The ratio of the time constant in charging an | vedclass

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

Question

(B) When the switch is closed (charging),the inductor $L$ is in series with the resistor $2R$. The resistor $R$ is in parallel with the series combina

Vedclass · Accepted Answer

$3 : 2$

Vedclass · Answer

(B) When the switch is closed (charging),the inductor $L$ is in series with the resistor $2R$. The resistor $R$ is in parallel with the series combination of $2R$ and $L$. The time constant for the charging circuit is determined by the equivalent resistance $R_{eq}$ seen by the inductor. When the battery is shorted (or considered ideal),the equivalent resistance is $R_{eq} = 2R + R = 3R$. Thus,the charging time constant is $	au_{1} = \frac{L}{3R}$.When the switch is opened (discharging),the inductor $L$ forms a closed loop with the resistors $2R$ and $R$ in series. The total resistance in this loop is $R_{eq} = 2R + R = 3R$. Thus,the discharging time constant is $	au_{2} = \frac{L}{3R}$.Wait,re-evaluating the circuit: When the switch is closed,the inductor is in series with $2R$. The resistor $R$ is in parallel with the branch containing $2R$ and $L$. The effective resistance seen by the inductor is $R_{eq} = 2R + 0 = 2R$ (since the battery is ideal,the parallel branch $R$ does not affect the time constant). So $	au_{1} = \frac{L}{2R}$.When the switch is opened,the inductor discharges through the series combination of $2R$ and $R$. So $R_{eq} = 2R + R = 3R$. Thus,$	au_{2} = \frac{L}{3R}$.The ratio $\frac{	au_{1}}{	au_{2}} = \frac{L/2R}{L/3R} = \frac{3}{2}$.

The ratio of the time constant in charging and discharging in the circuit shown in the figure is

Similar Questions

An inductor coil stores $U$ energy when $I$ current is passed through it and dissipates energy at the rate of $P$. The time constant of the circuit,when this coil is connected across a battery of zero internal resistance,is

In a decaying $L-R$ circuit,the time after which energy stored in the inductor reduces to one-fourth of its initial value is

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 that contains three resistors with resistance $R = 2.0 \, \Omega$, two inductors with inductance $L = 2.0 \, mH$, and an ideal battery with $emf$ $E = 9 \, V$. The current $i$ just after the switch $S$ is closed will be .... $A$.

An ideal coil of $10 \ H$ is connected in series with a resistance of $5 \ \Omega$ and a battery of $5 \ V$. $2 \ s$ after the connection is made,the current flowing in the circuit in amperes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module