Question

The potential difference V and the current i flowing through an instrument in an ac circuit of frequency f are given by V = 5 cos t volts and I = 2 sin t amperes (where = 2f). The power dissipated in the instrument is -
(1)Zero
(2)10 W
(3)5 W
(4)2.5 W

Answer 1

Answer on Question #40685, Physics, Electric Circuits

The potential difference $V$ and the current $i$ flowing through an instrument in an ac circuit of frequency $f$ are given by $V = 5\cos \omega t$ volts and $I = 2\sin \omega t$ amperes (where $\omega = 2\pi f$). The power dissipated in the instrument is (1) Zero (2) 10 W (3) 5 W (4) 2.5 W

Solution

The potential difference $V$ can be expressed as

$$V = 5 \cos \omega t = 5 \sin \left(\omega t + \frac {\pi}{2}\right),$$

because $\cos a = \sin \left(a + \frac{\pi}{2}\right)$ for any $a$.

The power dissipated in the instrument is

$$P = V _ {r m s} I _ {r m s} \cos \varphi ,$$

where rms means the root mean square, $\cos \varphi$ - the power factor, where $\varphi$ is the phase angle between the voltage and current.

Since

$$\varphi = \frac {\pi}{2}.$$

therefore

$$\cos \varphi = \cos \frac {\pi}{2} = 0.$$

The power dissipated in the instrument is

$$P = V _ {r m s} I _ {r m s} \cdot 0 = 0.$$

Answer: (1) Zero.

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