A 3-microfarad capacitor is connected in series with 6-microfarad capacitor. When a 300-volt potential difference is applied across this combination, the total energy stored in the two capacitor is
B. 0.18 J
C. 0.27 J
D. 0.41 J
E. 0.81 J
(GR0177 #10)
Solution:C in Series:
1/Cnet = 1/C1 + 1/C2 = ( 1/3 + 1/6 ) ( 1/10-6 ) = 1/2 ∙ 10-6
Cnet = 2 ∙ 10-6 FU = ½ QV = ½ CnetV2 = ½ (2 ∙ 10-6) (300)2 = 9 (10-6) (104) = 0.09 J
Answer: A
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