Question

An inductor of $10\text{mH}$ is connected to a $20\text{V}$ battery through a resistor of $10k\Omega$ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after $1\mu s$ is $\frac{x}{100}\text{mA}$. Then $x$ is equal to (Take $e^{-1}=0.37$)

• A. 74.00
• B. 74
• C. 74.0

Answer 1

Answer: (A), (B), (C)

$I_{max}=I_0=\frac{V}{R}=\frac{20}{10\times {10}^3}=2\text{mA}$
For LR - decay circuit
$I=I_0{e}^{\frac{-Rt}{L}}$
$=2e^{\left(\frac{-10\times {10}^3\times 1\times {10}^{-6}}{10\times {10}^{-3}}\right)}$
$=2e^{-1}$
$I=2\times 0.37\text{mA}$
$I=\frac{74}{100}\text{mA}$
$\therefore x=74$