# Basic Transient Question

Discussion in 'Homework Help' started by Spoon, May 18, 2010.

1. ### Spoon Thread Starter Member

May 18, 2009
12
0
Okay this is a 'basic' transient question but I'm not sure what to do for 8.3 and 8.4

QUESTION 8 [4 MARKS]
A 1F capacitor is connected in series with a 1Ω resistor across a 10V DC supply. A switch is also connected in series with another 1Ω resistor, which is connected in parallel with the capacitor.

Initially, the switch is open, however at t=0 the switch is closed.
8.1. Find the initial steady state voltage across the capacitor before the switch is closed. [1]

I got this to be 10V ???

8.2. Find the final steady state voltage across the capacitor once the switch is closed. [1]

I got 5V ???

8.3. How much time would have to pass once the switch is closed, for the voltage to reach 136.8% of its
final value. [1]

How would I go about doing this and 8.4?

8.4. How much energy is retained in the capacitor at this time? [1]

2. ### mik3 Senior Member

Feb 4, 2008
4,846
67
You are right about 8.1 and 8.2.

For 8.3, fins the time it needs to drop from the initial 10V to 6.84V (136.8% of 5V) after the switch closes.

For 8.4, use E=0.5*C*V^2

Spoon likes this.
3. ### Spoon Thread Starter Member

May 18, 2009
12
0
8.4 i guess is just (1/2)CV^2

I thought of doing 8.3 as 5(1.368) = 10(1 - e^(-T/RC)
RC =1
T = 1.15 seconds but this doesn't seem to be correct somehow?

4. ### Spoon Thread Starter Member

May 18, 2009
12
0
never refreshed to see your reply before i posted, is my way of doing 8.3 correct?...

Last edited: May 18, 2010
5. ### mik3 Senior Member

Feb 4, 2008
4,846
67
Because you have an initial charge on the capacitor you have to use this formula:

Vc=Vfinal(1-exp(-t/RC)+Vinitial*exp(-t/RC)

6. ### Spoon Thread Starter Member

May 18, 2009
12
0
I have just looked at the provided solution and it says 1.15 seconds, the way you provided seems to get me 0.999.... Am not sure which one is correct... Or maybe I have misunderstood your formula

7. ### mik3 Senior Member

Feb 4, 2008
4,846
67
When the switch is closed the time constant is not 1 but 0.5 because the resistance seen by the capacitor is the parallel combination of the two 1R resistors.

8. ### Spoon Thread Starter Member

May 18, 2009
12
0
according to a computer simulation, the time taken is 0,5 seconds. I reworked RC to be 0.5, and I don't get the 0.5 seconds as the answer though??

9. ### mik3 Senior Member

Feb 4, 2008
4,846
67
I get 0.5 sec.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Mik3 is quite correct.

Sometimes it helps to draw pictures and write things down together ....

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