Question

In the circuit shown here, the point $\text{'}C\text{'}$ is kept connected to point $\text{'}A\text{'}$ till the current flowing through the circuit becomes constant. Afterward, suddenly, point $\text{'}C\text{'}$ is disconnected from point $\text{'}A\text{'}$ and connected to point $\text{'}B\text{'}$ at time $t=0$. The ratio of the voltage across resistance and the inductor at $t=\frac{L}{R}$ will be equal to:

• A. $0$
• B. $1$
• C. $-1$
• D. $\infty$

Answer 1

The correct option is C

$-1$

Step 1: Calculate Current at $t=0$

Let the emf of the battery be E. From the question, point $C$ of the circuit is connected to point $A$.

So the inductor is connected to the DC source. When a DC source is applied across an inductor, it behaves as a short circuit after the steady-state is reached.

Current in the circuit at this instant is given by

$I=\frac{E}{R}$

Current through the LR circuit is given by

$i\left(t\right)=i_0{e}^{\frac{-t}{\tau }}$ $\left(1\right)$

As the current cannot suddenly change through an inductor, so the current at the time instant $t=0$ will be equal to the current through the inductor before closing the circuit, that is,

$i_0=I=\frac{E}{R}$ $\left(2\right)$

$\therefore i\left(t\right)=\frac{E}{R}{e}^{\frac{-t}{\tau }}$ $\left(3\right)$

Step 2: Calculate Current at $t=\frac{L}{R}$

We know that the time constant of an LR circuit is given by

$\tau =\frac{L}{R}$ $\left(4\right)$

According to the question, the given instant of time is

$t=\frac{L}{R}$ $\left(5\right)$

From $\left(4\right)$ and $\left(5\right)$

$t=\tau$

Putting this in $\left(3\right)$ we get

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

$\Rightarrow i\left(\frac{L}{R}\right)=\frac{E}{Re}$ $\left(6\right)$

Step 3: Calculate Voltage through Resistor

Now, we know from the Ohm’s law that the voltage across the resistance is

$V_R=IR$

Putting $\left(6\right)$ above, we get
$V_R=\frac{E}{Re}\times R$

$V_R=\frac{E}{e}$ $\left(7\right)$

Step 4: Calculate Voltage through Inductor

We also know that the voltage across an inductor is given by

$V_L=L\frac{di\left(t\right)}{dt}$

From $\left(3\right)$

$V_L=L\frac{d}{dt}\left(\frac{E}{R}{e}^{-\frac{t}{\tau }}\right)$

$V_L=-\frac{EL}{R\tau }{e}^{-\frac{t}{\tau }}$

Putting the value of the time constant from $\left(4\right)$

$V_L=-\frac{ELR}{RL}{e}^{-\frac{Rt}{L}}$

$\Rightarrow V_L=-E{e}^{-\frac{Rt}{L}}$

At the time $t=\frac{L}{R}$ we get the voltage across the inductor as

$V_L=-E{e}^{-\frac{RL}{LR}}$

$\Rightarrow V_L=-\frac{E}{e}$ $\left(8\right)$

Step 5: Calculate the required ratio

By dividing $\left(7\right)$ by $\left(8\right)$ we get

$\frac{V_L}{V_R}=-1$

Hence, the correct answer is option (C).