Question

Three capacitors of capacitance 12 μF,6 μF, and 4 μF are first joined in series and then in parallel. The ratio of their equivalent capacitance will be:
A. 2:3 B. 1:11 C. 1:3 D. 11:11

Answer 1

Explanation:

When in series the eq. Capacitance is 1/12+1/6+1/4

When connected in parallel the eq. Capacitance becomes the sum of all the capacitance

Hence its ratio becomes 1:11

Answer 2

⠀⠀⠀⠀$\purple{\mathbb{ANSWER}}$

C1 = 4μF, C2 = 6μF, C3 = 12μF

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

(1) when in series (1/C) = [(1/C1) + (1/C2) + (1/C3)]

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

(1/C) = [(1/C1) + (1/C2) + (1/C3)]

(1/C) = [(1/4) + (1/6) + (1/12)]

(1/C) = [(3 + 2 + 1)/(12)]

∴ (1/C) = (1/2) C = 2μF

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

(2) when in parallel C' = C1 + C2 + C3 ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

C' = C1 + C2 + C3

C' = 4 + 6 + 12 C' = 22μF

∴ (C/C') = (2/22) = (1/11)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

∴ $\green{\tt{(B)1:11✓}}$