A capacitor of capacity C1 = 1µF can with stand the maximum voltage V1 = 16 KV while another capacitor C2 = 2µF can with stand the maximum voltage V2 = 4KV. The maximum voltage of capacitance connected in series will be:
Correct answer: B. 12 KV
- A. 10 KV
- B. 12 KV
- C. 6 KV
- D. 4 KV
Explanation
When capacitors are connected in series, the total maximum voltage they can withstand is determined by the least maximum energy storage capacity among them. Since the energy stored by a capacitor is given by E = 1/2 CV2, we calculate the maximum energy for each capacitor and determine the series limit. For C1: E1 = 1/2 * 1µF * (16KV)2 and for C2: E2 = 1/2 * 2µF * (4KV)2. The capacitor with lower energy capacity will limit the overall voltage. Here, both capacitors can withstand a combined maximum voltage of 12 KV when connected in series. Thus, option B is correct.
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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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