Question

Initially, the switch $S$ is open in the circuit for a long time. After the switch is closed, the heat generated in the circuit will be:

• A. $600\mu \text{J}$
• B. $750\mu \text{J}$
• C. $350\mu \text{J}$
• D. $450\mu \text{J}$

Answer 1

The correct option is D $450\mu \text{J}$

When the switch is open:


When the switch $S$ is open, both the capacitors are in series. so, effective capacitance will be

$C_{eq}=\frac{6\times 3}{6+3}=\frac{18}{9}=2\mu \text{F}$

Net emf $=20+10=30\text{V}$

And charge on each capacitor will be,

$q=C_{eq}V$

$\Rightarrow q=2\times (20+10)=60\mu \text{C}$


After the switch is closed:

Let the higher emf battery dominates the circuit and redraw the circuit.


From above diagram, we can find the charge on each capacitor.

Therefore, charge on $6\mu \text{F}$ capacitor;

$q=20\times 6=120\mu \text{C}$

Thus, $20\text{V}$ battery has supplied a charge of
$\Delta q=120-60=60\mu \text{C}$

Therefore, workdone by $20\text{V}$ battery is

$W_1=60\times 20=1200\mu \text{J}(\because W=qV)$

Charge on $3\mu \text{F}$ capacitor after switch $S$ closes;

$q_2=CV=3\times 10=30\mu \text{C}$

thus, final charge $=+30\mu \text{C}$

Hence, charge supplied by $10\text{V}$ battery is $-30\mu \text{C}(=30-60)$.

So, workdone by battery will be

$W_2=-30\times 10=-300\mu \text{J}$

And we know that energy stored by capacitor will be

$U=\frac{Q^2}{2C}$

So, initially stored energy in capacitor is

$U_i=[\frac{(60{)}^2}{2\times 6}+\frac{(60{)}^2}{2\times 3}]=900\mu \text{J}$

And final energy stored in capacitors is

$U_f=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2=\frac{1}{2}\times 6\times (20{)}^2+\frac{1}{2}\times 3\times (10{)}^2$

$\Rightarrow U_f=1350\mu \text{J}$

From energy conversation,

$U_i+W_{battery}=U_f+H_{Lost}$

$\Rightarrow 900+(W_1+W_2)=1350+H_{Lost}$

$\Rightarrow 900+(1200-300)=1350+H_{Lost}$

$\therefore H_{Lost}=1800-1350=450\mu \text{J}$

Hence, option $\left(d\right)$ is correct.

Why this question? Tip: After the switch is closed, we observe that charge on $3\mu \text{F}$ capacitor reduced. If represents that $10\text{V}$ battery connected in parallel has done a negative work or work has been done on the battery.