In an A.C. circuit,V and I are given by:V = 1 | vedclass

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

Question

(C) The instantaneous voltage is $V = V_0 \sin(\omega t)$,where $V_0 = 100 	ext{ V}$.The instantaneous current is $I = I_0 \sin(\omega t + \phi)$,whe

Vedclass · Accepted Answer

$2.5 	ext{ W}$

Vedclass · Answer

(C) The instantaneous voltage is $V = V_0 \sin(\omega t)$,where $V_0 = 100 	ext{ V}$.The instantaneous current is $I = I_0 \sin(\omega t + \phi)$,where $I_0 = 100 	ext{ mA} = 0.1 	ext{ A}$ and $\phi = \pi/3$.The average power dissipated in an $A.C.$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} 	ext{ V}$.$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} 	ext{ A}$.Substituting the values:$P = \left(\frac{100}{\sqrt{2}}ight) 	imes \left(\frac{0.1}{\sqrt{2}}ight) 	imes \cos(\pi/3)$.$P = \frac{100 	imes 0.1}{2} 	imes \frac{1}{2} = \frac{10}{2} 	imes 0.5 = 5 	imes 0.5 = 2.5 	ext{ W}$.

In an $A.C.$ circuit,$V$ and $I$ are given by:
$V = 100 \sin(100 t) \text{ V}$
$I = 100 \sin(100 t + \pi/3) \text{ mA}$
The power dissipated in the circuit is:

Similar Questions

In a series $LCR$ circuit,$R = 18 \ \Omega$ and the impedance $Z = 33 \ \Omega$. An $RMS$ voltage of $220 \ V$ is applied across the circuit. The true power consumed in the $AC$ circuit is: (in $W$)

In an $ac$ circuit,the power factor

Can the instantaneous power output of an $AC$ source ever be negative? Can the average power output be negative?

The power dissipated in a resistor of resistance $R$ when an alternating current with a peak value of $I_p$ flows through it is:

In an $A.C.$ circuit,the instantaneous values of $\text{e.m.f}.$ and current are $e = 10 \sin(50 \pi t) \text{ V}$ and $I = 10 \sin(50 \pi t + \frac{\pi}{3}) \text{ A}$. The average power consumed in the circuit is $... \text{ W}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In an $A.C.$ circuit,$V$ and $I$ are given by:$V = 100 \sin(100 t) \text{ V}$$I = 100 \sin(100 t + \pi/3) \text{ mA}$The power dissipated in the circuit is: