The time constant of an LR circuit is 10 , s. | vedclass

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

Question

(D) The time constant $	au$ of an $LR$ circuit is given by $	au = \frac{L}{R}$.Initially,$\frac{L}{R} = 10$,which implies $L = 10R$  --- $(1)$.When

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(D) The time constant $	au$ of an $LR$ circuit is given by $	au = \frac{L}{R}$.Initially,$\frac{L}{R} = 10$,which implies $L = 10R$  --- $(1)$.When a resistance of $10 \, \Omega$ is added in series,the new resistance becomes $(R + 10) \, \Omega$. The new time constant is $\frac{L}{R + 10} = 2$,which implies $L = 2(R + 10) = 2R + 20$ --- $(2)$.Equating $(1)$ and $(2)$:$10R = 2R + 20$$8R = 20$$R = 2.5 \, \Omega$.Substituting $R$ into equation $(1)$:$L = 10 	imes 2.5 = 25 \, H$.

The time constant of an $LR$ circuit is $10 \, s$. When a resistance of $10 \, \Omega$ is connected in series with the existing circuit,the time constant becomes $2 \, s$. The self-inductance of the circuit is ..... $H$.

Similar Questions

Find $V_A - V_B$ in steady state. (in $, V$)

$A$ $5\,cm$ long solenoid having $10\,\Omega$ resistance and $5\,mH$ inductance is joined to a $10\,V$ battery. At steady state,the current through the solenoid in ampere will be.....

Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:
$(a)$ Just after the closing of the key.
$(b)$ $A$ long time after the closing of the key.

The figure shows a circuit containing three resistors ($9 \ \Omega$ each) and two inductors ($4 \ mH$ each). The reading of the ammeter at the moment the switch $K$ is turned $ON$ is $...... \ A$.

In the given circuit,the switch is in position $a$. For what value of $R$ will the circuit have a time constant of $10\,\mu s$ in $K\Omega$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module