In the arrangement of the capacitors shown in the figure, each C 1 capacitor has capacitance of 3µF and each C 2 capacitor has capacitance of 2µF then,
(i) Equivalent capacitance of the network between the points a and b is:
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(i)
⇒ C 1 ' = 1µF
C 2 ' = C 2 + C 1 ' = 3µF ⇒ C eq = 1µF
(ii)
C eq = 1µF Q = C eq V = 900µF
charge on nearest capacitor = 900µF
(iii)
from point potential method
V c – V d = 100V
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