A circuit containing a \( \frac{2}{\pi}\,\mathrm{H} \) inductor, a \( \frac{40}{\pi}\,\mu\mathrm{F} \) capacitor and a \( 50\,\Omega \) resistor in series are connected to an a.c. source of emf \( v = 230\sin 314t \). The amplitude of the current in the circuit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3.25 A
In a series LCR circuit the peak current is the peak emf divided by the impedance \( Z = \sqrt{R^{2} + (X_C - X_L)^{2}} \). From \( v = 230\sin 314t \): \( V_0 = 230\,\mathrm{V} \) and \( \omega = 314\,\mathrm{rad\,s^{-1}} \), with \( 314 = 100\pi \).
\( \displaystyle X_L = \omega L = 100\pi \times \frac{2}{\pi} = 200\,\Omega \)
\( \displaystyle X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \frac{40}{\pi}\times10^{-6}} = \frac{1}{4\times10^{-3}} = 250\,\Omega \)
\( \displaystyle Z = \sqrt{50^{2} + (250 - 200)^{2}} = 50\sqrt{2} \approx 70.7\,\Omega \)
\( \displaystyle I_0 = \frac{V_0}{Z} = \frac{230}{70.7} \approx 3.25\,\mathrm{A} \)