Question

An inductor ($L=100mH$), a resistor ($R=100\Omega$) and a battery ($E=100V$) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit $1ms$ after the short circuit is:

• A. $1/eA$
• B. $eA$
• C. $0.1A$
• D. $1$

Answer 1

Answer: (A)

$1/eA$

energy stored in Inductor $=\frac{1}{2}L(\frac{E}{R}{)}^2$
$I_o=\frac{E}{R}$
$i=I_o{e}^{\frac{-tR}{L}}$
$=\frac{100}{100}\times e^{\frac{-{10}^{-3}\times {10}^2}{100\times {10}^{-3}}}$
$i=e^{-1}A$