Question

A coil of inductance $300mH$ and resistance $2\Omega$ is connected to a source of voltage $2V$. The current reaches half of its steady-state value in.

• A. $0.15sec$
• B. $0.3sec$
• C. $0.05sec$
• D. $0.1sec$

Answer 1

The correct option is D

$0.1sec$

Step1: Given data.

Inductance of the coil, $L=300mH=300\times {10}^{-3}H$

Resistance, $R=2\Omega$

supply voltage, $e=2V$

Step2: Finding the time $t$ at which current reaches half of its steady-state.

We know that,

The charging of inductance is given by

$I=I_0\left[1-e^{-\frac{Rt}{L}}\right]$

Where, $I=$induced current, $I_0=$steady state current.

According to question:

Induced current reaches the half of its steady state current.

Therefore,

$I=\frac{I_0}{2}$ …..$\left(i\right)$

Now,

$\Rightarrow$ $I=I_0\left[1-e^{-\frac{Rt}{L}}\right]$

$\Rightarrow$ $\frac{I_0}{2}=I_0\left[1-e^{-\frac{Rt}{L}}\right]$ $\left[\because fromequation\left(i\right)\right]$

$\Rightarrow$ $\frac{1}{2}=1-e^{-\frac{Rt}{L}}$

$\Rightarrow$ $e^{-\frac{Rt}{L}}=2^{-1}$ …..$\left(ii\right)$

Apply log both side in equation $\left(ii\right)$ we get.

$\Rightarrow$$\frac{Rt}{L}=\ln 2$

$\Rightarrow$ $t=\frac{L}{R}\ln 2$

$\Rightarrow$ $t=\frac{300\times {10}^{-3}}{2}\times 0.693$

$\Rightarrow$ $t=0.1sec$

Hence, option D is correct.