First slide

Capacitance

Question

Five capacitors 10 $μ$ F capacity each are connected to a D.C. potential of 100 volts as shown in fig. The equivalent capacitance between the points A and B will be equal to

Moderate

A

$40 μF$
B

$20 μF$
C

$30 μF$
D

$10 μF$

Solution

The given circuit is equivalent to Wheat stone's bridge. when the bridge is balanced, the upper two condensers between A and B will be in series. Their resultant capacitance would be
$\frac{1}{C^{'}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$
or $C^{'} = 5 μF$
Similarly, the lower two condensers between A and B are also in series. Their resultant capacitance would be
$\frac{1}{C^{''}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$
or $C^{''} = 5 μF$
The capacitor C' and C'' will then be in parallel. Hence the effective capacitance between A and B will be $C = C^{'} + C^{''} = 5 + 5 = 10 μF$

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