Question

Three capacitors $C_1,C_2$ and $C_3$ of capacitance $4\mu \text{F}$, $3\mu \text{F}$ and $6\mu \text{F}$ are charged upto a potential difference of $5\text{V},6\text{V}$ and $4\text{V}$ respectively. If terminal $\text{a}$ is connected with $\text{f}$, terminal $\text{e}$ is connected with $\text{d}$ and $\text{b}$ is conncted with $\text{c}$, then find the amount of heat produced in the circuit?

• A. $20\mu \text{J}$
• B. $2\mu \text{J}$
• C. $152\mu \text{J}$
• D. $150\mu \text{J}$

Answer 1

The correct option is D $150\mu \text{J}$

Let initial charges on caapcitors $C_1,C_2$ and $C_3$ are $q_1,q_2$ and $q_3$ respectively.
$q_1=4\times 5=20\mu \text{C}$,
$q_2=6\times 3=18\mu \text{C}$,
$q_3=6\times 4=24\mu \text{C}$.


Let $q$ amount charge flows in the circuit,


Applying $\text{KVL}$, to the loop $\text{PQRS}$,

$\frac{24-q}{6}+\frac{18-q}{3}+\frac{20-q}{4}=0$

$\Rightarrow 48-2q+72-4q+600-3q=0$

$\Rightarrow 180=9q$

$\therefore q=20\mu \text{C}$

Now the charges on capcitors $C_1,C_2$ and $C_3$ respectively,

$q_1^{\prime }=20-20=0$

$q_2^{\prime }=|18-20|=2\mu \text{C}$

$q_3^{\prime }=24-20=4\mu \text{C}$

We know that,
Heat produced $\left(H\right)=U_i-U_f$

Where, initial stored energy, $U_i=U_1+U_2+U_3$

$\Rightarrow U_i=\frac{q_1^2}{2C_1}+\frac{q_2^2}{2C_2}+\frac{q_3^2}{2C_3}$

$\Rightarrow U_i=\frac{{20}^2}{2\times 4}+\frac{{18}^2}{2\times 3}+\frac{{24}^2}{2\times 6}$

$\Rightarrow U_i=50+54+48=152\mu \text{J}$

Final stored energy,
$U_f=U_1^{\prime }+U_2^{\prime }+U_3^{\prime }$

$\Rightarrow U_f=\frac{0^2}{2\times 4}+\frac{2^2}{2\times 3}+\frac{4^2}{2\times 6}$

$\Rightarrow U_f=2\mu \text{J}$

$\therefore H=U_i-U_f=152-2=150\mu \text{J}$

Hence, option (d) is correct option.