You visited us 1 times! Enjoying our articles? Unlock Full Access!

Electric Power

The potential...

Question

The potential difference $V$ and the current $I$ flowing through an instrument in an AC circuit are given by $V = 5 cos ω t v o l t, I = 2 sin ω t A m p$ . The power dissipated in the circuit is

5

watt

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10

watt

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.5

watt

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $0$
The voltage can be written as,

$V = 5 (sin ω t + \frac{π}{2})$

The angle between the voltage and the current is $90^{\circ}$ .

The power dissipated in the circuit is given as,

$P = V_{r m s} I_{r m s} cos ϕ$

$= V_{r m s} I_{r m s} cos 90^{\circ}$

= $0$

Thus, the power dissipated in the circuit is $0$ .

Suggest Corrections

Similar questions

V

I

5

ω t

I =

s i n ω t

ω = 2 π f

V = 5 c o s ω t

i = 2 s i n ω t

V = 10 \sqrt{2} sin ω t

I = 2 \sqrt{2} cos ω t

p . d . V

I

a . c .

V = 2 cos ω t V

I = 2 sin ω t A

V

I

V = 100 sin (100 t) V

I = 100 sin (100 t + \frac{π}{3}) m A

Join BYJU'S Learning Program