A series LCR circuit containing a resistance | vedclass

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

Question

(C) At resonance,the impedance of the circuit is purely resistive,so $X_L = X_C$. The current $I$ in the circuit is given by $I = \frac{V_R}{R} = \fra

Vedclass · Accepted Answer

$0.2\,mH$ and $0.03125\,\mu F$

Vedclass · Answer

(C) At resonance,the impedance of the circuit is purely resistive,so $X_L = X_C$. The current $I$ in the circuit is given by $I = \frac{V_R}{R} = \frac{60}{120} = 0.5\,A$.The voltage across the inductor is $V_L = I X_L = I \omega_0 L$. Substituting the known values:$40 = 0.5 	imes (4 	imes 10^5) 	imes L$$L = \frac{40}{0.5 	imes 4 	imes 10^5} = \frac{40}{2 	imes 10^5} = 2 	imes 10^{-4}\,H = 0.2\,mH$.At resonance,$V_C = V_L = 40\,V$. Since $V_C = I X_C = \frac{I}{\omega_0 C}$,we have:$C = \frac{I}{\omega_0 V_C} = \frac{0.5}{(4 	imes 10^5) 	imes 40} = \frac{0.5}{160 	imes 10^5} = \frac{0.5}{1.6 	imes 10^7} = 0.3125 	imes 10^{-7}\,F = 0.03125 	imes 10^{-6}\,F = 0.03125\,\mu F$.

$A$ series $LCR$ circuit containing a resistance of $120\,\Omega$ has a resonance frequency of $4 \times 10^5\, rad\, s^{-1}$. The voltages,at resonance,across the resistance and inductance are $60\,V$ and $40\,V$ respectively. The values of $L$ and $C$ respectively are:

Similar Questions

$A$ resonant frequency of a circuit is $f$. If the capacitance is made four times the initial value,then the resonant frequency will become

At resonance,the value of current in a series $L-C-R$ circuit is: (Symbols have their usual meanings.)

In the circuit shown in the figure,an a.c. source gives voltage $V = 20 \cos (2000 t)$. The impedance and r.m.s. current respectively will be:

The graph shows the variation of current $I$ with frequency $f$ for a series $R-L-C$ network. Keeping $L$ and $C$ constant,if resistance $R$ decreases:

Vedclass Test Series

Exam Paper Generator

Online Exam Module