An e.m.f. e=E_0 cos omega t is applied to a c | vedclass

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

Question

(B) The average power dissipated in an $LCR$ series circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.Given $X

Vedclass · Accepted Answer

$\frac{E_0^2}{10 R}$

Vedclass · Answer

(B) The average power dissipated in an $LCR$ series circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.Given $X_L = 3R$ and $X_C = R$,the net reactance is $X = X_L - X_C = 3R - R = 2R$.The impedance of the circuit is $Z = \sqrt{R^2 + X^2} = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5}$.The root mean square voltage is $V_{rms} = \frac{E_0}{\sqrt{2}}$.The root mean square current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_0}{\sqrt{2} \cdot R\sqrt{5}} = \frac{E_0}{R\sqrt{10}}$.The power factor is $\cos \phi = \frac{R}{Z} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}}$.Substituting these into the power formula: $P = V_{rms} I_{rms} \cos \phi = \left( \frac{E_0}{\sqrt{2}} \right) \left( \frac{E_0}{R\sqrt{10}} \right) \left( \frac{1}{\sqrt{5}} \right) = \frac{E_0^2}{R \sqrt{2} \cdot \sqrt{10} \cdot \sqrt{5}} = \frac{E_0^2}{R \sqrt{100}} = \frac{E_0^2}{10R}$.

An e.m.f. $e=E_0 \cos \omega t$ is applied to a circuit containing $L, C$ and $R$ in series where $X_L=3 R$ and $X_C=R$. The average power dissipated in the circuit is

Similar Questions

In an $ac$ circuit,the power factor

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

In an $AC$ circuit,$I = 5 \sin \left( 100t - \frac{\pi}{2} \right)$ and $V = 200 \sin 100t$. The power consumed is

$A$ circuit has a resistance of $12 \; \Omega$ and an impedance of $15 \; \Omega$. The power factor of the circuit will be

The power factor of an $AC$ circuit containing peak current $2 \ A$ and peak voltage $1 \ V$ is $1/2$. The angle between voltage and current is: (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module