Question

Two capacitors with capacity C1 and C2 are charged to potential V1 and V2 respectively and then connected in parallel. Calculate the common potential across the combination, the charge on each capacitor, the electrostatic energy stored in the system and the change in the electrostatic energy from its initial value.

Answer 1

Two capacitors with capacity C1 and C2 are charged to potential V1 and V2 respectively and then connected in parallel. Calculate the common potential across the combination, the charge on each capacitor, the electrostatic energy stored in the system and the change in the electrostatic energy from its initial value.

Capacitors connected in parallel

Q1 = C1 * V1

Q2 = C2 * V2

Capacitors connected in parallel

Total capacitance = C1 + C2

Total Charge =  Q1 + Q2

Potential Difference =  Total Charge / Total Capacitance

=> Potential Difference = (Q1 + Q2) / (C1 + C2)

=> Potential Difference = (C1V1 + C2V2) / (C1 + C2)

Energy = (1/2)CV²

earlier energy =  (1/2) ( C1V1² + C2V2²)

in parallel

= (1/2) (C1 + C2) ((C1V1 + C2V2) / (C1 + C2))²

= (1/2) (C1V1 + C2V2)² / (C1 + C2)

Answer 2

Given:

Capacity of the first capacitor $\[{=C_1}\]$

Capacity of the second capacitor $\[ = {C_2}\]$

Potential of the first capacitor $\[ = {V_1}\]$

Potential of the second capacitor$\[ = {V_2}\]$

To Calculate: The common potential across the combination, the charge on each capacitor, the electrostatic energy stored in the system and the change in the electrostatic energy from its initial value.

Solution:

As the capacity of the two capacitors is $\[{C_1}\]$ and $\[{C_2}\]$ and the potential is $\[{V_1}\]$ and $\[{V_2}\]$.

Therefore, the charge on each capacitor can be given as:

$\[{Q_1} = {C_1}{V_1}\]$

$\[{Q_2} = {C_2}{V_2}\]$

Now, the total charge before parallel combination can be given as:

$\[{Q} = {C_1}{V_1} + {C_2}{V_2}\]$               ... (i)

In parallel combination, the potential remains common. Therefore, the total charge across the parallel combination is given as:

$\[Q = ({C_1} + {C_2})V\]$                  ... (ii)

From equations (i) and (ii), we have,

$\[{C_1}{V_1} + {C_2}{V_2} = ({C_1} + {C_2})V\]$

$\[V = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}\]$

Hence, the common potential across the combination is $\[V = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}\]$.

The charge on each capacitor can be given as:

$\[{Q_1} = {C_1}{V_1}\]$

$\[{Q_2} = {C_2}{V_2}\]$

The total electrostatic energy before the combination is given as:

$\[{E_1} = \frac{1}{2}{C_1}{V_1}^2 + \frac{1}{2}{C_2}{V_2}^2\]$                  ... (iii)

The total electrostatic energy after the combination is given as:

$\[{E_2} = \frac{1}{2}({C_1} + {C_2}){V^2}\]$                      

$\[{E_2} = \frac{{{{({C_1}{V_1} + {C_2}{V_2})}^2}}}{{2({C_1} + {C_2})}}\]$                         ... (iv)

Hence, the electrostatic energy stored in the system is $\[{E_2} = \frac{{{{({C_1}{V_1} + {C_2}{V_2})}^2}}}{{2({C_1} + {C_2})}}\]$.

The change in the electrostatic energy from its initial value can be given as:

$\[\begin{gathered} E' = {E_1} - {E_2} \hfill \\ E' = \frac{{{C_1}{C_2}{{({V_1} - {V_2})}^2}}}{{2({C_1} + {C_2})}} \hfill \\ \end{gathered} \]$

Hence, the change in the electrostatic energy from its initial value is $\[E' = \frac{{{C_1}{C_2}{{({V_1} - {V_2})}^2}}}{{2({C_1} + {C_2})}}\]$.

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