The potential differences across the resistan | vedclass

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

Question

(A) Given: $V_{R} = 80 \, V$,$V_{C} = 40 \, V$,$V_{L} = 100 \, V$.The power factor of an $L-C-R$ circuit is given by $\cos \phi = \frac{R}{Z}$.Since $

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(A) Given: $V_{R} = 80 \, V$,$V_{C} = 40 \, V$,$V_{L} = 100 \, V$.The power factor of an $L-C-R$ circuit is given by $\cos \phi = \frac{R}{Z}$.Since $V = IZ$ and $V_{R} = IR$,we can write $\cos \phi = \frac{V_{R}}{V}$.The total voltage $V$ in an $L-C-R$ circuit is $V = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$.Substituting the given values:$V = \sqrt{80^{2} + (100 - 40)^{2}}$$V = \sqrt{80^{2} + 60^{2}}$$V = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, V$.Now,calculate the power factor:$\cos \phi = \frac{V_{R}}{V} = \frac{80}{100} = 0.8$.

The potential differences across the resistance,capacitance,and inductance are $80 \, V$,$40 \, V$,and $100 \, V$ respectively in an $L-C-R$ circuit. The power factor of this circuit is:

Similar Questions

An alternating voltage of $V = V_{0} \sin \omega t$ is applied across a circuit. As a result, a current $I = I_{0} \sin (\omega t - \frac{\pi}{2})$ flows in it. The power consumed per cycle is . . . . . . .

In an $AC$ circuit,the instantaneous values of emf and current are $e = 200 \sin(314t) \text{ V}$ and $I = \sin(314t + \frac{\pi}{3}) \text{ A}$. The average power consumed in watt is

Define average power (true power) in an $AC$ circuit.

In an $ac$ circuit,the current is given by $i = 5 \sin \left( 100 t - \frac{\pi}{2} \right)$ and the $ac$ potential is $V = 200 \sin (100 t) \text{ volts}$. Then the power consumption is $....... \text{ watts}$.

If the power factor changes from $0.5$ to $0.25$ because impedance changes from $Z_1$ to $Z_2$, then $Z_1 = x Z_2$. The value of $x$ is (Resistance remains constant).

Vedclass Test Series

Exam Paper Generator

Online Exam Module