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

An emf of $15 V$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of currents at time $t=\infty$ and $t=1 s$ after closing the switch is:

$A$ solenoid has an inductance of $60 \, H$ and a resistance of $30 \, \Omega$. If it is connected to a $100 \, V$ battery,how long will it take for the current to reach $\frac{e-1}{e} \approx 63.2\%$ of its final value?

$Henry/ohm$ can be expressed in

The figure shows three circuits with identical batteries,inductors,and resistors. Rank the circuits according to the currents through the battery just after the switch is closed,greatest first.

Vedclass Test Series

Exam Paper Generator

Online Exam Module