Two capacitors of 3 μF and 6 μF are connected in series and charged to 120 V. The potential difference across the 3 μFcapacitor is:
Correct answer: A. 80 V
- A. 80 V
- B. 40 V
- C. 60 V
- D. 120 V
Explanation
In series, Ceq = (3 × 6)/(3 + 6) = 2 μF. Charge Q = CeqV = 2 × 10⁻⁶ × 120 = 2.4 × 10⁻⁴ C. Voltage across 3 μF capacitor = Q/C = 2.4 × 10⁻⁴ / 3 × 10⁻⁶ = 80 V.
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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