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Find the Charge Appearing on Each of the Three Capacitors Shown in Figure .

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प्रश्न

Find the charge appearing on each of the three capacitors shown in figure .

बेरीज
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उत्तर

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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पाठ 31: Capacitors - Exercises [पृष्ठ १६५]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 31 Capacitors
Exercises | Q 8 | पृष्ठ १६५

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