The equivalent capacity of the infinite network shown in the figure (across AB) is :
(Capacity of each capacitor is 1 μF)
A
∞
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B
1μF
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C
(√3−12)μF
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D
(√3+12)μF
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Solution
The correct option is C(√3−12)μF If we remove one set from infinite it is still equal to infinite series, assume the equivalent capacitance is Ceff, capacitance across OE can be replaced by Ceff. now 1μF and Ceff are in parallel, their equivalent capacitance is Ceff+1. Now 2 capacitors of 1μF and Ceff+1 are in series. 1Ceff=11+11+11+Ceff 1Ceff=2+11+Ceff 1Ceff=2+2Ceff+11+Ceff 1+Ceff=3Ceff+2C2eff 2C2eff+2Ceff−1=0 ⇒Ceff=√3−12