Question

In the LR circuit shown, what is the variation of the current I as a function of time? The switch is closed at time t=0 sec.

• A. $\frac{V}{R}(1-e^{\frac{Rt}{L}})$
• B. $\frac{V}{R}{e}^{\frac{Rt}{L}}$
• C. $-\frac{V}{R}{e}^{\frac{Rt}{L}}$
• D. None

Answer 1

The correct option is A $\frac{V}{R}(1-e^{\frac{Rt}{L}})$

The quantity is shown in figure.

Applying KVL on loop CGDE

$I^{\prime }R=2V$

$I^{\prime }=\frac{2V}{R}$

$-3V+(I+I^{\prime })R+L\frac{d(I+I^{\prime })}{dt}+2V=0$

$IR+\frac{2V}{R}R+L\frac{d}{dt}(I+\frac{2V}{R})=V$

$IR+2V+L\frac{dI}{dt}=v$

$L\frac{dI}{dt}=-IR-V$

$-\frac{L}{R}\frac{dI}{dt}=+I+\frac{V}{R}$

$-\frac{dI}{(-I-\frac{V}{R})}=-\frac{R}{L}dt$

$-\frac{dI}{(-i-\frac{V}{R})}=-\frac{R}{L}dt$

$\ln (-i-\frac{V}{R})=-\frac{R}{L}t+c$

$att=0i=-i^{\prime }=\frac{-2V}{R}$

$\ln (\frac{2V}{R}-\frac{V}{R})=c$

$c=\ln \frac{V}{R}$

Now,

$\ln (-i-\frac{V}{R})=-\frac{R}{L}t+\ln \frac{V}{R}$

$-i-\frac{V}{R}=\frac{V}{R}\left(e^{-\frac{R}{L}t}\right)$

$i=-\frac{V}{R}(e^{-\frac{R}{L}t}+1)$