Three capacitors of 4 μF each are connected: two in parallel, that combination in series with the third. The equivalent capacitance is:
A8.0 μF
B2.67 μF
C12.0 μF
D1.33 μF
Answer & Solution
Correct answer: B. 2.67 μF
Two in parallel: $4+4 = 8$ μF. In series with the third 4 μF: $1/C = 1/8 + 1/4 = 3/8$, so $C = 8/3 \approx 2.67$ μF. (Series with a larger capacitor pulls the equivalent below the smaller of the two.)
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