Question

Two parallel plate capacitors $\text{A}$ and $\text{B}$ having capacitance $2\mu \text{F}$ and $4\mu \text{F}$ are charged seprately upto $20\text{V}$ and $10\text{V}$ respectively. Now the positive plate of $\text{A}$ is connected to the positive plate of $\text{B}$. Similarly negative plate of $\text{A}$ is connected with negative plate of $\text{B}$. Find how much heat is produced in the circuit?

• A. $\frac{200}{3}\mu \text{J}$
• B. $\frac{400}{3}\mu \text{J}$
• C. $\frac{1600}{3}\mu \text{J}$
• D. $600\mu \text{J}$

Answer 1

The correct option is A $\frac{200}{3}\mu \text{J}$

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

$\text{Before connection:}$

Energy stored in the first capacitor,

$U_1=\frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}\times \left(2\right)\times {20}^2=400\mu \text{J}$

Energy stored in the second capacitor,

$U_2=\frac{1}{2}{C}_2{V}_2^2=\frac{1}{2}\times \left(4\right)\times {10}^2=200\mu \text{J}$

So, total stored energy,

$U_i=U_1+U_2=400+200=600\mu \text{J}$

$\text{After connection:}$

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

$Q_1^{\prime }=2V$ and $Q_2^{\prime }=4V$

Using principle of conservation of charge,

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

$\Rightarrow 2\times 20+4\times 10=2V+4V$

$\Rightarrow V=\frac{80}{6}=\frac{40}{3}\text{V}$


Now, the final stored energy,

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

$\Rightarrow U_f=\frac{1}{2}\times \left(6\right)\times {\left(\frac{40}{3}\right)}^2$

$\Rightarrow U_f=\frac{1600}{3}\mu \text{J}$

So, the heat produced

$=U_i-U_f$

$=600-\frac{1600}{3}$

$=\frac{200}{3}\mu \text{J}$

Hence, option (a) is the correct answer.