From the picture right down below, I don't understand. why the Ic current be negative?
if it's a series circuit I think it should has the same value and same direction for currents.
for this example, Find the Vc(t) and IL(t) for t>0 (after the switch is opened).

 

Hi,

A quick look tells me that the inductor has some initial current through it when the switch is opened, and the cap has some voltage, so the curent flow is the same as it was before the switch was open and that is down through the inductor..
Make sense to you?

In theory we have a few main things at play here:
1. The indudctor has some current through it before the switch is opened and its direction is downward.
2. The cap is charged to some voltage level positive on top.
3. When the switch is open, the inductor wants to keep the current flowing in the same direction as before. The current in the cap now follows that.

 

From the picture right down below, I don't understand. why the Ic current be negative?
if it's a series circuit I think it should has the same value and same direction for currents.
for this example, Find the Vc(t) and IL(t) for t>0 (after the switch is opened).

View attachment 141410View attachment 141411View attachment 141412

The issue is the definition of the two currents. The diagram explicitly defines the inductor current to be downward through the inductor. The diagram does not define the direction of the capacitor current directly, but rather does so indirectly via the passive sign convention. The diagram defines the voltage polarity across the capacitor and the passive sign convention states that the current through a device enters the positive terminal of the defined voltage across the device. Thus the direction of the capacitor current is also downward through the capacitor. Hence the inductor current and the capacitor current will always be equal and opposite once the switch opens.