^\circ = \frac{X_C}{R} \)
\( X_L = X_C \)
With all three elements present the reactances cancel, so the L-C-R circuit is at resonance and its impedance equals the resistance.
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = R = 100\,\Omega \)
\( \displaystyle I = \frac{V}{Z} = \frac{220\,\mathrm{V}}{100\,\Omega} \)
\( I = 2.2\,\mathrm{A} \)
The source voltage is an rms value, so this is the rms current.