Question

In the circuit arrangement shown in figure, the switch $S$ is closed at $t=0$. Find the current in the inductance as a function of time? Does the current through $10\Omega$ resistor vary with time or remains constant.

Answer 1

At $t=0$,equivalent resistance of an inductor in infinite and at $t=\infty$, equivalent resistance is zero.
$\therefore$ Initial current through inductor $=0$ and
Final current through inductor $=\frac{36}{10}=3.6A$
To find equivalent time constant we will have to short circuit the battery and find net resistance across inductor.
$R_{net}=\frac{10\times 5}{10+5}=\frac{10}{3}\Omega$
${\tau }_L=\frac{L}{R_{net}}=\frac{3}{10}ms$
Current through inductor will increase exponentially from $0$ to $3.6A$
$\therefore$ $i=3.6(1-e^{-t/{\tau }_L})$ where ${\tau }_L=300\mu s$
Current through $10\Omega$ will vary with time.