Three capacitors with capacitances C1 = 1.50 nF,C2 = 2.50 nF, and C3 = 5.50 nf a
ID: 1502729 • Letter: T
Question
Three capacitors with capacitances C1 = 1.50 nF,C2 = 2.50 nF, and C3 = 5.50 nf are wired to a battery with V = 16.0 v, as shown in the figure what is the potential drop across capacitor C2? A 4.00 103 nf parallel plate capacitor is connected to a 14.0 V battery and charged What is the charge Q on the positive plate of the capacitor? What is the electric potential energy stored in the capacitor? The 4.00 103-nF capacitor is then disconnected from the 12.0-V battery and used to charge two uncharged capacitors, a 100.-nF capacitor, a 200.-nF capacitor, connected in series. After charging, what is the potential difference across each of the three capacitors?Explanation / Answer
3) Net capacitance of circuit = 2.434 nF
=> Net charge from battery = 2.434 * 16 = 38.94 nC
=> potential drop across C2 = 16 - (38.94/3.50)
= 4.874 V
4) a) charge on positive plate = 4000 * 14 * 10-9 = 5.6 * 10-5 C
b) Electric potential energy = 1/2 * 4000 * 10-9 * 14 * 14 = 3.92 * 10-4 J
c) potential difference across 4000 nF = 65.57 * 12/4000
= 0.196 V
potential difference across 200 nF = 65.57 * 12/200
= 3.93 V
potential difference across 100 nF = 65.57 * 12/100
= 7.868 V
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