For an AC circuit V = 15 sin omega t and I = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: $V = 15 \sin \omega t$ and $I = 20 \cos \omega t$.We can rewrite the current equation as $I = 20 \sin(\omega t + 90^\circ)$.The phase diffe

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given: $V = 15 \sin \omega t$ and $I = 20 \cos \omega t$.We can rewrite the current equation as $I = 20 \sin(\omega t + 90^\circ)$.The phase difference $\phi$ between voltage and current is $90^\circ$.The average power consumed in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.Since $\phi = 90^\circ$,the power factor $\cos \phi = \cos 90^\circ = 0$.Therefore,the average power $P = V_{rms} I_{rms} 	imes 0 = 0 \ W$.

For an $AC$ circuit $V = 15 \sin \omega t$ and $I = 20 \cos \omega t$,the average power consumed in this circuit is ..... $W$.

Similar Questions

On which of the following does the power consumed in an $L-C-R$ series $AC$ circuit depend?

For the series $LCR$ circuit connected with a $220 \ V$, $50 \ Hz$ a.c. source as shown in the figure, the power factor is $\frac{\alpha}{10}$. The value of $\alpha$ is . . . . . .

The power factor of an $ac$ circuit having resistance $(R)$ and inductance $(L)$ connected in series and an angular velocity $\omega$ is:

Two electric circuits $A$ and $B$ are shown in the figure. The ratio of power factor of circuit $B$ to that of circuit $A$ is

In an $ac$ circuit with voltage $V$ and current $I$,the power dissipated is

Vedclass Test Series

Exam Paper Generator

Online Exam Module