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 ?

 

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?

 

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 ?

 

Your figure for It is okay. Remember that the current divides into the parallel branches.

 

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

 

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:

\(12\frac{4.5}{4+4.5} = 6.35 V\)

Or, using your approach:
\(I = \frac{12}{8.5} = 1.41 mA\\
x = 12 - 1.41(4) = 6.35 V\)


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

 

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

 

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

 

What is meant by Vab(0+)?

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

 

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.

 

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.

 

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)

 

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)

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.

 

Indeed it is, any idea on my last question ?

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 ?

 

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.

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

 

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.

 

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.

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.

 

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 ?

 

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 ?

Yes, Is is not the same with the switch open and closed.

With the switch closed Is increases.

 

In that case how can I calculate it ?
I thought about source transformations but I don't see how I can do that.