Question

An inductor of 12 mH and a resistor of 4 K$\Omega$ are connected in series across a battery of 240 V through a switch. After closing the switch the current in the circuit starts growing. When the current is 15 mA, the potential difference across the inductor will be:

• A. 60 V
• B. 120 V
• C. 180 V
• D. 240v

Answer 1

The correct option is C 180 V

$i=\frac{E}{R}(1-e^{\frac{-Rt}{L}})$
$15\times {10}^{-3}=\frac{240}{4\times {10}^3}(1-e^{\frac{-4t\times {10}^3}{12\times {10}^{-3}}})$
$e^{\frac{-{10}^{6}t}{3}}=\frac{3}{4}$
$V=E{e}^{\frac{-Rt}{L}}$
$=E\times \frac{3}{4}$
$=240\times \frac{3}{4}$
$=180V$