1 / 11

Physics 2102 Lecture: 12 MON FEB

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.

djackie
Download Presentation

Physics 2102 Lecture: 12 MON FEB

An Image/Link below is provided (as is) to download presentation 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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Physics 2102 Jonathan Dowling Physics 2102 Lecture: 12 MON FEB Capacitance II 25.4–5

  2. V = VAB = VA –VB C1 Q1 VA VB A B C2 Q2 VC VD D C V = VCD = VC –VD Ceq Qtotal V=V Capacitors in Parallel: V=Constant • An ISOLATED wire is an equipotential surface: V=Constant • Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge! • VAB = VCD = V • Qtotal = Q1 + Q2 • CeqV = C1V + C2V • Ceq = C1 + C2 • Equivalent parallel capacitance = sum of capacitances

  3. Q1 Q2 B C A C1 C2 Capacitors in Series: Q=Constant Isolated Wire: Q=Q1=Q2=Constant • Q1 = Q2 = Q = Constant • VAC = VAB + VBC Q = Q1 = Q2 • SERIES: • Q is same for all capacitors • Total potential difference = sum of V Ceq

  4. C1 Q1 C2 Q2 Qeq Q1 Q2 Ceq C1 C2 Capacitors in parallel and in series • In parallel : • Cpar = C1 + C2 • Vpar= V1 = V2 • Qpar= Q1 + Q2 • In series : 1/Cser = 1/C1 + 1/C2 Vser= V1  + V2 Qser= Q1 = Q2

  5. Parallel: Circuit Splits Cleanly in Two (Constant V) C1=10 F C2=20 F C3=30 F 120V Example: Parallel or Series? • Qi = CiV • V = 120V = Constant • Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C • Q3 = (30 F)(120V) = 3600 C What is the charge on each capacitor? • Note that: • Total charge (7200 mC) is shared between the 3 capacitors in the ratio C1:C2:C3— i.e. 1:2:3

  6. Series: Isolated Islands (Constant Q) Example: Parallel or Series C2=20mF C3=30mF C1=10mF What is the potential difference across each capacitor? • Q = CserV • Q is same for all capacitors • Combined Cser is given by: 120V • Ceq = 5.46 F (solve above equation) • Q = CeqV = (5.46 F)(120V) = 655 C • V1= Q/C1 = (655 C)/(10 F) = 65.5 V • V2= Q/C2 = (655 C)/(20 F) = 32.75 V • V3= Q/C3 = (655 C)/(30 F) = 21.8 V Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3) (largest C gets smallest V)

  7. 10 F 10F 10F 10V 5F 5F 10V Example: Series or Parallel? Neither: Circuit Compilation Needed! In the circuit shown, what is the charge on the 10F capacitor? • The two 5F capacitors are in parallel • Replace by 10F • Then, we have two 10F capacitors in series • So, there is 5V across the 10 F capacitor of interest • Hence, Q = (10F )(5V) = 50C

  8. Energy U Stored in a Capacitor • Start out with uncharged capacitor • Transfer small amount of charge dq from one plate to the other until charge on each plate has magnitude Q • How much work was needed? dq

  9. General expression for any region with vacuum (or air) volume = Ad Energy Stored in Electric Field of Capacitor • Energy stored in capacitor: U = Q2/(2C) = CV2/2 • View the energy as stored in ELECTRIC FIELD • For example, parallel plate capacitor: Energy DENSITY = energy/volume = u =

  10. 10mF (C1) 20mF (C2) Example • 10mFcapacitor is initially charged to 120V. • 20mF capacitor is initially uncharged. • Switch is closed, equilibrium is reached. • How much energy is dissipated in the process? Initial charge on 10mF = (10mF)(120V)= 1200mC After switch is closed, let charges = Q1 and Q2. Charge is conserved: Q1 + Q2 = 1200mC • Q1 = 400mC • Q2 = 800mC • Vfinal= Q1/C1 = 40 V 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

  11. Any two charged conductors form a capacitor. • Capacitance : C= Q/V • Simple Capacitors:Parallel plates: C = e0 A/d Spherical : C = 4pe0 ab/(b-a)Cylindrical: C = 2pe0 L/ln(b/a) • Capacitors in series: same charge, not necessarily equal potential; equivalent capacitance 1/Ceq=1/C1+1/C2+… • Capacitors in parallel: same potential; not necessarily same charge; equivalent capacitance Ceq=C1+C2+… • Energy in a capacitor: U=Q2/2C=CV2/2; energy densityu=e0E2/2 • Capacitor with a dielectric: capacitance increasesC’=kC Summary

More Related