Question

A $2\mu \text{F}$ capacitor is charged to a potential difference of $10\text{V}$. Another $4\mu \text{F}$ capacitor is charged to a potential difference of $20\text{V}$. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of another. What is the amount of heat dissipated in the circuit ?

• A. $600\mu \text{J}$
• B. $300\mu \text{J}$
• C. $450\mu \text{J}$
• D. $900\mu \text{J}$

Answer 1

The correct option is A $600\mu \text{J}$

Before connection:


Initial charges on capacitors are,

$Q_1=C_1{V}_1=20\mu \text{C}$

$Q_2=C_2{V}_2=80\mu \text{C}$

When opposite polarity plates are connected, each plate will have a charge
$Q=80-20=60\mu \text{C}\text{(Charge conservartaion)}$


Let the potential difference be $V^{\prime }$ across plates after connection completes.

Thus, the equivalent capacitor has $60\mu \text{C}$ charge on it.

$\Rightarrow 60\mu \text{C}=C_{eq}\times V^{\prime }$

Since, capacitors make parallel combination

$\Rightarrow 60\mu \text{C}=(4+2)\mu \text{F}\times V^{\prime }$

$\Rightarrow V^{{}^{\prime }}=10\text{V}$

Initial potential energy of system,

$U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$

$\Rightarrow U_i=(\frac{1}{2}\times 2\times 100)+(\frac{1}{2}\times 4\times 400)$

$\Rightarrow U_i=900\mu \text{J}$

Final potential energy stored in the system is,

$U_f=\frac{1}{2}{C}_{eq}(V^{\prime }{)}^2$

$\Rightarrow U_f=\frac{1}{2}\times 6\times 100=300\mu \text{J}$

Heat dissipated in circuit:

$H=U_i-U_f$

$\Rightarrow H=900-300=600\mu \text{J}$

Hence, $\left(b\right)$ is correct.

$$\begin{array}{c}\text{Why this question?}\\ \text{Tip:In such problems the system after connection}\\ \text{can be visualized as an equivalent capacitor}\\ \text{with +ve plate indicating where net charge is}\\ \text{coming to be positive.}\end{array}$$