Question Detail
Question
A 2-µF and a 1-µF capacitor are connected in series and charged from a battery. They store charges P and Q, respectively. When disconnected and charged separately using the same battery, they have charges R and S, respectively. Then:
Correct Answer
Option A is the correct answer.
Detailed Explanation
Let C1 = 2 µF and C2 = 1 µF. Let Vbattery be the voltage of the battery.Scenario 1: Capacitors in seriesIn a series connection, both capacitors store the same amount of charge. Therefore, P = Q.The equivalent capacitance in series is Ceq = (C1C2) / (C1 + C2) = (2 µF × 1 µF) / (2 µF + 1 µF) = 2/3 µF.The total charge stored from the battery is Qtotal = Ceq × Vbattery = (2/3)Vbattery. So, P = Q = (2/3)Vbattery.Scenario 2: Capacitors charged separatelyWhen charged separately by the same battery, both capacitors are charged to the full battery voltage Vbattery.Charge R for C1: R = C1 × Vbattery = 2 µF × Vbattery = 2Vbattery.Charge S for C2: S = C2 × Vbattery = 1 µF × Vbattery = Vbattery.Comparing the charges:R = 2VbatteryS = VbatteryP = Q = (2/3)VbatteryClearly, R > S (2Vbattery > Vbattery).Also, S > P (Vbattery > (2/3)Vbattery).And P = Q.Combining these, we get R > S > P = Q.
Hint
Remember that capacitors in series hold the same charge, and when charged separately by the same voltage source, their charges are proportional to their individual capacitances and the battery voltage.