Question

In the given circuit, the AC source has ? = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

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

Answer 1

Answer: (C)

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

Answer 2

Circuit 1
$\, \, \, \, \, \, \, \, \, X_c = \frac{1}{\omega C} = 100 \Omega$
$\therefore \, \, \, \, \, \, \, \, Z_1 = \sqrt{(100)^2 + (100)^2}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 100 \sqrt{2} \, \Omega$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \phi_1 = cos^{-1} \bigg( \frac{R_1}{Z_1}\bigg) = 45^{\circ}$
In this circuit current leads the voltage.
$\, \, \, \, \, \, \, \, \, \, I_1 = \frac{V}{Z_1} = \frac{20}{100 \sqrt{2}} = \frac{1}{5 \sqrt{2}} A$
$\, \, \, \, \, \, \, \, \, V_{100 \Omega} = (100) I_1 = (100) \frac{1}{5 \sqrt{2}} V$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 10 \sqrt{2} V$
Circuit 2
$\, \, \, \, \, \, \, \, \, \, \, \, X_L = \omega L = (100)(0.5) = 50 \Omega$
$\, \, \, \, \, \, \, \, \, \, \, \, \, Z_2 = \sqrt{(50)^2 + (50)^2} = 50\sqrt{2} \Omega$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \phi_1 = cos^{-1} \bigg( \frac{R_1}{Z_1}\bigg) = 45^{\circ}$
In this circuit voltage leads the current.
$\, \, \, \, \, \, \, \, \, \, I_2 = \frac{V}{Z_2} = \frac{20}{50 \sqrt{2}} = \frac{\sqrt{2}}{5} A$
$\, \, \, \, \, \, \, \, \, V_{50 \Omega} = (50) I_2 = 50 \bigg( \frac{ \sqrt{2}}{5}\bigg) = 10\sqrt{2} V$
Further, $I_1$ and $I_2$ have a mutual phase difference of $90^{\circ}$
$I = \sqrt{I^2 , + I^2_2} = 0.34$