The switch in the following circuit was in position 'a' for long time , before passing to position b at t=0.

Find the initial current.

Find the initial Voltages of the capacitor , the inductor and the resistors.

What i tried

At t=0- , the circuit will look like this :

Since the switch was closed for a longtime , the current going through the capacitor will become 0 , while the voltage of the inductor will become 0.

Applying KVL : 28 +14I +20=0 , Hence I = -3A = I0

The voltage through the capacitor will be equal to the voltage through the 4K resistor + 28v

vc(t=0) = 4(-(-3))-28=-16V

and finally VL(t=0) = 0 since the inductor is short circuited.

However when the switch passes to the position b , the KVL equation doesn't apply anymmore: -16+8(-3)+0+20 is different than zero.

Thanks in advance.

The switch in the following circuit was in position 'a' for long time , before passing to position b at t=0.

Find the initial current.

Find the initial Voltages of the capacitor , the inductor and the resistors.

What i tried

At t=0- , the circuit will look like this :

Since the switch was closed for a longtime , the current going through the capacitor will become 0 , while the voltage of the inductor will become 0.

Applying KVL : 28 +14I +20=0 , Hence I = -3A = I0

The voltage through the capacitor will be equal to the voltage through the 4K resistor + 28v

vc(t=0) = 4(-(-3))-28=-16V

and finally VL(t=0) = 0 since the inductor is short circuited.

However when the switch passes to the position b , the KVL equation doesn't apply anymore: -16+8(-3)+0+20 is different than zero.

Thanks in advance.