Question

The capacitors shown in the figure are in steady state. If at $t=0$ switch $S$ is closed, then find the heat generated in the circuit.

• A. $4200\mu \text{J}$
• B. $1200\mu \text{J}$
• C. $1800\mu \text{J}$
• D. $2100\mu \text{J}$

Answer 1

The correct option is D $2100\mu \text{J}$

Let $V$ be the potential at point $\text{P}$ as shown in the figure.


Applying the law conservation of charges on the given capacitor network we get,

$4(V-30)+3(V-30)+5(V-0)+9(V-0)=0$

$\Rightarrow 21V-120-90=0$

$\Rightarrow V=10\text{volts}$


Therefore,
Charge of $9\mu \text{F}$ capacitor is $Q_9=9\mu \times 10=90\mu \text{C}$

Charge of $5\mu \text{F}$ capacitor is $Q_5=5\mu \times 10=50\mu \text{C}$

Charge of $4\mu \text{F}$ capacitor is $Q_4=4\mu \times (30-10)=80\mu \text{C}$

Charge of $3\mu \text{F}$ capacitor is $Q_3=3\mu \times (30-10)=60\mu \text{C}$

Thus, charge flow in the circuit $Q=Q_9+Q_5=Q_3+Q_4=140\mu \text{C}$

Workdone by the battery $W=Q\times 30=140\mu \times 30=4200\mu \text{J}$

Energy stored in $4\mu \text{F}$ capacitor $E_1=\frac{1}{2}\times 4\mu \times (20{)}^2=800\mu \text{J}$

Energy stored in $3\mu \text{F}$ capacitor $E_2=\frac{1}{2}\times 3\mu \times (20{)}^2=600\mu \text{J}$

Energy stored in $9\mu \text{F}$ capacitor $E_3=\frac{1}{2}\times 9\mu \times (10{)}^2=450\mu \text{J}$

Energy stored in $5\mu \text{F}$ capacitor $E_4=\frac{1}{2}\times 5\mu \times (10{)}^2=250\mu \text{J}$

So, $H=W-(E_1+E_2+E_3+E_4)$

$\Rightarrow H=4200-(800+600+450+250)\mu \text{J}$

​​​​​​​$\Rightarrow H=2100\mu \text{J}$


Hence, option (b) is the correct answer.