In LCR series circuit, an alternating e.m.f. 'e' and current 'i' are given by the equations $e = 100 sin (100 t)$ volt, $i = 100 sin (100 t + \frac{π}{3}) m A$ . The average power dissipated in the circuit will be.

100 W

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10 W

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5 W

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2.5 W

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Solution

The correct option is D $2.5 W$
Comparing the given e.m.f equation with $e = e_{o} sin w t$
We get peak value of e.m.f $e_{o} = 100$ volt
Thus r.m.s value of e.m.f $e_{r m s} = \frac{e_{o}}{\sqrt{2}} = \frac{100}{\sqrt{2}}$
Similarly, on comparing the given current equation with $i = i_{o} sin (w t + ϕ)$
We get peak value of current $i_{o} = 100$ $m A$
Thus r.m.s value of current $i_{r m s} = \frac{i_{o}}{\sqrt{2}} = \frac{100}{\sqrt{2}}$ $m A$
Also we get $ϕ = \frac{π}{3}$
Thus average power dissipated $< P >= e_{r m s} i_{r m s} cos ϕ$
$∴$ $< P >= \frac{100}{\sqrt{2}} . \frac{100}{\sqrt{2}} \times 10^{- 3} . \frac{1}{2}$
$⟹$ $< P >= 2.5 W$

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In LCR series circuit, an alternating e.m.f. 'e' and current 'i' are given by the equations e=100sin(100t) volt, i=100sin(100t+π3)mA. The average power dissipated in the circuit will be.

In LCR series circuit, an alternating e.m.f. 'e' and current 'i' are given by the equations $e = 100 sin (100 t)$ volt, $i = 100 sin (100 t + \frac{π}{3}) m A$ . The average power dissipated in the circuit will be.