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In the circuit shown, C(1) = C(5) = C(6)...

In the circuit shown, `C_(1) = C_(5) = C_(6) = 6.0 muF and C_(2) = C_(3) = C_(4) = 4.0 muF`. What is the net charge stored on the capacitors and charge on `C_(4)` only ?

Text Solution

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As is clear fro Fig. `C_(2), C_(3), C_(4)` are parallel. Therefore, `C_(p1) = C_(2) + C_(3) + C_(4)`
`= 4+4+4 = 12 muF`
Again, `C_(5), C_(6)` are in parallel. Therefore
`C_(p2) = C_(5) + C_(6) = 6+6 = 12 muF`
The equialent circuit is redrawn as shown in FIg. As `C_(1) , C_(p1), C_(p2)` are in series, therefore net capacitance of the circuit.

`(1)/(C ) = (1)/(6) + (1)/(12) + (1)/(12) = (4)/(12) = (1)/(3), C = 3 muF`
`:.` Charge stored in the system,
`q = CV = 3xx12 = 36 muC`
Now, charge of `36 muC` on `C_(p1)` is divided equally among three capacitors, `C_(2) , C_(3), C_(4)` as their capacities are equal.
`:.` Charge on `C_(4)` alone `= (36)/(3) = 12 muC`
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