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Question

Two capacitors have an equivalent capacitance of 20μF when connected in parallel and 4.8μF when connected in series. The capacitance of these capacitors are

A
8μF & 12μF
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B
10μF & 16μF
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C
12μF & 18μF
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D
6μF & 8μF
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Solution

The correct option is A 8μF & 12μF
Let the capacitance of capacitors be C1 and C2.

Given, Cp=20μF and Cs=4.8μF

In parallel combination,

C1+C2=20μF .......(1)

And for series combination

C1C2C1+C2=4.8μF .......(2)

From (1) and (2) we get,

C1C220=4.8C1C2=96

C2=96C1

Putting the value of (C2) in eq.(1)

C1+96C1=20

C2120C1+96=0

C2112C18C1+96=0

(C112)(C18)=0

C1=8μF or 12μF

This implies either
(C1=8μF & C2=12μF) or (C1=12μF & C2=8μF)

Hence, option (a) is correct.

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