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The net capacitance in the circuit above is


Question

The net capacitance in the circuit above is

Options

A)
80µF
B)
6.0µF
C)
4.0µF
D)
2.0µF

The correct answer is D.

Explanation:

For capacitance i n parallel, 2µF and 2µF are in parallel,
their equivalence is 2µF and 2µF = 4µF

The 4µF generated is now in series with the remaining 4µF.

The net capacitance for series connection is

\(\frac{1}{C}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) = 1 + \(\frac{1}{4}\) = \(\frac{2}{4}\)

C = \(\frac{4}{2}\)

= 2µF


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Dicussion (1)

  • For capacitance i n parallel, 2µF and 2µF are in parallel,
    their equivalence is 2µF and 2µF = 4µF

    The 4µF generated is now in series with the remaining 4µF.

    The net capacitance for series connection is

    \(\frac{1}{C}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) = 1 + \(\frac{1}{4}\) = \(\frac{2}{4}\)

    C = \(\frac{4}{2}\)

    = 2µF

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