In the figure below,the switches S_1 and S_2 | vedclass

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

Question

(D) Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. The current $I$ in the middle wire is $I = I_2 - I_1$.For th

Vedclass · Accepted Answer

Vedclass · Answer

(D) Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. The current $I$ in the middle wire is $I = I_2 - I_1$.For the left loop: $V - I_1 R - L \frac{dI_1}{dt} = 0 \implies I_1(t) = \frac{V}{R}(1 - e^{-(R/L)t})$.For the right loop: $V - I_2 R - 2L \frac{dI_2}{dt} = 0 \implies I_2(t) = \frac{V}{R}(1 - e^{-(R/2L)t})$.The current in the middle wire is $I(t) = I_2(t) - I_1(t) = \frac{V}{R} [e^{-(R/L)t} - e^{-(R/2L)t}]$.To find the maximum current,set $\frac{dI}{dt} = 0$:$\frac{dI}{dt} = \frac{V}{R} [-\frac{R}{L} e^{-(R/L)t} + \frac{R}{2L} e^{-(R/2L)t}] = 0$.$\frac{1}{L} e^{-(R/L)t} = \frac{1}{2L} e^{-(R/2L)t} \implies e^{-(R/2L)t} = \frac{1}{2}$.Taking the natural logarithm on both sides: $-\frac{R}{2L} T = \ln(1/2) = -\ln 2 \implies T = \frac{2L}{R} \ln 2$.Substituting $T$ back into the expression for $I(t)$:$I_{\max} = \frac{V}{R} [e^{-(R/L) \cdot (2L/R) \ln 2} - e^{-(R/2L) \cdot (2L/R) \ln 2}] = \frac{V}{R} [e^{-2 \ln 2} - e^{-\ln 2}] = \frac{V}{R} [\frac{1}{4} - \frac{1}{2}] = -\frac{V}{4R}$.The magnitude is $I_{\max} = \frac{V}{4R}$.Thus,statements $(B)$ and $(D)$ are true.

Similar Questions

$A$ $5\,cm$ 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.....

The resistance and inductance of a series $RL$ circuit are $5 \, \Omega$ and $20 \, H$ respectively. At the instant of closing the switch,the current is increasing at the rate of $4 \, A/s$. The supply voltage is ... $V$.

An inductor coil carries a steady-state current of $2.0 \ A$ when connected across an ideal battery of $emf$ $4.0 \ V$. If its inductance is $1.0 \ H$,find the time constant of the circuit in $sec$.

$Henry/ohm$ can be expressed in

$A$ bulb of $100\, W$ is connected in parallel with an ideal inductance of $1\, H$. This arrangement is connected to a $90\, V$ battery through a switch. On pressing the switch,the

Vedclass Test Series

Exam Paper Generator

Online Exam Module