A series combination of 10 capacitors, each of value ' C 1 ' is charged by a source of potential difference ' 4 V '. When another parallel combination of 8 capacitors, each of va

Question

A series combination of 10 capacitors, each of value ' C 1 ​ ' is charged by a source of potential difference ' 4 V '. When another parallel combination of 8 capacitors, each of value ' C 2 ​ ' is charged by a source of potential difference ' V '. The relation between C 1 ​ and C 2 ​ , so that the energies in the two combinations will be same is

TestHub · Accepted Answer

For the series combination of 10 capacitors, each of capacitance $C_{1}$ :

Equivalent capacitance, $C_{e q 1} = \frac{C _{1}}{10}$

Potential difference, $V_{1} = 4 V$

Energy stored, $U_{1} = \frac{1}{2} C_{e q 1} V_{1}^{2} = \frac{1}{2} (\frac{C _{1}}{10}) (4 V)^{2} = \frac{16 C _{1} V ^{2}}{20} = \frac{4}{5} C_{1} V^{2}$

For the parallel combination of 8 capacitors, each of capacitance $C_{2}$ :

Equivalent capacitance, $C_{e q 2} = 8 C_{2}$

Potential difference, $V_{2} = V$

Energy stored, $U_{2} = \frac{1}{2} C_{e q 2} V_{2}^{2} = \frac{1}{2} (8 C_{2}) V^{2} = 4 C_{2} V^{2}$

Given that the energies in the two combinations are the same ( $U_{1} = U_{2}$ ):

$\frac{4}{5} C_{1} V^{2} = 4 C_{2} V^{2}$

$\frac{4}{5} C_{1} = 4 C_{2}$

$C_{2} = \frac{C _{1}}{5}$

Answer: $C_{2} = C_{1} /5$

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