Two capacitors $C_{1} = 5.2 μ F \pm 0 . 1 μ F a n d c_{3} = 12 . 2 μ F a r e$ joined (i)I n series (ii) In paraUel Find the net capacitance in these two cases.

A. 2.8%, 1.23%

B. 3.6%, 1.31%

C. 3.4%, 1.3%

D. 3.9%, 1.15%

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Correct Answer : D.

I n p a r a l l e l c = c_{1} + c_{2} a n d

\frac{∆ c}{c} \times 100 = \frac{∆ c_{1} + ∆ c_{2}}{c_{1} + c_{2}} \times 100 = \frac{0.2 x 100}{17.4}

I n s e r i e s c = \frac{c_{1} c_{2}}{c_{1} + c_{2}} a n d

\frac{∆ c}{c} \times 100 = (\frac{∆ c_{1}}{c_{1}} + \frac{∆ c_{2}}{c_{2}} + \frac{∆ c_{1} + ∆ c_{2}}{c_{1} + c_{2}}) 100

= (\frac{0.1}{5.2} + \frac{0.1}{12.2}) \times 100 + 1.15 = 3.9 %

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