Chapter 19, Problem 10Q

To determine

When the capacitors will store more energy?

Answer to Problem 10Q

Solution:

The capacitor will store more energy when they are connected in parallel connection.:

Explanation of Solution

Given:

Three capacitors of capacitance, $C$

Formula used:

The energy stored by the capacitor is given as

       $E=\frac{1}{2}C{V}^2$

Here,

       $E$ is energy

       $C$ is total capacitance of the circuit

       $V$ is the voltage

Calculation:

The energy stored by the capacitor is given as

       $E=\frac{1}{2}C{V}^2$

Since, in both series and parallel connection the voltage across the capacitor is same. So, the only deciding parameter is capacitance$C$, the connection in which the capacitance is maximum will store more energy.

In series, connection the overall capacitance of the circuit is calculated as

       $\begin{array}{c}\frac{1}{C_{\text{overall}}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}\\ C_{\text{overall}}=0.33C\end{array}$………… (1)

In parallel connection the overall capacitance of the circuit is calculated as

       $\begin{array}{c}{C}_{\text{overall}}=C+C+C\\ =3C\end{array}$………… (2)

Therefore it’s clear from Eq. (1)and Eq. (2)that the capacitor will store more energy when they are connected in parallel connection.