Capacitors in Parallel and in Series Problems And Solutions
Answer: (a) The total capacity for the series of capacitors arranged in series (for C 1 and C 2) is 1/C 12 = 1/C 1 + 1/C 2 = 1/(0.1 μF) + 1/(0.2) μF = 3/0.2 μF Then C 12 = 1/15 μF C AB = C 12 + C 3 = 1/15 μF + 0.3 μF = 11/30 μF (b) C 3 capacitors are arranged in parallel with the capacitor (C 1 + C 2) = C 12, then the potential difference for capacitor C 3 namely V 3 …
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