In an LCR circuit R = 100 Omega. When capaci | vedclass

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

Question

(B) When $C$ is removed,the circuit becomes an $RL$ circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = \pi /3$,so $

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) When $C$ is removed,the circuit becomes an $RL$ circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = \pi /3$,so $	an(\pi /3) = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{R} \implies X_L = R\sqrt{3}$.When $L$ is removed,the circuit becomes an $RC$ circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_C}{R}$.Given $\phi = \pi /3$,so $	an(\pi /3) = \frac{X_C}{R} \implies \sqrt{3} = \frac{X_C}{R} \implies X_C = R\sqrt{3}$.Since $X_L = X_C$,the circuit is in a state of resonance.In an $LCR$ circuit at resonance,the impedance $Z = R$.Given $R = 100 \ \Omega$,therefore $Z = 100 \ \Omega$.

In an $LCR$ circuit $R = 100 \ \Omega$. When capacitance $C$ is removed,the current lags behind the voltage by $\pi /3$. When inductance $L$ is removed,the current leads the voltage by $\pi /3$. The impedance of the circuit is......$\Omega$.

Similar Questions

Statement-$I$: $A$ capacitor can be used in an $a.c.$ circuit in place of a choke coil.
Statement-$II$: $A$ capacitor blocks $d.c.$ and allows $a.c.$ to pass.

$A$ choke coil is preferred to a $Rheostat$ in $AC$ circuits because:

In the given $AC$ circuit,when switch $S$ is at position $1$,the source $emf$ leads the current by $\pi / 6$. Now,if the switch is at position $2$,then

If $C, R, L$ and $I$ denote capacity, resistance, inductance and electric current respectively, the quantities having the same dimensions of time are :
$(1)$ $C R$
$(2)$ $\frac{L}{R}$
$(3)$ $\sqrt{L C}$
$(4)$ $L I^2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module