Consider the network in Fig. E7.6. If the switch opens at t = 0, find the output voltage v...

Apply KVL to the loop .

…… (2)

Apply KVL to the loop .

Substitute for in equation (2).

The current , which is same as the loop current is, .

At , the switch opens and the inductor is replaced by a current source. The current flowing throw the inductor cannot change in zero time. Hence, .

Redraw the Figure E7.6 at .

Figure 2:

In Figure 2, same current flow the circuit as it is a series circuit.

Calculate the voltage across the resistor .

After a long time the switch is opened the circuit goes to steady state. In steady state replace the inductor by a short circuit.

Redraw the Figure E7.6 at .

Figure 3:

Calculate the voltage across resistor at .

Redraw the Figure E7.6 to calculate the equivalent resistance by looking into open terminals of the inductor and replacing the voltage sources by short circuit.

Figure 4

In Figure 4, resistors and are in series. Hence,

Calculate the time constant.

Substitute for 6 V , for and for .

Thus, the voltage for is, .