Question

What amount of heat will be generated in the circuit shown in figure, after the switch '$\text{S}$' is shifted from position '$1$' to position '$2$' $?$

• A.
  $\frac{1}{2}C{\epsilon }_2^2$
• B.
  $\frac{1}{2}C[{\epsilon }_1^2-{\epsilon }_2^2]$
• C.
  $\frac{1}{2}C[{\epsilon }_1^2+{\epsilon }_2^2]$
• D.
  $\frac{1}{2}C{\epsilon }_2{\epsilon }_1^2$

Answer 1

The correct option is A

$\frac{1}{2}C{\epsilon }_2^2$
When the switch is at position '$1$', the capacitor and two batteries will form a closed loop.

$\text{Let}{\epsilon }_1>{\epsilon }_2$


Applying $\text{KVL}$ in closed loop.

$\Rightarrow +{\epsilon }_1-{\epsilon }_2-\frac{q}{C}=0$

$\Rightarrow q=({\epsilon }_1-{\epsilon }_2)C...\left(1\right)$

Thus initially energy stored is,

$U_i=\frac{q^2}{2C}=\frac{1}{2}C({\epsilon }_1-{\epsilon }_2{)}^2$

When switch '$\text{S}$' is at position '$2$', the battery with $\text{EMF}$, ${\epsilon }_2$ gets short circuited and the equivalent circuit could be drawn as,



Let the battery ${\epsilon }_1$ supplied extra charge '$Q$' after switch moved to position '$2$'.

Thus,

$(Q+q)={\epsilon }_1\times C=C{\epsilon }_1$

$\Rightarrow {\epsilon }_1=\frac{Q+q}{C}$

$\Rightarrow {\epsilon }_1=\frac{Q+({\epsilon }_1-{\epsilon }_2)C}{C}$

$\Rightarrow {\epsilon }_{1}C=Q+{\epsilon }_{1}C-{\epsilon }_{2}C$

$\therefore Q=C{\epsilon }_2...\left(2\right)$

Final energy stored in capacitor.

$U_f=\frac{(Q+q{)}^2}{2C}$

Work done by battery ${}^{\prime }{\epsilon }_1^{\prime }$ in second case;

$\text{W}=Q_{supplied}\times {\epsilon }_1$

$\Rightarrow W=Q{\epsilon }_1$

$\Rightarrow W=\left(C{\epsilon }_2\right){\epsilon }_1=C{\epsilon }_1{\epsilon }_2$

From energy conservation, amount of heat generated

$H=W-(U_f-U_i)=W-\Delta U$

$\Rightarrow H=C{\epsilon }_1{\epsilon }_2-[\frac{(Q+q{)}^2}{2C}-\frac{q^2}{2C}]$

Subsitituting value of $Q$ & $q$ from $\left(1\right)$ & $\left(2\right)$;

$\Rightarrow H=C{\epsilon }_1{\epsilon }_2-[\frac{(C{\epsilon }_2+C{\epsilon }_1-C{\epsilon }_2{)}^2}{2C}-\frac{C^2({\epsilon }_1-{\epsilon }_2{)}^2}{2C}]$

$\Rightarrow H=\frac{1}{2}C{\epsilon }_2^2$

(expending & simplifying)

$\therefore H=\frac{1}{2}C{\epsilon }_2^2$

Hence, option (a) is correct answer.

Why this question? We can see that the heat dissipated is only depended on the battery which gets short circuited.