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Four capacitors with capacitances C(1) =...

Four capacitors with capacitances `C_(1) = 1 muF, C_(2) = 1.5 muF, C_(3) = 2.5 muF` and `C_(4) = 0.5 muF` are connected as shown in Fig, to a 30 voltg source. The potentail difference between points `a` and `b` is

A

`5 V`

B

`9 V`

C

`10 V`

D

`13 V`

Text Solution

Verified by Experts

The correct Answer is:
D
In Fig, let `q muC` be charge on `C_(1)` and `C_(2)`
`:. (q)/(C_(1)) + (q)/(C_(2)) = 30` or `(q)/(1) + (q)/(1.5) = 30`
`q = 18 mu C`
`V_(A) - V_(a) = (q)/(C_(1)) = (18)/(1) = 18 V`
Similarly, `(q')/(C_(3)) + (q')/(C_(4)) = 30` or `(q')/(2.5) + (q')/(0.5) = 30`
`q = (25)/(2) muC`
`V_(A) - V_(b) = (q')/(C_(3)) = (25)/(2xx2.5) = 5`
`V_(a) - V_(b) = (V_(A) - V_(b)) - (V_(A) - V_(a)) = 5-18`
`= -13` volt
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