Three capacitors of equal capacitance are arranged as shown in the figure, with a voltage source across the combination. If the voltage drop across C₁ is $10.0 \mathrm {~V} \text {, }$ what is the voltage drop across $C _ { 3 } ?$ A) 20 V B) 10.0 V C) 40 V D) 30 V

Question

Quizplus · Accepted Answer

Since the capacitors are connected in series, the charge on each capacitor is equal. Let the capacitance of each capacitor be C, then the total charge on the combination is Q = CΔV where ΔV is the voltage drop across the combination. Since the voltage drop across C₁ is 10 V, the voltage drop across the other two capacitors (C1 and C3) must be equal to maintain conservation of energy. Thus, the remaining voltage drop is 10 V as well. Now, the potential difference across C2 and C3 is (2/3)ΔV = (2/3) * 10 V = 6.67 V. Since the voltage across C3 is the sum of the voltage across C2 and the potential difference across C3, its voltage drop is 6.67 V + 10 V = 16.67 V. Finally, the voltage drop across C3 is half of the potential difference across the entire combination, which is (1/2)*ΔV = (1/2)*20 V = 10 V. Thus, the correct answer is A) 20 V.

Three Capacitors of Equal Capacitance Are Arranged as Shown in the Figure