The instantaneous current and voltage of an a.c. circuit are given by
i = 10 sin 314 t A and
v = 50 sin (314 t + $π$ /2)V.
What is the power dissipation in the circuit?

Solution

Given,
Instantaneous current, i = 10 sin 314 t A

Instantaneous voltage, V = 50 sin (314 t + $π$ /2) V

Since,
$I = I_{O} \sin ωt and, V = V_{O} \sin (ωt + ϕ)$
Therefore, from the above equations we get

Peak current, peak voltage and $ϕ$ as,
$i_{0} = 10, V_{0} = 50 and ϕ = \frac{π}{2}$

Power dissipation in the circuit is given by
$E_{v} = \frac{E_{0}}{\sqrt{2}} and I_{0} = \frac{I_{o}}{\sqrt{2}}$

Therefore,

$P = (\frac{50}{\sqrt{2}}) (\frac{10}{\sqrt{2}}) \cos \frac{π}{2} = 0$
$[∵ \cos \frac{π}{2} = 0]$

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The instantaneous current and voltage of an a.c. circuit are given by
i = 10 sin 314 t A and
v = 50 sin (314 t + $π$ /2)V.
What is the power dissipation in the circuit?

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The instantaneous current and voltage of an a.c. circuit are given byi = 10 sin 314 t A andv = 50 sin (314 t + π/2)V.What is the power dissipation in the circuit?

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The instantaneous current and voltage of an a.c. circuit are given by
i = 10 sin 314 t A and
v = 50 sin (314 t + $π$ /2)V.
What is the power dissipation in the circuit?