# Equivalent capacitance of capacitors

• Jahnavi
In summary: The problem is that this wire is connecting the 1μF capacitor to ground instead of R .This is causing the net capacitance to be wrong .In summary, the net capacitance between P and Q is 244/15 μF when it should be 3μF.
Jahnavi

## 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 ?

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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.

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Jahnavi said:

## Homework Statement

View attachment 221625

## 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.

Jahnavi
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

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 .
As you can see there is an extra wire in the picture in the OP .

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• capacitors.png
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## What is the concept of equivalent capacitance?

The concept of equivalent capacitance refers to the total capacitance of a circuit when multiple capacitors are connected together. It is a measure of the combined effect of all the individual capacitors in the circuit.

## How is equivalent capacitance calculated?

Equivalent capacitance can be calculated by using the formula C_eq = C_1 + C_2 + ... + C_n, where C_eq is the equivalent capacitance and C_1, C_2, etc. are the individual capacitances of the capacitors in the circuit. This formula is applicable for capacitors connected in parallel.

## What is the equivalent capacitance for capacitors connected in series?

For capacitors connected in series, the equivalent capacitance can be calculated using the formula 1/C_eq = 1/C_1 + 1/C_2 + ... + 1/C_n. This is because the effective capacitance decreases as capacitors are connected in series.

## How does the arrangement of capacitors affect the equivalent capacitance?

The arrangement of capacitors, whether in parallel or in series, affects the equivalent capacitance. When capacitors are connected in parallel, the equivalent capacitance increases, while in series, it decreases. This is because the effective area of the plates and the distance between them are altered by the arrangement.

## What is the practical application of understanding equivalent capacitance?

Understanding equivalent capacitance is essential in designing and analyzing circuits, especially in electronics and electrical engineering. It allows engineers to determine the overall capacitance of a circuit and choose the appropriate capacitors for specific applications.

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