Physics 2102 Jonathan Dowling. Physics 2102 Lecture: 12 MON FEB. Capacitance II. 25.4–5. V = V AB = V A –V B. C 1. Q 1. V A. V B. A. B. C 2. Q 2. V C. V D. D. C. V = V CD = V C –V D. C eq. Q total. V=V. Capacitors in Parallel: V=Constant.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Physics 2102 Lecture: 12 MON FEB
V = VAB = VA –VB
V = VCD = VC –VD
Q = Q1 = Q2
1/Cser = 1/C1 + 1/C2
Vser= V1 + V2
Qser= Q1 = Q2
Parallel: Circuit Splits Cleanly in Two (Constant V)
What is the charge on each capacitor?
Series: Isolated Islands (Constant Q)
What is the potential difference across each capacitor?
Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3)
(largest C gets smallest V)
Neither: Circuit Compilation Needed!
In the circuit shown, what is the charge on the 10F capacitor?
General expression for any region with vacuum (or air)
volume = Ad
Initial charge on 10mF = (10mF)(120V)= 1200mC
After switch is closed, let charges = Q1 and Q2.
Charge is conserved: Q1 + Q2 = 1200mC
Also, Vfinal is same:
Initial energy stored = (1/2)C1Vinitial2 = (0.5)(10mF)(120)2 = 72mJ
Final energy stored = (1/2)(C1 + C2)Vfinal2= (0.5)(30mF)(40)2 = 24mJ
Energy lost (dissipated) = 48mJ