[Confuse]Capacitor and resistor in DC circuit,

[IIK*JII]

[Confuse!]Capacitor and resistor in DC circuit, please help :)

Homework Statement

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.

Homework Equations

V=IR
Q=CV
Potential difference between P and Q = V_P - V_Q

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 R_total = 5Ω
From ohm's law >> V=IR
I get I_total = 10/5 = 2A
So, Voltage at 3Ω is V_3Ω = 2*3 =6 V is the V_Q

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

However, the answer is 3V! What did I understand something wrong.. Please help I really stuck with this problem

Thanks a lot, for answer this is my first post on PF

 

[ehild]

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

ehild

 

[gneill]

Homework Statement

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.

Homework Equations

V=IR
Q=CV
Potential difference between P and Q = V_P - V_Q

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 R_total = 5Ω
From ohm's law >> V=IR
I get I_total = 10/5 = 2A
So, Voltage at 3Ω is V_3Ω = 2*3 =6 V is the V_Q

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

However, the answer is 3V! What did I understand something wrong.. Please help I really stuck with this problem

Thanks a lot, for answer this is my first post on PF

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

 

[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)

 

[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.

 

[IIK*JII]

Thank you gneill and ehild
I really appreciate your help

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

 

[gneill]

Thank you gneill and ehild
I really appreciate your help

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

It is unlikely that V_QR 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.

 

[ehild]

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

Show your work, please.

ehild

 

[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 Q_total= 15uC

I got V_@3uF = 5V and V_@1uF//2uF = V_QR=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, V_R=V_@2Ω??

 

[ehild]

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

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.

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

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

 

[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+q_1uF+q_2uF=0 ...(1)
From Q=CV
-3(10-U)+U+2(U-6)=0
U = 7 and U=V_{Q and ref. node}
Thus,, V_PQ = V_P -V_Q = 10-7 = 3V

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

Thanksss

 

[ehild]

I do not know the name of the method. It is just applying the definitions and logic. And it works .