The switch in the network in Fig. E7.14 opens at t = 0. Find i(t) for t > 0.Figure E7.1...

Calculate the value of the current .

Therefore, the value of the current is .

Inductor does not allow sudden changes in current.

Therefore, the value of the current is equal to the value of the current .

The value of the voltage is . Since, the current doesn’t flow in that branch.

Capacitor does not allow sudden changes in voltage. Hence, the voltage across the capacitor at is equal to the voltage across the capacitor at .

At , the switch is open. Hence, redraw the circuit diagram at as shown in Figure 2.

Apply Kirchhoff’s Voltage Law in the Loop.

Take derivative with respect to .

…… (1)

Consider the characteristic equation of the differential equation.

Solve for the characteristic equation.

The roots of the characteristic equation are and .

The roots the real and unequal, hence the circuit exhibits over damped characteristics.

The solution for the differential equation is

…… (2)

Substitute for .

Substitute for .

…… (3)

Therefore, the expression for the constant is .

Recall equation (2).

Take derivative with respect to .

Substitute for .

Substitute for and for .

Therefore, the value of the constant is .

Recall equation (3).

Substitute for .

Therefore, the value of the constant is .

Recall equation (2).

Substitute for and for .

Therefore, the value of the current is .