Question

A capacitor of capacitance $10\mu F$ is charged up to a potential difference of 2V and then the cell is removed. Now it is connected to a cell of emf 4 V and is charged fully. In both cases the polarities of the two cells are in same directions. Total heat produced in the complete charging process is

Answer 1

Step 1: Finding initial and final energy forced

Let $u_{i}and{u}_f$ be initial energy stored in capacitance.

$\therefore u_i=\frac{1}{2}C{V}_1^2$
$=\frac{1}{2}\times 10\mu F\times (2V{)}^2$
$=20\mu J$

Also,

$u_f=\frac{1}{2}\times C\times V_2^2$
$=\frac{1}{2}\times 10\mu F\times (4V{)}^2$
$=80\mu J$
$\Delta U=u_f-u_i=60\mu J$

Step 2: Work done by cell

Let $q_{1}and{q}_2$ be charge on capacitor when 2V and 4V battery is connected respectively

$q_1=CV$
$=10\mu F\times 2V$
$=20\mu C$
$q_2=CV$
$=10\mu F\times 4V$
$=40\mu C$
Charge flow through final cell of 4 V, $\Delta q_2=20\mu c$ from +ve terminal of cell to +ve plate

Work done by cell, $W_{cell}=\Delta q\times V=$
$20\times 4=80\mu J$

4 energy draw form the cell $=20\mu c\times 4V=80\mu J$

Step 3: Heat produced

We have the relation
$W_{cell}=\Delta U+Heat$

4 Heat produced $=W_{cell}-\Delta U$
$=(80-60)\mu J$
$=20\mu J$

Hence correct option is (B).