Question 678: Electrostats & Capacitance

Question

A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4V. When another parallel  combination of n2 capacitors,  each of value C2, 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 C2. in terms of C1, 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

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

Electrostats & Capacitance

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