Reducing Equivalent Resistors and Capacitors

In summary, the conversation is about reducing a circuit with resistors and capacitors to its simplest form. The person asking for help has provided a diagram of their attempt at reducing the circuit and is asking for confirmation if it is correct or if there is a simpler way to reduce it. They also ask for help in calculating the impedance of a circle in the circuit.
  • #1
fatcat39
40
0

Homework Statement



Reduce the circuit as much as possible. All resistors have R and all capacitors have same C.

Homework Equations





The Attempt at a Solution



http://photos-a.ak.facebook.com/photos-ak-sctm/v183/24/75/1238100168/n1238100168_30112052_7338.jpg



And this is my reduced one (I only reduced two capacitors in series and two parallel resistors)

http://photos-b.ak.facebook.com/photos-ak-sctm/v183/24/75/1238100168/n1238100168_30112053_7721.jpg

The reduced capacitors (the ones in the right corner of that little square) =

1/C_eq = 1/C + 1/C = 2/C
C_eq = .5C

1/R_eq = 1/R + 1/R = 2/R
R_eq = .5R


(Ignore the random squiggly at the bottom)

Anyway - that's as far as I got. I'm not sure if the two capacitors in the center are considered parallel? And if so, is that other capacitor on the far left considered in parallel with them too? (even though there are resistors between them?)
 
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  • #2
Can you attach the images to your post? I really don't feel like joining facebook just to look at your problem. :smile:
 
  • #3
oh, sorry - they were supposed to show up!
 
  • #4
Yeah, the two caps in the middle are in parallel, but the other one you are asking about would not be. Think of parallel as whether or not the wires are connected at both ends. Think of series as whether or not two elements share a common connection. (A fancy name for connection is a node.)
 
  • #5
Is this as far as it can be reduced?

http://photos-c.ak.facebook.com/photos-ak-sf2p/v164/24/75/1238100168/n1238100168_30112286_5295.jpg

Hmm...is this it? It seems like something simpler could be achieved somehow...*sigh*
 
Last edited by a moderator:
  • #6
Hello,
Can anyone give me a hint how to calculate the impedance of the following circle ?
Thank you
 

Attachments

  • impedance.png
    impedance.png
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Related to Reducing Equivalent Resistors and Capacitors

1. How do I calculate the equivalent resistance of resistors in series?

To calculate the equivalent resistance of resistors in series, simply add up the individual resistances. This is because in a series circuit, the current passing through each resistor is the same, and the total voltage drop across the circuit is equal to the sum of the voltage drops across each resistor.

2. What is the formula for the equivalent capacitance of capacitors in parallel?

The formula for calculating the equivalent capacitance of capacitors in parallel is Ceq = C1 + C2 + C3 + ... , where C1, C2, C3, etc. are the individual capacitances. This is because in a parallel circuit, the voltage across each capacitor is the same, and the total charge stored in the circuit is equal to the sum of the charges stored in each capacitor.

3. Can resistors be combined in both series and parallel circuits?

Yes, resistors can be combined in both series and parallel circuits. In a combination circuit, the equivalent resistance is calculated by first combining resistors in series and then combining the resulting equivalent resistors in parallel.

4. How does adding a capacitor in parallel affect the equivalent capacitance?

When a capacitor is added in parallel to an existing circuit, the equivalent capacitance increases. This is because the total charge stored in the circuit increases, while the voltage across each capacitor remains the same. The formula for calculating the equivalent capacitance in this case is Ceq = C1 + C2, where C1 is the original capacitance and C2 is the capacitance of the added capacitor.

5. Why is it important to reduce equivalent resistors and capacitors in a circuit?

Reducing equivalent resistors and capacitors in a circuit is important because it simplifies the circuit and makes it easier to analyze and understand. Additionally, reducing equivalent components can also help in calculating the total power dissipated in the circuit and can lead to more efficient circuit designs.

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