Finding unknown capacitor in complex network

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Jahnavi
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Homework Statement


Circuit diagram of a capacitor network between terminals A and B, with unknown capacitor C in series at A, and branches labelled 2 μF, 6 μF, 4 μF, 12 μF, 2 μF, 1 μF, 2 μF and 8 μF.


Homework Equations

The Attempt at a Solution



First I calculated equivalent capacitance between R and M . The two 2μF capacitors are in series . Their combination is in parallel with the 1μF capacitor . Their equivalent capacitance is 2μF .

Now I found equivalent capacitance between R and T .The previously calculated 2μF between R and M is in series with 8μF capacitor .The equivalent capacitance between R and T is 8/5 μF.

Now I found equivalent capacitance between P and Q . Using nodal analysis I found it to be 44/3 μF .

The net capacitance between P and Q i.e 44/3 μF and net capacitance between R and T i.e 8/5 μF are in parallel . So, net capacitance between P and T is 244/15 μF .

Across A and B this 244/15 μF is in series with the unknown capacitor C . Their net capacitance has to be 3μF . Calculations give me value of C =732/199 μF . This is incorrect .

What is the mistake ?
 
Attachments
  • Circuit diagram of a capacitor network between terminals A and B, with unknown capacitor C in series at A, and branches labelled 2 μF, 6 μF, 4 μF, 12 μF, 2 μF, 1 μF, 2 μF and 8 μF.
    capacitors.jpg
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Jahnavi said:
Now I found equivalent capacitance between P and Q . Using nodal analysis I found it to be 44/3 μF .
You can use delta-star transformation for that. It's quicker.
Diagram titled Start Delta Transformation for Capacitors: left, a star network of capacitors C_A, C_B, C_C meeting at a central node, double-headed arrows between it and a triangle network with C_CA, C_BC, C_AB; below, formulas for C_A, C_B, C_C and C_AB, C_BC, C_CA as ratios of products and sums of capacitances.
 
Attachments
  • Diagram titled Start Delta Transformation for Capacitors: left, a star network of capacitors C_A, C_B, C_C meeting at a central node, double-headed arrows between it and a triangle network with C_CA, C_BC, C_AB; below, formulas for C_A, C_B, C_C and C_AB, C_BC, C_CA as ratios of products and sums of capacitances.
    download.png
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Jahnavi said:

Homework Statement


View attachment 221625

Homework Equations

The Attempt at a Solution



First I calculated equivalent capacitance between R and M . The two 2μF capacitors are in series . Their combination is in parallel with the 1μF capacitor . Their equivalent capacitance is 2μF .

Now I found equivalent capacitance between R and T .The previously calculated 2μF between R and M is in series with 8μF capacitor .The equivalent capacitance between R and T is 8/5 μF.

Now I found equivalent capacitance between P and Q . Using nodal analysis I found it to be 44/3 μF .

The net capacitance between P and Q i.e 44/3 μF and net capacitance between R and T i.e 8/5 μF are in parallel . So, net capacitance between P and T is 244/15 μF .

Across A and B this 244/15 μF is in series with the unknown capacitor C . Their net capacitance has to be 3μF . Calculations give me value of C =732/199 μF . This is incorrect .

What is the mistake ?
Your solution is correct. Try to give C = 3.68 μF.
The capacitance between P and Q is easy to calculate with parallel and series capacitors. The 2μF and 6μF capacitors are parallel, connected in series with the 4 μF capacitor, and the whole in parallel with the 12 μF capacitor.
 
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ehild said:
The capacitance between P and Q is easy to calculate with parallel and series capacitors. The 2μF and 6μF capacitors are parallel, connected in series with the 4 μF capacitor, and the whole in parallel with the 12 μF capacitor

Wow ! This is so nice .

I couldn't make out that it could be simplified in this way .

For me , the difficult part in this question was finding capacitance between P and Q . You made it look so easy :smile:
 
ehild said:
Your solution is correct. Try to give C = 3.68 μF.

This is a previous year exam question . But the circuit in the question is actually wrong . This is why I was not getting the correct answer 48 μF

The original question is given below .
Circuit diagram of a capacitor network between terminals A and B, with capacitors labeled 2μF, 6μF, 4μF, 12μF, 2μF, 1μF, 2μF and 8μF, and an unlabeled capacitor C at the top.
As you can see there is an extra wire in the picture in the OP :oldgrumpy: .
 
Attachments
  • Circuit diagram of a capacitor network between terminals A and B, with capacitors labeled 2μF, 6μF, 4μF, 12μF, 2μF, 1μF, 2μF and 8μF, and an unlabeled capacitor C at the top.
    capacitors.png
    16.4 KB · Views: 1,050