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# The net capacitance in the circuit above is

### Question

The net capacitance in the circuit above is

A)
80µF
B)
6.0µF
C)
4.0µF
D)
2.0µF

### 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

## 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