An LCR series circuit with a resistance of 10 | vedclass

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

Question

(D) In an $LCR$ series circuit, when the capacitor is removed, the circuit becomes an $LR$ circuit. The phase angle is given by $	an \phi = \frac{X_L

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) In an $LCR$ series circuit, when the capacitor is removed, the circuit becomes an $LR$ circuit. The phase angle is given by $	an \phi = \frac{X_L}{R}$. Given $\phi = 60^o$, we have $	an 60^o = \frac{X_L}{R} \Rightarrow X_L = R \sqrt{3}$.When the inductor is removed, the circuit becomes an $RC$ circuit. The phase angle is given by $	an \phi = \frac{X_C}{R}$. Given $\phi = 60^o$, we have $	an 60^o = \frac{X_C}{R} \Rightarrow X_C = R \sqrt{3}$.Since $X_L = X_C$, the circuit is in resonance when both components are present.At resonance, the impedance $Z = R = 100 \, \Omega$.The average power dissipated is $P = \frac{V^2}{R} = \frac{200^2}{100} = \frac{40000}{100} = 400 \, W$.

Similar Questions

$A$ series $LR$ circuit is connected to an ac source of frequency $\omega$ and the inductive reactance is equal to $2R$. $A$ capacitance of capacitive reactance equal to $R$ is added in series with $L$ and $R$. The ratio of the new power factor to the old one is

In a series $LCR$ circuit,the voltages across $R, L$ and $C$ are shown in the figure. The voltage of the applied source is ........ $V$. (in $, V$)

In a parallel $AC$ circuit with $R_1 = R_2 = R$ and $X_L = X_C = X$. An $AC$ source $V = V_0 \sin(\omega t)$ is connected across the circuit as shown in the figure. The $RMS$ value of the potential difference $V_A - V_B$ will be:

In an $L-C-R$ circuit,the capacitance is changed from $C$ to $4 C$. For the same resonant frequency,the inductance should be changed from $L$ to

In a series $LCR$ circuit, $R = 10 \ \Omega$ and the impedance $Z = 30 \ \Omega$. An $r.m.s.$ voltage of $210 \text{ V}$ is applied across the circuit. The true power consumed in the $AC$ circuit is: (in $\text{ W}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module