If the charge on a capacitor is increased by 4C the energy stored in it increases by 44%. The original charge on the capacitor is?
10C
20C
5C
15C
A
10C
B
15C
C
20C
D
5C
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Solution
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The energy stored in the capacitor is given by U=q22C ⇒U∝q2 ∴U1U2=(q2q1)2 U2U1−1=(q2q1)2−1 ⇒U2−U1U1=q22q21−1 ∵U2−U1U1×100=(q22q21−1)100 ⇒44=(q22q21−1)×100 ⇒q22q21=0.44+1=1.44 ∴q2=1.2q1 Now, q2−q1=4C ∴0.2q1=4 ∴q1=40.2=20C.
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