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A Bridge Circuit Capacitor

  • Thread starter Kaoi
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  • #1
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A Bridge Circuit Capacitor [Solved]

EDIT: Problem has been solved.

Homework Statement


"The circuit has been connected as shown in the figure for a 'long' time.

(See attached diagram)

What is the magnitude of the electric potential across the capacitor? Answer in units of V."

Given:
[tex]R_{1} = 32 \Omega[/tex]
[tex]R_{2} = 1 \Omega[/tex]
[tex]R_{3} = 40 \Omega[/tex]
[tex]R_{4} = 26 \Omega[/tex]
[tex]C = 16 \mu F[/tex]
[tex]\Delta V = 27 V[/tex]

Unknown:
[tex]V_{C} = ?[/tex]

Homework Equations



[tex]C = \frac{Q}{V}[/tex]

The Attempt at a Solution



Oh, boy. Once again, if I could only grasp the conceptual parts of the problem, I could finish the mathematical parts with no problem.

I was stumped at the first look at the circuit-- I didn't even know what type of circuit it was until I Googled it.

My main question, though, is this: How, exactly, is current split across a bridge circuit? Is it parallel or series? How would one reduce it to an equivalent resistor?

The other question I didn't get was that it asks for the voltage of the capacitor, yet they give current instead of charge with no quantitative time specified with which to find the charge.

It's a shame, too, because I'm usually really good with circuit diagrams. :rolleyes:
 

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Answers and Replies

  • #2
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Alright. I don't know if you knew this already or not, but the key to looking at questions that are asking about what happens in this circuit after a long time is to see whether or not it is DC, which yours is. DC has great steady states because after a "long time" capacitors act like open circuits and inductors act like short circuits. You must have gotten this though, seeing has how you are asking about how to reduce the resistors.

Now you know that resistors are in series if they share the same node (i.e. the front end of one is connected to the back end of another). The resistors will be in parallel if the backs connect together and the fronts connect together.

If you want to find the current, you probably ought to only reduce series combos. That way, the current will stay the same through the elements, but the math will be easier (no info lost, and more time gained).

Now, if you want to find the steady-state solution for the capacitor, just go back to what I said in the beginning. The capacitor acts like an open circuit, so find the potential difference by figuring out the voltages at the opposing parts. If you know the current through the two paths, it should be pretty easy.

Give 'er another try. :)
 
  • #3
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So, if the capacitor acts like an open circuit, then it's a parallel circuit with two resistors in series on each branch, like the diagram I've attached? If that's true, then:

[tex]\Delta V = \Delta V_{13} = \Delta V_{24}[/tex]

[tex]R_{13} = 32 \Omega + 40 \Omega = 72 \Omega[/tex]

[tex]R_{24} = 1 \Omega + 27 \Omega = 27 \Omega[/tex]

[tex]\Delta V = IR[/tex]

[tex]I_{13} = \frac{\Delta V}{R} = \frac{27 V}{72 \Omega} = 0.375 A[/tex]

[tex]I_{24} = \frac{\Delta V}{R} = \frac{27 V}{27 \Omega} = 1 A[/tex]

Then, to figure out the voltage for each resistor:
[tex]\Delta V = IR[/tex]

[tex]\Delta V_{1} = I_{13}R_{1} = (0.375 A)(32 \Omega) = 12 V[/tex]

[tex]\Delta V_{2} = I_{24}R_{2} = (1 A)(1 \Omega) = 1 V[/tex]

[tex]\Delta V_{3} = I_{13}R_{3} = (0.375 A)(40 \Omega) = 15 V[/tex]

[tex]\Delta V_{4} = I_{24}R_{4} = (1 A)(26 \Omega) = 26 V[/tex]

You lost me at "the voltages at the opposing parts," though. I'm not quite sure what you mean by that.
 

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  • #4
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It might help to remember that you should really talk about potential difference across each resistor. This is the potential that is dropped as charge moves through the resistor.
The potential at the top of R1 and R3 is 27V assuming they are connected to the positive. Take the potential difference for R1 away from this value and you have the potential between R1 and R2. This is the point where one end of the capacitor is connected. Do the same for the other end and you can get the pd across C because you know the potential at each end.
 
  • #5
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I think I see what you're saying...

So:

Diff. between the original potential and the 32 [tex]\Omega[/tex] resistor is (27 - 12) = 15 V.
Diff. between the original potential and the 1 [tex]\Omega[/tex] resistor is (27 - 1) = 26 V.

So does that means the difference is (26 - 15) = 11 volts across the capacitor, or am I oversimplifying it?
 
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  • #6
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No I've not explained what I mean clearly enough.
Looking at the diagram you supplied it is possible to mark potentials at points(nodes) on the circuit.
Where the wire from the cell meets the wire that connects the 32ohm and 1 ohm resistor there is 0 volts potential. ie at this point each coulomb of charge has zero energy.
The 32ohm resistor has a 12 Volt potential difference across it. So at its right hand end the potential is 12 volts. ie at this point a coulomb of charge has 12 Joules of energy. This is the potential at the top of the capacitor because it also connects to this node.
The 1 ohm resistor has a 1 volt pd. So its right hand end has a potential of 1volt and so does the lower end of the capacitor.
You now have the potential at each end of the capacitor. Therefore you can work out the potential difference across it.
Ive attached your circuit with these potentials added. You should be able to see how potential differences you worked out fit with the potentials I have added.
 

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  • #7
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Oh, I marked the battery backwards. :rofl: I was thinking of electron flow instead of the "positive-negative" flow.

But the way I see it, for example, across the top branch:

-Charge leaves battery (positive terminal) with 27 J of potential energy to expend.
-Charge expends 12 J of energy across the 32 [tex]\Omega[/tex] resistor.
-Capacitor is here, with a potential of 15 J per charge (15 volts) remaining.
-Charge expends 15 J of energy across the 40 [tex]\Omega[/tex] resistor.
-Charge returns to battery (negative terminal), gains 27 J of charge traveling across the battery.

So the upper end of the capacitor has 15 volts potential difference remaining, and the lower end has 26 volts potential difference remaining after the travel across their respective resistors... or at least, that's what I think.

And that's where I got (26 - 15) = 11 from.

EDIT: 11 volts is the correct answer. Silly me, I entered it at the wrong number. So embarrassed.

Problem solved! :D
 
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  • #8
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Oh, I marked the battery backwards. :rofl: I was thinking of electron flow instead of the "positive-negative" flow.

But the way I see it, for example, across the top branch:

-Charge leaves battery (positive terminal) with 27 J of potential energy to expend.
-Charge expends 12 J of energy across the 32 [tex]\Omega[/tex] resistor.
-Capacitor is here, with a potential of 15 J per charge (15 volts) remaining.
-Charge expends 15 J of energy across the 40 [tex]\Omega[/tex] resistor.
-Charge returns to battery (negative terminal), gains 27 J of charge traveling across the battery.

So the upper end of the capacitor has 15 volts potential difference remaining, and the lower end has 26 volts potential difference remaining after the travel across their respective resistors... or at least, that's what I think.

And that's where I got (26 - 15) = 11 from.

EDIT: 11 volts is the correct answer. Silly me, I entered it at the wrong number. So embarrassed.

Problem solved! :D
If the battery is diconnected, how long does it take to discharge the capacitor?
 

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