An a.c. e.m.f. of peak value 230 V and frequ | vedclass

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

Question

(A) For the current in an $LCR$ series circuit to be maximum,the circuit must be in resonance.At resonance,the inductive reactance equals the capaciti

Vedclass · Accepted Answer

$4 \ \mu F, \quad 20 \ A$

Vedclass · Answer

(A) For the current in an $LCR$ series circuit to be maximum,the circuit must be in resonance.At resonance,the inductive reactance equals the capacitive reactance: $X_L = X_C$.Given $L = 2.5 \ H$ and $f = 50 \ Hz$,the inductive reactance is $X_L = 2 \pi f L = 2 \pi (50) (2.5) = 250 \pi \ \Omega$.Using $\pi^2 = 10$,we have $\pi = \sqrt{10} \approx 3.162$,so $X_L = 250 	imes 3.162 = 790.5 \ \Omega$.However,the resonance condition is $X_C = X_L$,so $\frac{1}{2 \pi f C} = 2 \pi f L$.$C = \frac{1}{4 \pi^2 f^2 L} = \frac{1}{4 	imes 10 	imes (50)^2 	imes 2.5} = \frac{1}{40 	imes 2500 	imes 2.5} = \frac{1}{250000} = 4 	imes 10^{-6} \ F = 4 \ \mu F$.At resonance,the impedance $Z = R = 11.5 \ \Omega$.The maximum current $I_{max} = \frac{V_{peak}}{Z} = \frac{230}{11.5} = 20 \ A$.Thus,the values are $4 \ \mu F$ and $20 \ A$.

Similar Questions

In a series $LCR$ resonant circuit,the capacitance is changed from $C$ to $3C$. To obtain the same resonant frequency,the inductance should be changed from $L$ to

An inductor of inductance $L$,a capacitor of capacitance $C$,and a resistor of resistance $R$ are connected in series to an $ac$ source. The quality factor of the circuit is

$A$ resonance circuit having inductance and resistance $2 \times 10^{-4} \ H$ and $6.28 \ \Omega$ respectively oscillates at $10 \ MHz$ frequency. The value of the quality factor of this resonator is .........
$[\pi = 3.14]$

In an $LCR$ circuit, the resonant frequency is $600 \ Hz$ and the half-power points are at $650 \ Hz$ and $550 \ Hz$. The quality factor is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module