The switch in FIGURE P30.76 has been open for a long time. It is closed at $t=0\mathrm{s}$. What is the current through the $20\mathrm{\Omega }$resistor

a. Immediately after the switch is closed?

b. After the switch has been closed a long time?

c. Immediately after the switch is reopened?

(a) When the switch is rapidly closed, the current does not always go to the inductor. In this scenario, both circuits are wired in series. The current flows in between two connected in series is similar. The comparable resistance of the combination is

$R_{\mathrm{eq}}=R_1+R_2=10\mathrm{\Omega }+20\mathrm{\Omega }=30\mathrm{\Omega }$

We use Ohm's law,

$I=\frac{V}{R_{\text{eq}}}=\frac{30\mathrm{V}}{30\mathrm{\Omega }}=1\mathrm{A}$

(b) The current that passes through the inductor varies over time. As the inductance becomes stable, the change in direction through it remains zero for just a long time. As a result, the voltage differential across inductor is actually zero.$dI/dt=0$

When the current through the inductor is zero, it performs like a straight wire. In this situation, the power decided to pass through the wire to which Joes is hooked. As a reason, there is no flow of charge through the resistor as a response.

$I=0\mathrm{A}$

(c)When the inductance behaves like a wire, the load current can just slide down; this power is measured by Ohm's law.

$I=\frac{V}{R}=\frac{30\mathrm{V}}{10\mathrm{\Omega }}=3\mathrm{A}$

As longer as the changeover continues to operate, this resistance will be abolished, and the current will indeed follow a path of each other to go. The contemporary flow, measured as a whole, will be

$I=3\mathrm{A}$