A 200 V ac source of angular frequency 1000 rad/s is connected to a series LCR circuit with C = 25 μF, R = 20 Ω and L = 40 mH. The potential difference across the resistance will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 200 V
First check whether the circuit is at resonance by comparing the two reactances.
\( X_L = \omega L = 1000\,\mathrm{rad\,s^{-1}} \times 40\times10^{-3}\,\mathrm{H} = 40\,\Omega \)
\( \displaystyle X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 25\times10^{-6}}\,\Omega = 40\,\Omega \)
Since \( X_L = X_C \), the impedance is purely resistive:
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = R = 20\,\Omega \)
\( \displaystyle I = \frac{V}{Z} = \frac{200\,\mathrm{V}}{20\,\Omega} = 10\,\mathrm{A} \)
\( V_R = IR = 10\,\mathrm{A} \times 20\,\Omega = 200\,\mathrm{V} \)
At resonance the voltages across L and C cancel each other, so the whole source voltage appears across R.