An L-C-R series circuit with 100 , Omega resi | vedclass

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

Question

(A) Given: Resistance $R = 100 \, \Omega$, Voltage $V = 100 \, V$.Case $1$: When the capacitor is removed, the circuit becomes an $L-R$ series circuit

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given: Resistance $R = 100 \, \Omega$, Voltage $V = 100 \, V$.Case $1$: When the capacitor is removed, the circuit becomes an $L-R$ series circuit. The current lags behind the voltage by $\phi = 45^{\circ}$.$	an \phi = \frac{X_L}{R} \implies 	an 45^{\circ} = \frac{X_L}{100} \implies 1 = \frac{X_L}{100} \implies X_L = 100 \, \Omega$.Case $2$: When the inductor is removed, the circuit becomes a $C-R$ series circuit. The current leads the voltage by $\phi = 45^{\circ}$.$	an \phi = \frac{X_C}{R} \implies 	an 45^{\circ} = \frac{X_C}{100} \implies 1 = \frac{X_C}{100} \implies X_C = 100 \, \Omega$.Since $X_L = X_C$, the circuit is in resonance. In resonance, the impedance $Z = R$.$Z = 100 \, \Omega$.The current flowing in the circuit is $I = \frac{V}{Z} = \frac{100}{100} = 1 \, A$.

Similar Questions

In a series $RLC$ circuit $(AC)$:
$L = 10 \, mH$
$C = 0.01 \, \mu F$
$R = 50 \, \Omega$
If the supply voltage is $V = 10 \sin \omega t$,find the power dissipated at resonance in $W$.

In an $L-C-R$ circuit,if the resistance increases,the quality factor:

The $Q$ factor of a series $LCR$ circuit with $L = 2 \ H$,$C = 32 \ \mu F$,and $R = 10 \ \Omega$ is:

$A$ series $L-C-R$ circuit containing a resistance $R$ has angular frequency $\omega$. At resonance,the voltages across the resistance and the inductor are $V_R$ and $V_L$ respectively. The value of the capacitance is:

$A$ series $LCR$ circuit is resonating with a source whose emf varies with time as described in diagram $-1$. If we replace the source by another source whose emf varies with time according to diagram $-2$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module