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QAIPMT-2010-Ph-2070: NEET 2010 Physics - Class 12 AIPMT-2010-Ph-2070 Solution: A series combination of \(n_1\) capacitors, each of value…
A series combination of capacitors, each of value , is charged by a source of potential difference 4V. When another parallel combination of capacitors, each of value , is charged by a source of potential difference V, it has the same (total) energy stored in it as the first combination has. The value of , in terms of , is then
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Step-by-Step Verified Solution
A series combination of n1 capacitors each of capacitance C 1 are connected to 4V source.
Total capacitance of the series combination of the capacitors is
1/C s = 1/C 1 + 1/C 1 + 1/C 1 .... upto n 1 terms = n 1 /C 1
_ C s = C 1 /n 1
Total energy stored in a series combination of the capacitors is
U s = (1/2) C s (4V) 2 = (1/2) (C 1 /n 1 ) (4V) 2 �(1)
Now a parallel combination of n 2 capacitors each of capacitance C 2 are connected to V source
Total capacitance of the parallel combination of capacitors is
C p = C 1 + C 2 + ... + upto n 2 terms = n 2 C 2
_ C p = n 2 C 2
Total energy stored in a parallel combination of capacitors is
U p = (1/2)C p V 2 = (1/2)(n 2 C 2 )(V) 2 ...(ii)
According to the given problem, U s = U p
(1/2) (C 1 /n 1 )(4V) 2 = (1/2)(n 2 C 2 )(V) 2
C 1 _ 16/n 1 = n 2 C 2
C 2 = 16 _ C 1 / (n 1 _ n 2 )