A resistor of resistance R, an inductor of \( (20/\pi)\,\mathrm{H} \) and a capacitor of \( (5/\pi)\,\mu\mathrm{F} \) are connected in series to a voltage source of 220 V, 50 Hz. The value of current in the circuit is 4 A. The impedance of the circuit is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 55\,\Omega \)
The impedance of an ac circuit is the ratio of the rms voltage to the rms current, whatever the elements are.
\( \displaystyle Z = \frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}} = \frac{220\,\mathrm{V}}{4\,\mathrm{A}} = 55\,\Omega \)
This agrees with the given elements. The reactances are
\( \displaystyle X_L = 2\pi f L = 2\pi \times 50 \times \frac{20}{\pi} = 2000\,\Omega \)
\( \displaystyle X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times \frac{5}{\pi}\times10^{-6}} = \frac{1}{5\times10^{-4}} = 2000\,\Omega \)
Since \( X_L = X_C \), the circuit is at resonance, so \( Z = R = 55\,\Omega \).