Question

In an AC circuit, if the current \(i = 2\sin\omega t\) ampere and the voltage \(V = 5\cos\omega t\) volt, then the power loss will be:

• A. zero
• B. 5 W
• C. 10 W
• D. 2\(\cdot\)5 W

Hint:

A sine current with a cosine voltage means a \(90^\circ\) phase difference; the power factor \(\cos 90^\circ = 0\).

Answer 1

The Correct Option is A

Step 1: Concept. The average power in an AC circuit is \(P = V_{rms}\,I_{rms}\cos\phi\), where \(\phi\) is the phase difference between the voltage and the current.
Step 2: Find the phase difference. Given \(i = 2\sin\omega t\) and \(V = 5\cos\omega t\). Rewrite the voltage as \(V = 5\sin(\omega t + 90^\circ)\), since \(\cos\omega t = \sin(\omega t + 90^\circ)\). So the voltage leads the current by \(\phi = 90^\circ\).
Step 3: Substitute the power factor. \(\cos\phi = \cos 90^\circ = 0\).
Step 4: Therefore \(P = V_{rms}\,I_{rms}\times 0 = 0\). The average power loss is zero, option (i). This is a purely reactive (wattless) circuit.
Why other options are wrong: The non-zero values (5 W, 10 W, 2.5 W) ignore the \(90^\circ\) phase difference; energy only oscillates back and forth and no net power is dissipated.
\[\boxed{P = 0}\]