Question

In Fig, take E0 = 12 V, R1 = 4.0 W, R2 = 8.0 W, R3 = 2.0 W, and L = 2.0 H. What is the current I2 (a) immediately after the switch is first closed and (b) a long time after the switch is closed? (c) After a long time the switch is again opened. Now what is I2?

The Problem with the picture is on Problem 36 in the website below

https://www-physics.ucsd.edu/students/courses/winter2010/physics2b/hw/hw9.pdf

Answer 1

a)

the inductor current is zero just after the switch is closed. At this instant,

the currents can be determined from the circuit with the inductance open-circuited, so that its branch can be removed. Thus,

I3=0, and I1=I2=E

Therefore, E=(R1+R2)I2

substitute the values in above,

         12 V=(4 W+8 W) I2

           I2=1 A.

(b) After the currents have been flowing a long time, they reach steady values then,

        dI=dt=0,

and the voltage across the inductance is zero. The currents can be found by short-circuiting the inductance

I1=E/(R1+R2*R3)

substitute the values,

I1=12 V/(4 W+8 W*2 W) =2.14 A,

I2=R3/(R2+R3)*I1

   =1/5*I1

   =0.429 A,

and I3=I1*R2=(R2+R3)=4/5*I1

          =1.71 A.

(c) When the switch is reopened, no current flows through the battery’s branch,

I1=0

which can be removed from the circuit to calculate

I2=I3

at this instant. The induced emf acts to keep the current flowing at its value in part (b) ,

so, I2=-I3

        =-1.71A