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

Answer: (A)

$\frac{E_0I_0}{2}cos\phi$

Answer 2

$e = E_0 sin ωt$
$I = I_0 sin (ωt – ɸ)$
Paverage over one cycle of a.c. is = Vrms Irms cos ɸ
Where ɸ is phase difference between V & I.
Here Vrms = $(\frac{E_0}{\sqrt2})$, Irms = $(\frac{I_0}{\sqrt2})$
ɸ = ɸ
Hence Paverage = $(\frac{E_0}{\sqrt2})$ × $(\frac{I_0}{\sqrt2})$ cos ɸ = $\begin {bmatrix} \frac{E_0I_0}{2} \end {bmatrix} cos ɸ$.