Question

In the circuit shown in the figure, the AC source gives a voltage \( V = 20\cos(2000t) \). Neglecting source resistance, the voltmeter and ammeter readings will be: 

• A. \( 0 \, {V}, \, 0.47 \, {A} \)
• B. \( 2.82 \, {V}, \, 1.41 \, {A} \)
• C. \( 1.41 \, {V}, \, 0.47 \, {A} \)
• D. \( 1.5 \, {V}, \, 8.37 \, {A} \)

Hint:

In AC circuits, use the formulas for RMS values and impedance to determine voltages and currents. Remember: - RMS Voltage = \( \frac{V_{{max}}}{\sqrt{2}} \), - RMS Current = \( \frac{I_{{max}}}{\sqrt{2}} \), - Impedance \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).

Answer 1

The Correct Option is B

Given: \[ R_1 = 8\Omega, \quad R_2 = 2\Omega, \quad L = 5 { mH} \] \[ C = 50\mu F, \quad V_0 = 20\cos(2000t) \] Impedance Calculation: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Step 1: Calculate Inductive Reactance \( X_L \) \[ X_L = \omega L = 2000 \times 5 \times 10^{-3} = 10\Omega \] Step 2: Calculate Capacitive Reactance \( X_C \) \[ X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 50 \times 10^{-6}} = 10\Omega \] Since \( X_L = X_C \), the impedance reduces to: \[ Z = R = 8 + 2 = 10\Omega \] Step 3: Calculate Maximum Current \( i_{\max} \) \[ i_{\max} = \frac{V_0}{Z} = \frac{20}{10} = 2A \] Step 4: Calculate RMS Current \( i_{{rms}} \) \[ i_{{rms}} = \frac{i_{\max}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = 1.41A \] Step 5: Calculate Voltage Across \( R_1 \) \[ V = R_1 i_{{rms}} = 1.41A \times 2 \] \[ V = 2.82V \]