# First Order Circuits

#### newzed

Joined Jan 11, 2012
16
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 #### Jony130

Joined Feb 17, 2009
5,314
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?

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

#### newzed

Joined Jan 11, 2012
16
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..

#### Jony130

Joined Feb 17, 2009
5,314
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
At time t = ∞ the switch is in close position, so current will be flow through the right resistor. I = E2/R = 5A

#### newzed

Joined Jan 11, 2012
16
ok i'm starting to understand... But why does Capacitor acts like voltage source?

#### Jony130

Joined Feb 17, 2009
5,314
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.