Gujarati
Hindi
1. Electric Charges and Fields
normal

A $2\,\mu F$ capacitor is charged to a potential $=10\ V$ . Another $4\,\mu F$ capacitor is charged to a potential $= 20\ V$ . 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 ?.........$\mu J$

A

$300$

B

$600$

C

$900$

D

$450$

Solution

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

Solution: As we know that,

By energy Conservation

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

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

$100+800=300+ H$

$H =900-300$

$=600 \mu J$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.