Question

A $2\mu F$ capacitor is charged to a potential = 10 V. Another $4\mu F$ capacitor is chargd to a potential = 20V. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. What heat is evolved in the circuit?

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

Answer 1

Answer: (B)

$600\mu J$

$Given:$ $C_1=2\mu F$ ; $C_2=4\mu F$ ; $V_1=10V$ ; $V_2=20V$

$Solution:$ As we know that,

$\text{By energy Conservation}$

$\frac{1}{2}{c}_1{v}_1^2+\frac{1}{2}{c}_2{v}_2^2$ $=\frac{1}{2}{c}_1{v}^2+\frac{1}{2}{c}_2{v}^2+H$

$\frac{1}{2}2(10{)}^2+\frac{1}{2}4(20{)}^2$ $=\frac{1}{2}(2+4)(10{)}^2+H$

$100+800=300+H$

$H=900-300$

$=600\mu J$

$So,the$ $correct$ $option:B$