Calculate the equivalent capacitance between the points A and B in the following Figure:

![a row of four capacitors 2, 3, 5 and 10 microfarad with bypass wires between A and B — Class 12 Physics electrostatics question](/figures/electrostatic-potential-and-capacitance/four-cap-r

Question

Calculate the equivalent capacitance between the points A and B in the following Figure:

![a row of four capacitors 2, 3, 5 and 10 microfarad with bypass wires between A and B — Class 12 Physics electrostatics question](/figures/electrostatic-potential-and-capacitance/four-cap-row-2-3-5-10uf-bypass-a-q-p-r-fig-pyq-93a160.svg)

Accepted Answer

>

> [!TLDR]
> The plain conducting wires short A to Q and P to R, collapsing C1, C2, C3 into one parallel combination (10 μF), which is then in series with C4 (10 μF), giving 5 μF overall.
>
> A plain connecting wire has no capacitance and no resistance, so any two nodes joined directly by such a wire are forced to the same potential — they become electrically one node.
>
> - The wire from A to Q makes A and Q the same node.
> - The wire from P to R makes P and R the same node.
>
> Now re-examine each capacitor's terminals using these identifications:
> - $C_1$ (2 μF) is between A and P.
> - $C_2$ (3 μF) is between P and Q — but Q = A, so $C_2$ is also between P and A.
> - $C_3$ (5 μF) is between Q and R — but Q = A and R = P, so $C_3$ is also between A and P.
>
> So $C_1$, $C_2$, and $C_3$ all sit between the *same* pair of nodes (A and P), i.e. they are in parallel: …

Q.Calculate the equivalent capacitance between the points A and B in the following Figure:

Concept understanding — Capacitors in Series and Parallel

Q.Calculate the equivalent capacitance between the points A and B in the following Figure:

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