1. Jun 4, 2012

IIK*JII

1. The problem statement, all variables and given/known data

The figure in attached file below shows a circuit in which a battery (electromotive force = 10V) is connected with three capacitors (capacitance = 1 microFarad, 2microFarad,3microFarad) and two resistors (resistance = 2Ω and 3Ω)

Determine the potential difference between points P and Q.
2. Relevant equations
V=IR
Q=CV
Potential difference between P and Q = VP - VQ

3. The attempt at a solution
I understand that capacitors are in this problem is fully charged so I can analyze them as open circuit. It means no current passes through three capacitors and I can solve this problem only voltage and resistance circuit

Thus, resistance 2Ω and 3Ω are connected in series and Rtotal = 5Ω
From ohm's law >> V=IR
I get Itotal = 10/5 = 2A
So, Voltage at 3Ω is V = 2*3 =6 V is the VQ

and VP is 10 V
Thus, I get potential difference is 4 V

Thanks a lot, for answer this is my first post on PF
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jun 4, 2012

ehild

There is some potential difference across the 2uF capacitor, so VQ is not 6 V.

ehild

3. Jun 4, 2012

Staff: Mentor

Hi IIK*JII, Welcome to Physics Forums.

You've calculated the final potential at the node where the resistors meet (let's call it node R) to be 6 V, but that is not node Q where all the capacitors meet; they are distinct nodes.

Nodes Q and R are separated by a 2 μF capacitor which will pass no current at steady state, but you cannot assume that the potential on that capacitor at steady state will be zero.

I think you're going to need a bigger hammer for this problem

4. Jun 4, 2012

willem2

Piont P isn't the same point as the point between the resistances, so it doesn't have to have the same potential as this point.

To solve this:

set the potential at point Q equal to V_Q.

Use V = qC for all three capacitors. You can compute all charges on the capacitors as a function of V_Q. You know all potentials for one of the sides of each capacitor.

If all the capacitors are initially uncharged, Kirchhoffs' current law: $\Delta I = 0$ for the point Q, is valid for the charges on the capacitors as well, since I = dq/dt, so $\Delta q = 0$ (where a charge on a capacitor is positive if the side of the capacitor connected to Q is more positive)

5. Jun 4, 2012

TSny

Something to always remember: Ohm's law, V = IR, does not give you the voltage at a point, it gives you the voltage change as you go from one side of the resistor to the other side.

6. Jun 4, 2012

IIK*JII

Thank you gneill and ehild

From your hint I think I should find the voltage across 2uF which is VQR not node Q,,right??
I calculated this and VQR is 5V

Now I understand that in fully charged capacitors I=0 but voltage can't be zero so voltage in 3uF, 2uF and 1uF remains....

7. Jun 4, 2012

Staff: Mentor

It is unlikely that VQR could be 5V; if node R is at 6V due to the resistor divider, then that would make VQ = 6+5 = 11V, which is higher than the source voltage!

Think about the hints that Willem2 gave in post #4.

8. Jun 4, 2012

ehild

9. Jun 4, 2012

IIK*JII

ehild...

from the figure 1uF is parallel with 2uF so C12= 3 uF. C12 is connected in series with 3uF. Thus, I got total capacitance is (3*3)/(3+3) = 1.5 uF

From Q=CV because of V=emf=10V so Qtotal= 15uC

I got V@3uF = 5V and V@1uF//2uF = VQR=5V

or I got wrong??

I have something wondering node R connected between 2 resistors (2Ω&3Ω) and because 3Ω
connect with capacitors so no current passes 3Ω ???,,, So, VR=V@2Ω??

10. Jun 5, 2012

ehild

NO, the capacitors are neither in series, nor in parallel.

Two things are connected in series if they have one pair of terminals in common, and nothing else is connected there.

Do not mix potential with potential difference. Point R is at a potential with respect to a reference point. There is a potential difference across the 2 ohm resistor which is equal the current through it multiplied by resistance.
The current flows where it can. The current flowing out from the positive terminal of the battery, can flow through the 2 ohm resistor, and reaching point R, flows further through the 3 ohm resistor toward the negative terminal of the battery, following the red line.

Yo need to operate with the charges on the capacitors. The three capacitor has got a common point, Q. The total charge on the connected plates must be zero: -q1+q2+q3=0.

The potential of point R is 6 V with respect to the negative terminal of the battery (which is the same as that of point O). If point Q is at potential U with respect O , the potential difference across the 1 μ F capacitor is U, that across the 3 μF capacitor is 10-U, and the PD across the 2μF capacitor is U-6. Find the charges in terms of U and use the charge neutrality at point Q.

ehild

Attached Files:

• rescap.JPG
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11. Jun 5, 2012

IIK*JII

Thank you!! million thanks to you ehild :)

I got it yeah

From now, I understand that because of capacitors are fully charged so ΔI = 0 (no current pass through all 3 capacitors like ehild picture) and from willem2's hint I=dq/dt so Δq = 0...

From KCL; -q@3uF+q1uF+q2uF=0 ...(1)
From Q=CV
-3(10-U)+U+2(U-6)=0
U = 7 and U=VQ and ref. node
Thus,, VPQ = VP -VQ = 10-7 = 3V

But this is the node voltage method,, right??

Thanksss

12. Jun 5, 2012