Question

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

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

Answer 1

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


Before connection , charges in each capacitor are given by $$Q=CV$$$Q_1=2\times 10=20\mu \text{C}$ and $Q_2=4\times 20=80\mu \text{C}$

And energy stored in capacitors before the connection is given by
$U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$
$U_i=\frac{1}{2}\times 2\times (10{)}^2+\frac{1}{2}\times 4\times (20{)}^2=900\mu \text{J}$

When the capacitors are connected,


Equivalent capacitance of circuit is ,$C_{eq}=2+4=6\mu \text{F}$

And the net charge of circuit is
$Q_{net}=-20+80=60\mu \text{C}$

Now the common voltage of capacitors is given by
$V=\frac{Q_{net}}{C_{eq}}$

$V=\frac{60}{2+4}=10\text{V}$

Energy stored in circuit is given by , $U_f=\frac{1}{2}{C}_{eq}{V}^2$

Substituting the data given in the question we get,

$U_f=\frac{1}{2}\times 6\times (10{)}^2=300\mu \text{J}$

Heat generated in circuit is
$=U_i-U_f=600\mu \text{J}$

Hece, option (b) is the correct answer.