Physics · Capacitors and capacitance, the combination of capacitors in series and parallel
When two capacitors of capacities and are connected in series and connected to t
When two capacitors of capacities \( 3 \mu F \) and \( 6 \mu F \) are connected in series and connected to \( 120 V, \) the potential difference across \( 3 \mu F \) is:
- A. \( 40 V \)
- B. \( 60 V \)
- C. \( 80 V \)
- D. \( 180 V \)
Step-by-step solution
In series, charge is same. Compute equivalent capacitance: 1/Ceq = 1/3 + 1/6 = 1/2 => Ceq = 2 μF. Total charge Q = Ceq*V = 2*120=240 μC. Voltage across 3 μF: V = Q/C = 240/3 = 80 V.
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