Question

A capacitor of capacitance C is charged to potential V
and another capacitor of capacitance 2C is charged to
potential 2V. Then they are joined across each other
with plates of same polarity together. The amount of
heat generated after connecting the two capacitors
together is​

Answer 1

Explanation:

As the two capacitors are connected with reverse polarity, So total charge

 Qt=Q2−Q1=(2C)(4V)−CV=7CV

After connecting them they are in parallel so total capacity is Ct=C+2C=3C

We know that after connecting the common potential is 

Vc=total capacitytotal charge=CtQt=3C7CV=(7/3)V

Heat produced H= energy loss=Ui−Uf

H=Ui−Uf=[21CV2+21(2C)(4V)2]−21CtVc2

=233CV2−2

Answer 2

Given:

Capacitance of capacitor 1 = C

Potential of capacitor 1 = V

Capacitance of capacitor 2 = 2C

Potential of capacitor 2 = 2V

To find:

Loss of heat after connecting two capacitors.

Solution:

As we known, the two capacitors are connected with same polarity. Therefore, total charge is

      $Q_t = Q_2 - Q_1$

here, $Q_2$ = charge on capacitor 2 = (2C)(2V)

         $Q_1$ = charge on capacitor 1 = CV

by putting all the values,

          $Q_t = (2C)(2V) - CV$

          $Q_t = 4CV - CV = 3CV$

Now, calculate the total capacitor, we get,

           $C_t = C_2 + C_1$

           $C_t = C + 2C$

           $C_t = 3C$

now, calculate the common potential as follows-

     $V_C = \frac{total charge}{total capacitor}$

     $V_C = \frac{Q_t}{C_t}$

     $V_C = \frac{3CV}{3C}$

     $V_C = V$

Heat loss will be calculated as follows,

Heat loss = $U_i - U_f$

                = $[\frac{1}{2}CV^2 + \frac{1}{2}(2C)(2V)^2] - [\frac{1}{2}C_tV_C^2]$

                = $[\frac{9}{2}CV^2] - [\frac{1}{2}(3C)(V)^2]$

                = $\frac{9}{2}CV^2 - \frac{3}{2}CV^2$

                = $\frac{6}{2}CV^2$

                = 3CV²

Therefore, heat loss is 3CV².