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The equivalent capacitance of the arrangement shown in figure (four 15 $\mu F$ capacitors with a source E) is: [add image]
A
30 $\mu F$
B
15 $\mu F$
C
25 $\mu F$
D
20 $\mu F$
Detailed Solution
Analysing the given network of four 15 $\mu F$ capacitors: two in series give an equivalent of $\dfrac{15\times15}{15+15}=7.5\ \mu F$; combined appropriately with the others as per the figure gives $C_{eq}=5+15=20\ \mu F$.