Question

A voltage source \( V_s = 5 \sin \left( 100t + \frac{\pi}{2} \right) \) volts is driving a current \[ I = 10 \sin \left( 100t + \frac{\pi}{6} \right) \, {mA} \] through a circuit. The average power dissipated in the circuit is ________mW.

• A. 0.0
• B. 12.5
• C. 25.0
• D. 50.0

Hint:

When calculating the average power in an AC circuit, ensure to use the RMS values for voltage and current and the phase difference between them.

Answer 1

The Correct Option is B

Step 1: Calculating RMS values
The RMS value of a sinusoidal waveform is given by: \[ V_{{rms}} = \frac{V_{{max}}}{\sqrt{2}}, \quad I_{{rms}} = \frac{I_{{max}}}{\sqrt{2}}. \] For the voltage source \( V_s = 5 \sin \left( 100t + \frac{\pi}{2} \right) \), the peak voltage \( V_{{max}} = 5 \) V, so: \[ V_{{rms}} = \frac{5}{\sqrt{2}} \approx 3.535 \, {V}. \] For the current \( I = 10 \sin \left( 100t + \frac{\pi}{6} \right) \), the peak current \( I_{{max}} = 10 \) mA, so: \[ I_{{rms}} = \frac{10}{\sqrt{2}} \approx 7.071 \, {mA} = 0.007071 \, {A}. \] Step 2: Calculating the phase difference
The phase difference \( \theta \) between the voltage and current is the difference in their phase angles: \[ \theta = \left( \frac{\pi}{2} - \frac{\pi}{6} \right) = \frac{\pi}{3}. \] Step 3: Calculating average power
Now we can substitute into the formula for average power: \[ P_{{avg}} = \frac{1}{2} \times 3.535 \times 0.007071 \times \cos \left( \frac{\pi}{3} \right). \] Since \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \), we get: \[ P_{{avg}} = \frac{1}{2} \times 3.535 \times 0.007071 \times \frac{1}{2} = 12.5 \, {mW}. \] Thus, the average power dissipated in the circuit is 12.5 mW.