Hello Everyone!
I have those two exercises from my textbook involving First Order Circuits with Capacitors and Inductors..

Now, in the first one the exercises wants a v(t) that is not the capacitor one..
For t>0 isn't the whole right part after the switch going away? So isn't it
v∞(t) = 6 ?
But v for t<0?

In the second one the inductor behaves like a short circuit to dc...
for t<0 isn't i=0 ?
for t>0 isn't it 0 again?

I'm kinda confused...
thanks for your help

 

Now, in the first one the exercises wants a v(t) that is not the capacitor one..
For t>0 isn't the whole right part after the switch going away? So isn't it
v∞(t) = 6 ?

After the switch is close the right resistor R will be shorted by the switch.
So yes at t=∞ v = 6V. But at t(0) voltage will we equal to 3V.

But v for t<0?

what is your guess

In the second one the inductor behaves like a short circuit to dc...
for t<0 isn't i=0 ?
for t>0 isn't it 0 again?

It seems to me that you are right

 

Answers in the book are in the file attached
Can you explain me why is it 3 at t(0) ?
And apparently it's not 0 for the second exercise..

 

Answers in the book are in the file attached
Can you explain me why is it 3 at t(0) ?

Before the switch is close t < 0 the voltage V1 = 6V and V2 = 3V

V1 = Eo*R/(2R) = 1/2Eo = 0.5Eo = 6V

V2 = E1 * R/(3R+R) = E1 * 1/4 = 0.25E1 = 3V

So the capacitor is charged to Vc = V1 - V2

At times equal t = 0 ( when the switch is close ) V2 = 0V (short by a switch) And the capacitor act lice a Voltage source at time t = 0.
So V1 must be equal Vc. And capacitor start charging to reach at t = ∞ 6V.

And apparently it's not 0 for the second exercise

Yes, your book is right. I made the mistake.
At time t = ∞ the switch is in close position, so current will be flow through the right resistor. I = E2/R = 5A

 

ok i'm starting to understand... But why does Capacitor acts like voltage source?

 

At time equal t = 0 voltage across capacitor is equal to 3V. So when we close the switch we connect already charged capacitor parallel to left resistor. And capacitor force the V1 voltage to drop from 6V to 3V. At that moment the circuit behaves very similarity to the circuit in which we replace the capacitor with voltage source.