Question

Two capacitors $\text{A}$ and $\text{B}$ with capacitances $15\mu \text{F}$ and $5\mu \text{F}$ are charged to potential difference of $5\text{V}$ and $9\text{V}$ respectively. Consider a wire connected to each cepacitor as shown (with free ends). An uncharged capacitor $5\mu \text{F}$ with lead wire falls on the free ends to complete the circuit. Find how much amount of heat produced in the circuit?

• A. $210\mu \text{J}$
• B. $180\mu \text{J}$
• C. $67.5\mu \text{J}$
• D. $390\mu \text{J}$

Answer 1

The correct option is A $210\mu \text{J}$

Initial charge on capacitors $\text{A}$ and $\text{B}$ are,

$Q_A=15\times 5=75\mu C$ and $Q_B=5\times 9=45\mu C$ respectively.

Let $q$ amount of charge flows through the circuit.


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

$\frac{75-q}{15}+\frac{45-q}{5}-\frac{q}{5}=0$

$\Rightarrow 75-q+135-3q-3q=0$

$\Rightarrow 210=7q$

$\therefore q=30\mu \text{C}$

Final charge on all three cepacitors as follows,

$Q{{}_A}^{{}^{\prime }}=75-q=75-30=45\mu \text{C}$

$Q{{}_B}^{{}^{\prime }}=45-q=45-30=15\mu \text{C}$

$Q{{}_C}^{{}^{\prime }}=30\mu \text{C}$

Heat produced $\left(H\right)=U_i-U_f$

Where,
Initial energy stored in the capacitors,

$U_i=U_A+U_B$

$\Rightarrow U_i=\frac{Q{{}_A}^2}{2C_A}+\frac{Q{{}_B}^2}{2C_B}$

$\Rightarrow U_i=\frac{(75{)}^2}{30}+\frac{(45{)}^2}{10}$

$\therefore U_i=187.5+202.5=390\mu \text{J}$

Final energy stored:

$U_f=U{{}_A}^{{}^{\prime }}+U{{}_B}^{{}^{\prime }}+U{{}_C}^{{}^{\prime }}$

$\Rightarrow U_f=\frac{(Q{{}_A}^1{)}^2}{2C_A}+\frac{(Q{{}_B}^1{)}^2}{2C_B}+\frac{(Q{{}_C}^1{)}^2}{2C_C}$

$\Rightarrow U_f=\frac{(45{)}^2}{30}+\frac{(15{)}^2}{10}+\frac{(30{)}^2}{10}$

$\therefore U_f=67.5+22.5+90=180\mu \text{J}$

So,

$H=U_i-U_f$
$=390-180$
$=210\mu \text{J}$

Hence, option (c) is correct answer.