Consider a circuit as in the attachment.
Assuming that the initial voltage of the capacitor is zero.
At the time t =0, K is closed we have Vc(0) = 0
But according to KVL:
Vc = E (at all times)
This is seems wrong to me. Is it right that KVL is not true in the transition time?

 

The problem is that yoiu have a circuit that can only exist on paper. You have an ideal voltage source that has no resistance and a capacitor that has no resistance and wires that have no resistance. None of those actually exist (well, let's leave superconductors out of the discussion) and most of the time the difference doesn't matter. But in this circuit that does matter because without some resistance the battery would have to dieliver an infinite amount of current for an infinitesimal amount of time just as the switch is closed in order to instantly change the charge on the capacitor. But with even a tiny resistance what you would see would be a large current flowing initially and all of the voltage dropped across that resistor. Then, over a short period of time, the current would die down to zero and the capacitor would charge up to E.

 

Thanks, I understand it.
Now I see that in most book often assume that voltage source, capacitor and wires all are ideal. Then at t = 0 the voltage across capacitor is 0V or E V?

 

Thanks, I understand it.
Now I see that in most book often assume that voltage source, capacitor and wires all are ideal. Then at t = 0 the voltage across capacitor is 0V or E V?

Well... since it's all theoretical, at t = -0 it's 0V, at t = +0 it's E, and at t = 0 then it's anything you want to call it.

 

Thanks, I understand it.
Now I see that in most book often assume that voltage source, capacitor and wires all are ideal. Then at t = 0 the voltage across capacitor is 0V or E V?

At t=0 it is undefined. At t=0- it is 0V and at t=0+ it is E.

 

Oh., ErnieM beat me to it. But now you have independent confirmation.