Two capacitors C1 = 12 mF and C2 = 4 mF are in group connections. Calculate the effective capacitance of the system when they are connected (a) in series (b) in parallel.
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Answered by
9
Answer ⇒ Series = 3mF and Parallel = 16 mF.
Explanation ⇒ Given,
C₁ = 12 mF, and C₂ = 4 mF.
(i). Using the formula for Series.
1/C = 1/C₁ + 1/C₂
∴ 1/C = 1/12 + 1/4
∴ 1/C = 1/12 + 3/12
∴ 1/C = 4/12
∴ 1/C = 1/3
∴ C = 3 mF.
(ii). Using the formula, when they are connected in parallel,
C = C₁ + C₂
∴ C = 12 + 4
∴ C = 16 mF.
Hope it helps.
Answered by
8
Answer:
(i). Using the formula for Series.
1/C = 1/C₁ + 1/C₂
∴ 1/C = 1/12 + 1/4
∴ 1/C = 1/12 + 3/12
∴ 1/C = 4/12
∴ 1/C = 1/3
∴ C = 3 mF.
(ii). Using the formula, when they are connected in parallel,
C = C₁ + C₂
∴ C = 12 + 4
∴ C = 16 mF.
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