Question 653: Electrostatics & Capacitance

Question

A series combination of n
1
capacitors, each of value C
1
, is charged by a source of potential difference 4V. When another parallel  combination of n
2
capacitors,  each of value C
2
, 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 C
2
. in terms of C
1
, is then

Accepted Answer

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
$$ \Rightarrow $$ 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
Ã¢ÂÂ¦(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
$$ \Rightarrow $$ 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
ÃÂ 16/n
1
= n
2
C
2
C
2
= 16 ÃÂ C
1
/ (n
1
ÃÂ n
2
)

Answer

$${{2{C_1}} \over {{n_1}{n_2}}}$$

Answer

$$16{{{n_2}} \over {{n_1}}}{C_1}$$

Answer

$$2{{{n_2}} \over {{n_1}}}{C_1}$$

Answer

$${{16{C_1}} \over {{n_1}{n_2}}}$$

Electrostatics & Capacitance

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