An ideal choke draws a current of 8 , A when | vedclass

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

Question

(A) Resistance of the resistor: $R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.Inductive reactance of the choke at $50 \, Hz$: $X_L = \frac{V}{I} =

Vedclass · Accepted Answer

$\frac{15}{\sqrt{2}} \, A$

Vedclass · Answer

(A) Resistance of the resistor: $R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.Inductive reactance of the choke at $50 \, Hz$: $X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.Since $X_L = 2 \pi f L$, we have $12.5 = 2 \pi 	imes 50 	imes L$, which gives $L = \frac{12.5}{100 \pi} = \frac{1}{8 \pi} \, H$.Now, for the new source of $40 \, Hz$, the new inductive reactance is $X_L' = 2 \pi f' L = 2 \pi 	imes 40 	imes \frac{1}{8 \pi} = 10 \, \Omega$.The impedance of the series $RL$ circuit is $Z = \sqrt{R^2 + X_L'^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \Omega$.The current in the circuit is $I = \frac{V'}{Z} = \frac{150}{10\sqrt{2}} = \frac{15}{\sqrt{2}} \, A$.

Similar Questions

An inductor of inductance $L$ and resistor $R$ are joined together in series and connected to a source of frequency $\omega$. The power dissipated in the circuit is

When $100 \ V$ $d.c.$ is applied across a solenoid,a current of $1 \ A$ flows in it. When $100 \ V$ $a.c.$ is applied across it,the current drops to $0.5 \ A$. If the frequency is $50 \ Hz$,the impedance and inductance are:

For a series $LCR$ circuit,which one of the following is a $CORRECT$ statement?

An ideal inductor of $\left(\frac{1}{\pi}\right) H$ is connected in series with a $300 \Omega$ resistor. If a $20 \ V, 200 \ Hz$ alternating source is connected across the combination,the phase difference between the voltage and current is

Calculate the capacitive reactance of a capacitor in order to run a bulb rated at $10 \, W, 60 \, V$ when connected in series to an $a.c.$ source of $100 \, V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module