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Q.

In the given circuit, charge Q₂ on the 2 $μ$ F capacitor changes as C is varied from 1 $μ$ F to 3 $μ$ F . Q₂ as a function of 'C' is given properly by : (figures are drawn schematically and are not scale)

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a

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b

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c

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d

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answer is B.

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Detailed Solution

$C_{e q} = \frac{3 C}{3 + C}$
Total charges $q = (\frac{3 C}{3 + C}) E$
Charge upon capacitor 2 $μ$ F ,
$Q_{2} = \frac{2}{3} \times \frac{3 C E}{(3 + C)} = \frac{2 C E}{3 + C} = \frac{2 E}{1 + \frac{3}{C}}$
If $C = 1 μF; Q_{2} = E$
Hence Q₂ is maximum at C = 3 $μ$ F

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Will be the best representation of the charge vs value of C

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