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1. Electric Charges and Fields
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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
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