A circuit containing a $\frac{2}{\pi}H$ inductor, a $\frac{40}{\pi}μF$ capacitor and a 50 Ω 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
Given:
$L=\frac{2}{\pi}\,H$, $C=\frac{40}{\pi}\,\mu F=\frac{40}{\pi}\times10^{-6}\,F$, $R=50\,\Omega$
Source: $v=230\sin(314t)$ → $\omega=314\,rad/s$
Inductive reactance:
$X_L=\omega L=314\cdot\frac{2}{\pi}=\frac{628}{\pi}\approx200\,\Omega$
Capacitive reactance:
$X_C=\frac{1}{\omega C}=\frac{1}{314\cdot\frac{40}{\pi}\cdot10^{-6}}=\frac{\pi}{314\cdot40\cdot10^{-6}}$
$X_C=\frac{\pi}{0.01256}\approx250\,\Omega$
Net reactance:
$X=X_L-X_C=200-250=-50\,\Omega$
Impedance:
$Z=\sqrt{R^2+X^2}=\sqrt{50^2+(-50)^2}=\sqrt{2500+2500}= \sqrt{5000}\approx70.71\,\Omega$
Current amplitude:
$I_0=\frac{V_0}{Z}=\frac{230}{70.71}\approx3.25\,A$
Answer: $I_0\approx3.25\,A$