Question

In the circuit shown in figure, $L=10\text{H, R}=5\Omega ,E=15\text{V}$. The switch S is closed at $t=0$.
At $t=2\text{s}$, the current in the circuit is:

• A. $3(1-\frac{1}{e})A$
• B. $3(1-\frac{1}{e^2})A$
• C. $3\left(\frac{1}{e}\right)A$
• D. $3\left(\frac{1}{e^2}\right)A$

Answer 1

The correct option is A $3(1-\frac{1}{e})A$

The current in LR circuit with time is given as
$i=\frac{E}{R}(1-e^{-\frac{Rt}{L}})$
$\Rightarrow i=\frac{15}{5}(1-e^{-\frac{5\times 2}{10}})$
$\Rightarrow i=3(1-\frac{1}{e})A$