V and I of a capacitor in RC circuit

Discussion in 'Homework Help' started by Hitman6267, May 1, 2010.

Apr 6, 2010
82
0

I'm having a little trouble with this so I'm going to explain my thought process.

Finding V(0) (the question in the attached picture)
ok so first thing, I know at t=0 the capacitor acts like an open circuit.
In other exercises usually we do one of two things. Say that the V of capacitor is equal to the voltage source because it has been connected for a long time or say that the V is equal to the voltage of something parallel to it. I can't find how any of them apply here.

So what's the first step ?

2. beenthere Retired Moderator

Apr 20, 2004
15,815
293
Remove the cap. With the switch open, what will be the potentials at A & B? Place the cap back in circuit, switch still open. Wait several seconds. What will the potentials be at A & B?

Apr 6, 2010
82
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This is my attempt at calculating the potential at A B with the cap removed.and the switch open

Calculating I debited from the source: I = V/Req = 12/8.5 = 1.41
Now I need to know the current in the branches so I can calculate the voltages of the resistors.

Current divider:
i= 2/3 * 1.41 = 0.94 that's for R1 and R2 = 1.41-0.94= 0.47

This is where something is off. If I calculate V (using V= RI) of all each resistors they all would have V=2.82. What is my mistake ?

Last edited: May 1, 2010
4. beenthere Retired Moderator

Apr 20, 2004
15,815
293
Your figure for It is okay. Remember that the current divides into the parallel branches.

Apr 6, 2010
82
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Sorry I didn't get what you said. You're saying that in principle my calculations are correct?
But they can't be. And how would I be able to calculate VAB from them

6. Ghar Active Member

Mar 8, 2010
655
73
It's a voltage source with a series resistance with two parallel voltage dividers.

The voltage dividers have total resistance each of 6+3 = 9 kilohms.

In parallel you get 4.5 kilohms.
The voltage across the voltage dividers therefore, is:

$12\frac{4.5}{4+4.5} = 6.35 V$

$I = \frac{12}{8.5} = 1.41 mA\\
x = 12 - 1.41(4) = 6.35 V$

Node A then, is:
$6.35\frac{6}{3+6} = 4.23 V$

Node B:
$6.35\frac{3}{3+6} = 2.12 V$

Voltage across the capacitor is node A - node B

Alternatively, the voltage dividers take half the supply current:
1.41 / 2 = 0.7 mA

Node A will be 0.7 * 6 = 4.2 V
Node B will be 0.7 * 3 = 2.1 V

Hitman6267 likes this.

Apr 6, 2010
82
0
wow, it's correct thank you
But why is node A 6.35 * 6/(3+6) and not 6.35 * 3(3+6)

8. Ghar Active Member

Mar 8, 2010
655
73
For Node A we have R2 = 6 kilohms as the output resistor while for node B we have R1 = 3 kilohms as the output resistor.

Also look at the current version... node A is 0.7 mA * 6 kilohms

FYI 6.35 / (3+6) = 0.7 mA

9. mik3 Senior Member

Feb 4, 2008
4,846
69
What is meant by Vab(0+)?

Does it mean just after the switch closes or after several minutes have passed?

Apr 6, 2010
82
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oh I was looking at the wrong resistors (the top ones) I should have looked at the bottom two.

Last question for today (I promise ) - I can't think any more any way
They want V(0.001) - same circuit -
So I'm using this equation:
V = Is R+ ( Vo - IsR) e^ -(t/RC)
V= (1)(4)+ (2.12 -4) e^- (0.001/(4*5))

What is wrong ?

@mik3 I'm thinking just after the switch closes.

11. mik3 Senior Member

Feb 4, 2008
4,846
69
If Vab(0+) is just after the switch closes, then Vab=0V because the uncharged capacitor acts a short circuit at the instant the switch closes.

Apr 6, 2010
82
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The 2.12 answer is correct so I don't know where that leaves us. But I don't see why the cap would be uncharged, it is connected to a voltage source (there are 2 voltage sources in the circuit)

13. mik3 Senior Member

Feb 4, 2008
4,846
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Ohh yes, I didn't realized that the right voltage source is after the switch.

Therefore, after a few calculations the answer is Vab=2.12V.

Apr 6, 2010
82
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Indeed it is, any idea on my last question ?

15. mik3 Senior Member

Feb 4, 2008
4,846
69
You have to explain what Is and R represent.

I don't know if your formula is correct.

The formula I know is:

Vc=V[1-exp(t/RC)]+Vo*exp(t/RC)

Vc=capacitor voltage
V=voltage source
RC=time constant
Vo=initial capacitor voltage

Apr 6, 2010
82
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Is is the current debited from source and R is the thevenin equivalent resistor.
I copied the formula from my book so it's correct.

17. mik3 Senior Member

Feb 4, 2008
4,846
69
I get an answer of 4V.

I think you mistake is the calculation of the time constant RC.

You have RC=4*5 but the correct is RC=4*5*10^-6, see the capacitor value.

Apr 6, 2010
82
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The answer 4 is not correct.
Here's something that might help.
1st question they asked. What is the value of iC(0+) in mA ? answer 0.47
2nd question they asked. What is the value of vab(0+) in V? answer 2.12

Now we need to calculate Vab(0.001).
I think my mistake other than the one you mentioned (because that was a typo, I used the correct value in my calculations) is the possibility of Is changing when the switch closes. Does it ?

19. mik3 Senior Member

Feb 4, 2008
4,846
69
Yes, Is is not the same with the switch open and closed.

With the switch closed Is increases.