A 20 uF capacitor is connected in a circuit to a 9.0 Vbattery. The capacitor is

Question

A 20 uF capacitor is connected in a circuit to a 9.0 Vbattery. The capacitor is then removed from the battery andconnected in a circuit to a second, initially uncharged, 52 uFcapacitor. 1) What is the final voltage across the two capacitors? 2) How much energy is finally stored in the twocapacitors? A 20 uF capacitor is connected in a circuit to a 9.0 Vbattery. The capacitor is then removed from the battery andconnected in a circuit to a second, initially uncharged, 52 uFcapacitor. 1) What is the final voltage across the two capacitors? 2) How much energy is finally stored in the twocapacitors?

Explanation / Answer

Initial charge Q = C V = 20 * 10E-6 * 9 = 1.8 * 10E-4 Capacitance of capacitors in series Cc =C1 C2 / (C1 + C2) = 20* 52 / (20 + 52) F = 14.4 F V = Q / Cc = 1.8 * 10E-4 / 14.4 * 10E-6 = 12.5V E = 1/2 Cc V2 = 1/2 * 14.4 * 10E-6 *12.52 = 1.125 * 10E-6 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.