For the given RL circuit,the maximum power lo | vedclass

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

Question

(D) From the given voltage equation $V = 20 \sin(10t + 30^\circ)$,the peak voltage is $V_0 = 20 	ext{ V}$ and the angular frequency is $\omega = 10 \

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) From the given voltage equation $V = 20 \sin(10t + 30^\circ)$,the peak voltage is $V_0 = 20 	ext{ V}$ and the angular frequency is $\omega = 10 	ext{ rad/s}$.The inductive reactance is $X_L = \omega L = 10 	imes 1 = 10 \, \Omega$.The power dissipated in an $RL$ circuit is given by $P = I_{rms}^2 R = \frac{V_{rms}^2 R}{R^2 + X_L^2} = \frac{V_0^2 R}{2(R^2 + X_L^2)}$.To find the maximum power,we differentiate $P$ with respect to $R$ and set it to zero,which gives the condition $R = X_L = 10 \, \Omega$.Substituting $R = 10 \, \Omega$ and $X_L = 10 \, \Omega$ into the power formula:$P_{\max} = \frac{V_0^2 R}{2(R^2 + X_L^2)} = \frac{20^2 	imes 10}{2(10^2 + 10^2)} = \frac{400 	imes 10}{2(100 + 100)} = \frac{4000}{400} = 10 	ext{ W}$.

For the given $RL$ circuit,the maximum power loss across the circuit can be $W$.

Similar Questions

$A$ resistor and a pure inductor are connected to an $AC$ supply of $120\,V$ and $50\,Hz$. The current in the circuit is $3\,A$. If the power consumed in the circuit is $108\,W$,then the resistance in the circuit is.....$\Omega $

In the figure,if $A$ and $B$ are identical bulbs,which bulb glows brighter?

With an alternating voltage source of frequency $f$,an inductor $L$,a capacitor $C$,and a resistor $R$ are connected in series. The voltage leads the current by $45^{\circ}$. The value of $L$ is $(\tan 45^{\circ} = 1)$.

$A$ capacitor of $\frac{2.5}{\pi} \mu F$ and a resistor of $3000 \, \Omega$ are connected in series with a $200 \, V, 50 \, Hz$ $AC$ source. What are the power factor and the power dissipated,respectively?

Vedclass Test Series

Exam Paper Generator

Online Exam Module