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 impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$.Given $X_L = 2R$,we have $Z^2 = R^2 + (2R)^2 = R^2 + 4R^2 = 5R^2$

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$.Given $X_L = 2R$,we have $Z^2 = R^2 + (2R)^2 = R^2 + 4R^2 = 5R^2$.The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi = \frac{R}{Z}$ is the power factor.Since $V_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_0}{\sqrt{2}Z}$,the power is $P = \frac{E_0}{\sqrt{2}} 	imes \frac{E_0}{\sqrt{2}Z} 	imes \frac{R}{Z} = \frac{E_0^2 R}{2 Z^2}$.Substituting $Z^2 = 5R^2$,we get $P = \frac{E_0^2 R}{2(5R^2)} = \frac{E_0^2 R}{10 R^2} = \frac{E_0^2}{10 R}$.

An e.m.f. $E=E_0 \cos \omega t$ is applied to a circuit containing $L$ and $R$ in series. If $X_L=2 R$,then the power dissipated in the circuit is

Similar Questions

The current $I$,potential difference $V_L$ across the inductor,and potential difference $V_C$ across the capacitor in the circuit as shown in the figure are best represented vectorially as:

$A$ $100 \;\mu\,F$ capacitor in series with a $40\; \Omega$ resistance is connected to a $110\; V, 60\; Hz$ supply.
$(a)$ What is the maximum current in the circuit?
$(b)$ What is the time lag between the current maximum and the voltage maximum?

If $E = 200\,V$,$R = 25\,\Omega$,$L = 2\,H$,and $C = 2\,\mu F$ and the frequency is variable,then the current at $f = 0$ and $f = \infty$ will be respectively:

In an $RL$ circuit,the current is $i = 100 \sin(314t) \ A$ and the voltage is $e = 200 \sin(314t + \pi/3) \ V$. If the resistance is $1 \ \Omega$,what is the reactance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module