What is the potential at the top of R2 in this circuit?

AI Thread Summary
The discussion focuses on calculating the charge on a capacitor in a circuit with resistors, a capacitor, and a battery. The key point is that at steady state, the capacitor current is zero, which implies that the current through resistor R2 is also zero. This leads to the conclusion that the potential at the top of the capacitor is the same as the potential at the top of resistor R3. The final relationship established is that the charge on the capacitor can be expressed as Q = C * I3R3, where I3 is the current flowing through R3. Understanding these relationships is crucial for solving the problem effectively.
marchcha
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Homework Statement


A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 41 Ω, R3 = 68 Ω and R4 = 154 Ω. The capacitance is C = 49 μF and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.
h11_RC_time.png

What is Q(∞), the charge on the capacitor after the switch has been closed for a very long time?

Homework Equations


Kirchhoff's Laws

The Attempt at a Solution


I attempted to find Q(inf) by setting up kirchhoffs voltage and loop equations as shown:
1. I1 = I4
2. 14 = I2 + I3
3. V - I1R1 - I3R3 - I4R4 = 0
4. V - I1R1 - I2R2 - (Q/C) - I4R4 = 0

I used the equalities in equation 1 to solve equation 3 in terms of I3. I then used I3 from equation 3 and plugged it into equation 2 to get an equation for I2. I then plugged in what I got for I2 into equation 4 as well as plugging in I4 for I1. The final equation I got for the value of Q is: Q = C(V - I4R4 - (I4 - (V - I4(R1 + R4))/R3) * R2 - I4R4). I had already found the value for I4 in a previous problem using V = IR when the switch is closed for a very long time. I4's value is 0.54A.

Thank you so much in advance!
 
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Hi marchcha,

You may be putting too much effort into this one :smile:

You're looking for the charge on the capacitor at steady state, Q(inf). What do you know about the value of the capacitor current ##I_c## at steady state? What does that tell you about the loop current for that loop?
 
gneill said:
Hi marchcha,

You may be putting too much effort into this one :smile:

You're looking for the charge on the capacitor at steady state, Q(inf). What do you know about the value of the capacitor current ##I_c## at steady state? What does that tell you about the loop current for that loop?
The capacitors current will be zero at steady state. I'm a little confused as to what this tells us about the loop current for that loop.
 
marchcha said:
The capacitors current will be zero at steady state. I'm a little confused as to what this tells us about the loop current for that loop.
Yes, ##I_c = 0## at steady state.

Is there any way to distinguish ##I_c## from the loop current? Aren't they identical? Just as the loop current for the first loop is identical to the currents flowing through the voltage source, R1, and R4.
 
gneill said:
Yes, ##I_c = 0## at steady state.

Is there any way to distinguish ##I_c## from the loop current? Aren't they identical? Just as the loop current for the first loop is identical to the currents flowing through the voltage source, R1, and R4.
So would I2 also be 0?
 
marchcha said:
So would I2 also be 0?
Yes. ##I2 = I_c = 0## at steady state.
 
gneill said:
Yes. ##I2 = I_c = 0## at steady state.
So then Q = C (V - I1R1 - I4R4)?
 
marchcha said:
So then Q = C (V - I1R1 - I4R4)?
Sure. Or Q = C*I3R3, to use your notation. You really don't need to invent so many different currents if there's just one loop current flowing...
 
gneill said:
Sure. Or Q = C*I3R3, to use your notation. You really don't need to invent so many different currents if there's just one loop current flowing...
Ok, then how do you find I3?
 
  • #10
marchcha said:
Ok, then how do you find I3?
Draw in the current flowing through loop 1. It's the only current flowing at steady state.

upload_2017-2-16_16-50-58.png
 
  • #11
gneill said:
Draw in the current flowing through loop 1. It's the only current flowing at steady state.

View attachment 113321
Ok so I3 is .0456A I don't understand how the capacitors charge has anything to do with the first loop though?
 
  • #12
marchcha said:
Ok so I3 is .0456A I don't understand how the capacitors charge has anything to do with the first loop though?
If ##I_c = 0##, what's the current through R2? Hence, what's the potential drop across R2?
 
  • #13
gneill said:
If ##I_c = 0##, what's the current through R2? Hence, what's the potential drop across R2?
Zero for both.
 
  • #14
marchcha said:
Zero for both.
So if the potential drop across R2 is zero, what's the difference between the potential at the top of the capacitor and the potential at the top of R3?
 
  • #15
gneill said:
So if the potential drop across R2 is zero, what's the difference between the potential at the top of the capacitor and the potential at the top of R3?
Would it be just V?
 
  • #16
marchcha said:
Would it be just V?
er, no. If the potential at the top of R3 is, say V3, and you "walk" from there across R2 which you say has a potential drop of zero, what's the potential at the other end of R2 (also the top of the capacitor)?
 
  • #17
gneill said:
er, no. If the potential at the top of R3 is, say V3, and you "walk" from there across R2 which you say has a potential drop of zero, what's the potential at the other end of R2 (also the top of the capacitor)?
Ah so they'd be the same and you can say Q/C = I3R3 the equation above. Thanks!
 
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