An LCR series circuit with 100 ,Omega resista | vedclass

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

Question

(D) Given: $R = 100 \,\Omega$,$V = 200 \,V$,$\omega = 300 \,rad/s$.When capacitance is removed,the circuit becomes an $LR$ circuit. The phase angle $\

Vedclass · Accepted Answer

$2 \,A, 400 \,W$

Vedclass · Answer

(D) Given: $R = 100 \,\Omega$,$V = 200 \,V$,$\omega = 300 \,rad/s$.When capacitance is removed,the circuit becomes an $LR$ circuit. The phase angle $\phi_1 = 60^o$ (current lags voltage). Wait,the problem states current leads by $60^o$ when capacitance is removed,which implies an $RC$ circuit. Let's re-evaluate: If $C$ is removed,we have $LR$ circuit,$	an(60^o) = \frac{X_L}{R} \implies X_L = R 	an(60^o) = 100\sqrt{3} \,\Omega$.If $L$ is removed,we have $RC$ circuit,$	an(60^o) = \frac{X_C}{R} \implies X_C = R 	an(60^o) = 100\sqrt{3} \,\Omega$.Since $X_L = X_C$,the circuit is at resonance when all components are connected.At resonance,$Z = R = 100 \,\Omega$.Current $I = \frac{V}{Z} = \frac{200}{100} = 2 \,A$.Power dissipated $P = I^2 R = (2)^2 	imes 100 = 400 \,W$.

Similar Questions

What is the reason we prefer $A.C.$ voltage over $D.C.$ voltage for power transmission?

Assertion: Ohm's law cannot be applied to $a.c.$ circuits.
Reason: Resistance offered by a capacitor for an $a.c.$ source depends upon the frequency of the source.

At a particular angular frequency,the reactance of a capacitor and that of an inductor are the same. If the angular frequency is doubled,the ratio of the reactance of the capacitor to that of the inductor will be:

$A$ resistor of resistance $R$ and an inductor of inductive reactance $R$ are connected in series to an $AC$ source. $A$ capacitor of capacitive reactance $2R$ is then connected in series with $L$ and $R$. The ratio of the power factors of the $LR$ and $LCR$ circuits is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module