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: Use the instantaneous-power average. Average power is \(P = \dfrac{1}{T}\int_0^T V\,i\,dt\) over one full cycle of period \(T\).
Step 2: Multiply the given expressions. \(V\,i = (5\cos\omega t)(2\sin\omega t) = 10\sin\omega t\cos\omega t = 5\sin 2\omega t\), using \(2\sin\theta\cos\theta = \sin 2\theta\).
Step 3: Average over a cycle. The mean value of \(\sin 2\omega t\) over a complete cycle is zero, so \(P = 5\times \langle \sin 2\omega t\rangle = 0\).
Step 4: Conclusion. Because the current and voltage are \(90^\circ\) out of phase, the circuit is wattless and the net power loss is zero, confirming option (i).
\[\boxed{P = \langle 5\sin 2\omega t\rangle = 0}\]