Question

In the figure below, the battery is ideal, E = 22.0 V, R2 = 30.0 2, R2= 18.012, and L = 3.20 H. Switch Sis closed for a long time. What is the potential difference across the inductor immediately after the switch is open, in V? I R} Ri M

Answer 1

For DC circuit, inductor behaves as a short circuit when battery has been connected for a long time.

So, when switch is closed,current through inductor=current through R1=voltage/resistance=22/30 A.

Now,current through inductor cannot change instantaneously. So,when the switch is open, the current through the inductor is still 22/30 A. This same current will flow through R1 and R2 as both R1 and R2 are in series with the inductor.

Now,using Kirchoff's voltage law, V-22/30*R1-22/30*R2=0,where V is the voltage across the inductor immediately after the switch is open.

=> V=22/30(R1+R2)=22/30*(30+18)=35.2 volts.

So,required potential difference=35.2 volts.