Question

The equivalent capacitance
of the capacitors shown in
figure is 12.4 mF. Find the
value of capacitance C.​

Answer 1

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Answer 2

We have given C1 = 7.22 μF, C2 = 4.25μF , C3 = 12μF , C4 = 8.35μF and the equivalent capacitance of all capacitances is 12.4μF

we have to find the capacitance C5 = C = ?

solution : from figure it is clear that,

C4 and C3 are in series combination.

so, 1/C' = 1/C4 + 1/C3 = 1/8.35 + 1/12

⇒C' = 8.35 × 12/(8.35 + 12)

= 4.924μF

now C' and C2 are joined in parallel combination,

so, C" = C' + C2 = 4.924 + 4.25 = 9.174μF

again, C" and C5 are joined in series combination,

so, C''' = C" × C5/(C" + C5)

= 9.174 × C/(9.174 + C)

now, C''' and C1 joined in parallel combination

so, C"" = C''' + C1

= 9.174C/(9.174 + C) + 7.22

here C"" is equivalent capacitance i.e., C"" = Ceq = 12.4μF

now, 12.4 = 7.22 + 9.174C/(9.174 + C)

⇒0.564 = C/(9.174 + C)

⇒0.564 × (9.174 + C) = C

⇒5.174 = 0.436C

⇒C ≈ 11.9 μF

Therefore the capacitance of C is 11.9μF (approx).