Q135 / 200 · NEET 2008 (Screening) · Physics

In an $A C$ circuit the emf (e) and the current (i) at any instant are given respectively by $e=E_0 \sin \omega t$ $i=I_0 \sin (\omega t-\phi)$ The average power in the circuit over one cycle of AC is

  • A
    $\frac{E_0 I_0}{2}$
  • B
    $\frac{E_0 I_0}{2} \sin \phi$
  • Correct answer
    C
    $\frac{E_0 I_0}{2} \cos \phi$
  • D
    $E_0 I_0$

Solution

Key Idea: The power is defined as the rate at which work is being done in the circuit. Power $=$ rate of work done in one complete cycle. $\begin{aligned} & P_{a \nu}=\frac{W}{T} \\ & P_{a \nu}=\frac{\left(E_0 I_0 \cos \phi\right) T / 2}{T} \\ & P_{a \nu}=\frac{E_0 I_0 \cos \phi}{2} \end{aligned}$ where $\cos \phi$ is called the power factor of an AC circuit.

Hence, the correct answer is option: 'C'.

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