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Sum of capacitors connected in Series and in Parallel

  • Thread starter kakadas
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  • #1
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Homework Statement


2z3xi04.jpg


Find the sum value of capacitors.

Homework Equations





The Attempt at a Solution



I tried to redraw the scheme like in the picture below, but my teacher told me it's not correctly redrawn. I hope You'll prove to me that my redrawn scheme is correct... If not I'm more than happy to know the both: correctly redrawn scheme and the right answer.

dwyl1g.jpg
 

Answers and Replies

  • #2
vela
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Looks fine to me the way you redrew it.
 
  • #3
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That looks correct to me.
Anyways, why did you spend time redrawing this when its clear which one are in series and parallel? :)
 
  • #4
CWatters
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Hint. Start at the right hand side of the original drawing. Simplify the three caps in series then see what you have.
 
  • #5
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looks correct.
 
  • #6
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Yep it's correct.
I like moving from the far end of the circuit to the source (from left to right now). It's easier for me.
 
  • #7
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That looks correct to me.
Anyways, why did you spend time redrawing this when its clear which one are in series and parallel? :)
Because I didn't know how those 4 parallel capacitors' sum will look like in the drawing...

Say I use the old drawing:

2z3xi04.jpg


Could Anyone tell me How will it look like after summing those 4 parallel capacitors?

This?:

zkpxd1.jpg



OR maybe This?:

2isbeoo.jpg


oh, and Please explain Your choice.
 
  • #8
vela
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Neither is correct as the circuit doesn't have four capacitors in parallel.
 
  • #9
vela
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If two elements are in parallel, they have to be connected to the same two nodes in the circuit. In the original circuit, there are six nodes: the four dots and the two corners on the right. You can see that no two elements are connected to the same two nodes. None of the elements is in parallel with another.

If two elements are in series, all of the current through one element has to flow through the other element. For this to happen, the two elements have to be connected end to end, and no other element can be connected to the node that connects the two elements because it will siphon off some of the current. In the original circuit, only the three capacitors on the right are in series. (Can you see that?) Using the formulas you have for combining capacitors, you'll find you can replace those three with an equivalent capacitance of 1/3 μF.

Now redraw the circuit with those three capacitors replaced by a single 1/3 μF capacitor. Are any capacitors in the new circuit in parallel or in series?
 
  • #10
PeterO
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If two elements are in parallel, they have to be connected to the same two nodes in the circuit. In the original circuit, there are six nodes: the four dots and the two corners on the right. You can see that no two elements are connected to the same two nodes. None of the elements is in parallel with another.

If two elements are in series, all of the current through one element has to flow through the other element. For this to happen, the two elements have to be connected end to end, and no other element can be connected to the node that connects the two elements because it will siphon off some of the current. In the original circuit, only the three capacitors on the right are in series. (Can you see that?) Using the formulas you have for combining capacitors, you'll find you can replace those three with an equivalent capacitance of 1/3 μF.

Now redraw the circuit with those three capacitors replaced by a single 1/3 μF capacitor. Are any capacitors in the new circuit in parallel or in series?
Actually C1 = 2 μF so that total should be 2/3 μF
 
  • #11
vela
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Yes, of course. I mixed up the two values. Thanks for catching that.
 

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