A capacitor of 2 uF is charged as shown in the
diagram. When the switch S is turned to position
2, the percentage of its stored energy dissipated is
(1) 80%
(3) 20%
(2) 0%
(4) 75%
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Energy stored in parallel plate capacitor
Consider a parallel plate capacitor is charged to potential (V), then small amount of work done to add another charge (dq) is
This work done is stored as energy stored in parallel plate capacitor.
...(1)
According to question,
We have given,
Substitute value of C in (1)
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Now,
Capacitor :
It is a device which is used to store electrical charge and release it at once.
(q = 2 × 10^-6 × V)
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Now, to find the final energy stored in the capacitor. Substitute value of V = V/5 Volt and C = 10 × 10^-6 = (2+8) 10^-6 in equation (1)
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Now, loss of energy in percentage =
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