Question

An alternating voltage $V=V_0\sin\omega t$ is applied across a circuit and as a result, a current $I=I_0\sin\left(\omega t+\frac{\pi}{2}\right)$ flows in it. The power consumed per cycle is

• A. \( I_0 V_0 \)
• B. \( 0.5 I_0 V_0 \)
• C. \( 0.7 I_0 V_0 \)
• D. \( 1.414 I_0 V_0 \)
• E. Zero

Hint:

When the phase difference is \(90^\circ\), the circuit is purely inductive or purely capacitive. In such cases, the current is called "wattless current" because no power is dissipated.

Answer 1

Answer: (E)

Concept: The average power \(P_{avg}\) consumed in an AC circuit over a complete cycle is given by: \[ P_{avg} = V_{rms} I_{rms} \cos \phi \] where \(\phi\) is the phase difference between the voltage and the current, and \(\cos \phi\) is known as the power factor.

Step 1: Identify the phase difference \(\phi\).
The given equations are: \[ V = V_0 \sin(\omega t) \] \[ I = I_0 \sin\left(\omega t + \frac{\pi}{2}\right) \] Comparing the arguments of the sine functions, the phase difference is: \[ \phi = \frac{\pi}{2} \text{ (or } 90^\circ\text{)} \]

Step 2: Calculate the power factor.
The power factor is: \[ \cos \phi = \cos\left(\frac{\pi}{2}\right) = 0 \]

Step 3: Calculate average power.
Substituting the power factor into the power equation: \[ P_{avg} = V_{rms} I_{rms} \times 0 = 0 \]