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Question
Find the charge appearing on each of the three capacitors shown in figure .
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Solution
Let us first find the equivalent capacitance. It can be observed from the circuit diagram that capacitors B and C are in parallel and are in series with capacitor A.
The equivalent capacitance can be calculated as follow :-
`1/C_(eq) = 1/C_A + 1/(C_B+C_C`
`1/C_(eq) = 1/8 + 1/(4+4) = 1/8+1/8`
`⇒ 1/C_(eq) = 2/8`
`⇒ C_(eq) = 4 "uF"`
Capacitors B and C are parallel and are in series with capacitor A. The equivalent capacitance of capacitors B and C is given by
(4 + 4) μF = 8 μF
It is the same as the capacitance of capacitor A. Therefore, equal potential difference will be there on capacitor A and the system of capacitors B and C.
Now,
Potential difference across capacitor A = 6 V
Thus,
Charge on capacitor A = (8 µF) × (6 V) = 48 µC
And,
Potential difference across capacitors B and C = 6 V
Charge on capacitor B = (4 µF) × (6 V) = 24 µF
Charge on capacitor C = (4 µF) × (6 V) = 24 µF
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