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Discussion Response Post to: Finding effective capacitance

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Soumen Goswami (2446)

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total posts: 1777
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(a)
Connect a battery to the two ends such that the +ve terminal is connected to the left end of the network, and the -ve end at the right end.
Let the charge on the 1

F capacitor be x and that on the 3

F capacitor be y. By symmetry of the circuit, the charges on the right 1

F and 3

F capacitors should be x and y.
Now the left plate of the top left 1

F capacitor, the right plate of the right top 3

F capacitor and the top plate of the 4

F capacitor form an isolated system. Hence the total charge of these three plates must be zero. Hence the charge on the top plate of the 4

F capacitor must be x-y.
In the first closed loop,
-x + -(x-y)/4 + y/3 = 0
15x = 7y -------------(1)
Working along the loop formed by the lower branches of the network and the battery,
-y/3 - x + E = 0
where E is the emf of the cell.
E = x + y/3 = 12x/7 (Using (1))
Total charge given by the cell = x + y = C_eqE
22x/7 = C_eq(12x/7)
C_eq = 11/6

All men dream but not equally. Those who dream by night in the dusty recesses of their minds wake in the day to find that it was vanity; but the dreamers of the day are dangerous men, for they may act their dream with open eyes to make it possible.

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