Two capacitors of capacitance 2µF each are connected in series to a power supply of 6 volt. The voltage across each capacitor is :
Correct answer: A. 6 volt
- A. 6 volt
- B. 1 volt
- C. 3 volt
- D. 2 volt
Explanation
When capacitors are connected in series, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of the individual capacitances (C1, C2, C3, ...): 1/C_total = 1/C1 + 1/C2 + 1/C3 + ... In this case, we have two capacitors of 2µF each connected in series. Plugging in the values, we get: 1/C_total = 1/2µF + 1/2µF Simplifying the expression: 1/C_total = 2/2µF 1/C_total = 1/µF C_total = µF So the total capacitance is 1µF. Now, the voltage across each capacitor in a series circuit is the same. Therefore, the voltage across each capacitor in this case is 6 volts.
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Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
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