Question

In an a.c. circuit the emf (e) and the current (i) at any instant core given respectively by \(e = E_0 sin ωt\), \(I = I_0 sin (ωt – ɸ)\). The average power in the circuit over one cycle of a.c. is.

• A. $\frac{E_0I_0}{2}cos\phi$
• B. $E_0I_0$
• C. $\frac{E_0I_0}{2}$
• D. $\frac{E_0I_0}{2}sin\phi$

Answer 1

The Correct Option is A

The problem requires us to determine the average power in an AC circuit given the expressions for emf and current at any instant. Let's solve it step-by-step:

1. Given expressions:
   • Emf at any instant: e = E_0 \sin \omega t
   • Current at any instant: I = I_0 \sin (\omega t - \phi)
2. The instantaneous power, which is the product of instantaneous emf and current, is given by: P(t) = e \cdot I = E_0 \sin \omega t \times I_0 \sin (\omega t - \phi)
3. Using the trigonometric identity: \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]
   • By applying this identity, we have: E_0 I_0 \sin \omega t \sin (\omega t - \phi) = \frac{E_0 I_0}{2} [\cos \phi - \cos(2\omega t - \phi)]
4. The average power over one complete cycle is the average value of P(t):
   • Due to the periodic nature of the cosine function, the average value of \cos(2\omega t - \phi) over one complete cycle is zero.
   • Therefore, the average power over one cycle is: \text{Average Power} = \frac{E_0 I_0}{2} \cdot \cos \phi
5. Therefore, the correct answer to the problem is: \frac{E_0 I_0}{2} \cos \phi

This explains the choice of the given correct answer: \frac{E_0 I_0}{2} \cos \phi. This expression represents the average power consumed in the AC circuit, accounting for the phase difference \phi between the voltage and current. Only real power is included, which is why \cos \phi (the power factor) appears in the formula.