An emf E = 6 cos(6000t) V is applied to an L | vedclass

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

Question

(A) Given: $E = 6 \cos(6000t) \ V$,$L = 4 \ mH = 4 	imes 10^{-3} \ H$,$R = 7 \ \Omega$.Comparing $E = E_0 \cos(\omega t)$ with the given equation,we

Vedclass · Accepted Answer

$0.24$

Vedclass · Answer

(A) Given: $E = 6 \cos(6000t) \ V$,$L = 4 \ mH = 4 	imes 10^{-3} \ H$,$R = 7 \ \Omega$.Comparing $E = E_0 \cos(\omega t)$ with the given equation,we get $E_0 = 6 \ V$ and $\omega = 6000 \ rad/s$.The inductive reactance is $X_L = \omega L = 6000 	imes 4 	imes 10^{-3} = 24 \ \Omega$.The impedance of the $L-R$ circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \ \Omega$.The amplitude of the current $I_0$ is given by $I_0 = \frac{E_0}{Z} = \frac{6}{25} = 0.24 \ A$.

An emf $E = 6 \cos(6000t) \ V$ is applied to an $L-R$ circuit of inductance $L = 4 \ mH$ and resistance $R = 7 \ \Omega$. The amplitude of the current in the circuit is . . . . . . (in $A$)

Similar Questions

In a series $L-C-R$ circuit,

In an $A.C.$ circuit,a resistance '$R$' is connected in series with an inductance '$L$'. If the phase angle between voltage and current is $45^{\circ}$,the value of inductive reactance will be $(\tan 45^{\circ} = 1)$.

For a series $LCR$ circuit with $R=100\,\Omega$,$L=0.5\,mH$ and $C=0.1\,pF$ connected across $220\,V-50\,Hz$ $AC$ supply,the phase angle between current and supplied voltage and the nature of the circuit is:

$A$ coil of inductance $0.1 H$ and resistance $110 \Omega$ is connected to a source of $110 V$ and $350 Hz$. The phase difference between the voltage maximum and the current maximum is

If the power factor is $1/2$ in a series $RL$ circuit with $R = 100\,\Omega$ connected to an $ac$ mains $(50\,Hz)$,find the value of $L$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module