Question

Two parallel plate capacitors $\text{A}$ and $\text{B}$ having capacitances $3\mu \text{F}$ and $15\mu \text{F}$ are charged seperately upto $10\text{V}$ and $20\text{V}$ respectively. Now positive plate of $\text{A}$ is connected to the negative plate of $\text{B}$ and vice-versa. Find how much heat produced in the circuit?

• A. $2025\mu \text{J}$
• B. $150\mu \text{J}$
• C. $3000\mu \text{J}$
• D. $1125\mu \text{J}$

Answer 1

The correct option is D $1125\mu \text{J}$

$\text{Heat produced=Initial energy stored}\left(U_i\right)-\text{Final energy stored}\left(U_f\right)$

$\text{Before connection:}$

Energy stored in the first capacitor,

$U_1=\frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}\times \left(3\right)\times {10}^2=150\mu \text{J}$

Energy stored in the second capacitor,

$U_2=\frac{1}{2}{C}_2{V}_2^2=\frac{1}{2}\times \left(15\right)\times {20}^2=3000\mu \text{J}$

So, total stored energy,

$U_i=U_1+U_2=150+3000=3150\mu \text{J}$

$\text{After connection:}$

Due to the charge distribution both the capacitors attain a common potential $V$ and let their charges become $Q_1^{\prime }$ and $Q_2^{\prime }$ respectively. So, using $Q=CV$, we get

$Q_1^{\prime }=3V$ and $Q_2^{\prime }=15V$

Using principle of conservation of charge,

$Q_1+Q_2=Q_1^{\prime }+Q_2^{\prime }$

$\Rightarrow -3\times 10+15\times 20=3V+15V$

$\Rightarrow V=\frac{270}{18}=15\text{V}$


Now, the final stored energy,

$U_f=\frac{1}{2}(C_1+C_2)V^2$

$\Rightarrow U_f=\frac{1}{2}(3+15)\times {15}^2$

$\Rightarrow U_f=2025\mu \text{J}$

So, the heat produced

$=U_i-U_f$

$=3150-2025$

$=1125\mu \text{J}$

Hence, option (d) is the correct answer.