Question

In the given circuit, the AC source has $\omega =100$ rad/s. Considering the inductor and capacitor to be ideal, the approximate r.m.s value (s) can be

• A. The current through the circuit, I is $0.3A$.
• B. The current through the circuit, I is $0.6A$.
• C. The voltage across $100\Omega$ resistor =$10\sqrt{2}$.
• D. The voltage across $50\Omega$ resistor =$10$.

Answer 1

Answer: (A), (C)

The voltage across $100\Omega$ resistor =$10\sqrt{2}$.

Considering r.m.s values


$x_L=\omega L=50$
$x_C=\frac{1}{\omega c}=100$

Impedance $Z_1=\sqrt{{X_L}^2+R^2}=100\sqrt{2}$
$Z_2=\sqrt{{X_C}^2+R^2}=50\sqrt{2}$

$I_1=\frac{V}{R}=\frac{20}{100\sqrt{2}}$ leads voltage by ${45}^{\circ }$
$I_2=\frac{20}{50\sqrt{2}}=\frac{2}{5\sqrt{2}}$ lags voltage by ${45}^{\circ }$

Net current
$I=\sqrt{I_1^2+I_2^2}=0.3A$ approx.

Voltage across $100\Omega$ resistor is $V=I_1\times 100\Omega =\frac{20}{100\sqrt{2}}\times 100=10\sqrt{2}$
Voltage across $50\Omega$ resistor is
$V=I_2\times 50\Omega =\frac{20}{50\sqrt{2}}\times 50=10\sqrt{2}$