A capacitor of \( 100\,\mu\mathrm{F} \), a resistor of resistance \( 50\,\Omega \) and an inductor of inductance 0.5 H are connected in series with a 110 V, 50 Hz AC source. The impedance of the circuit is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 134.6\,\Omega \)
In a series LCR circuit the impedance combines the resistance with the net reactance \( X_L - X_C \).
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} \)
\( \omega = 2\pi f = 2\pi \times 50 = 314\,\mathrm{rad\,s^{-1}} \)
\( C = 100\,\mu\mathrm{F} = 1\times10^{-4}\,\mathrm{F} \)
\( X_L = \omega L = 314 \times 0.5 = 157\,\Omega \)
\( \displaystyle X_C = \frac{1}{\omega C} = \frac{1}{314 \times 10^{-4}} = 31.8\,\Omega \)
\( X_L - X_C = 157 - 31.8 = 125.2\,\Omega \)
\( \displaystyle Z = \sqrt{50^{2} + 125.2^{2}} = \sqrt{2500 + 15675} = \sqrt{18175} \approx 134.8\,\Omega \)
This is approximately \( 134.6\,\Omega \); the small difference comes from rounding \( \pi \) and the reactances.