# Chapter 10 EMT1150 Introduction to Circuit Analysis

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1 Chapter 10 EM1150 Introduction to Circuit Analysis Department of Computer Engineering echnology Fall 2018 Prof. Rumana Hassin Syed

2 Chapter10 Capacitors Introduction to Capacitors he Electric Field Capacitance Capacitors in Series and Parallel ransients in Capacitive Networks

3 Introduction Always compare with resistors wo-terminal device Symbol

4 Intro CND.. Capacitor does not dissipate energy as does the resistor but store it. Capacitor displays its true characteristics only when a change in the voltage or current is made in the network.

5 Electric field Electric filed/electrostatic field is a field of force generated by the electrical charges. It is represented by electric flux lines, which indicate the strength of the electric field at any point around any charged body. he denser the lines of flux, the stronger is the electric field.

6 Capacitor A capacitor is constructed simply of two parallel conducting plates separated by insulating material. Insulating material is called dielectric material.

7 Electric field in capacitor Note the fringing that occurs at the edges as the flux lines originating from the points farthest away from the negative plate strive to complete the connection E = V d E: Volt/m V: Volt d: m

8 Capacitance is a measure of a capacitor s ability to store charge on its plates. Unit: Farad (F) A capacitor has a capacitance of 1 farad if 1 coulomb of charge is deposited on the plates by a potential difference of 1 volt cross the plates. C Q V C = farad (F) Q = coulombs (C) V = volts (V)

9 Example1: If 40V are applied across a 470µF capacitor, find the charge on the plates. Q = CV = 470μF 40V = 18.8 (mc)

10 Capacitance he capacitance of any capacitor is due primarily to three factors: Permittivity (Dielectric material) Distance Area C = ε A d C: Farad (F) ɛ : Permittivity (F/m) A: m 2 d: m

11 Permittivity is the measure of the resistance that is encountered when forming an electric field in a medium. In general, permittivity of other materials can compare to the permittivity of vacuum. ε r = ε ε 0

12 ε 0 = F/m C = ε 0 ε r A d

13 Example2: Determine the capacitance of each capacitor on the right side. (a) C = 3(5µF) =15 (µf) (b) C = (0.1µF)/2 =0.05 (µf)

14 (c) C = 2.5 (20µF) =50 (µf) (d) C = 4 5 (1000pF) /(1/8) =0.16 (µf)

15 Example3:For the capacitor in the figure: a. Determine the capacitance. b. Determine the electric field strength between the plates if 450 V are applied across the plates. c. Find the resulting charge on each plate. ( a). C o o A d 12 F m / 0.01m m V 450V 3 ( b ) V / m 3 d m ( d ). C Q CV Q V V 9 C 26.55nC F

16 ype of capacitors Fixed capacitor Electrolytic capacitors Ceramic (disc) capacitor Film/foil polyester capacitor Mica capacitors Oil-filled capacitor Variable capacitor

17 Capacitors in Series Apply KVL to CP E = V 1 + V 2 + V 3 Similar format to parallelconnected resistor V = Q C Q C = Q 1 C 1 + Q 2 C 2 + Q 3 C 3 1 C = 1 C C C 3 CP Q = Q 1 = Q 2 = Q 3 C = C 1C 2 C 1 + C 2

18 Capacitors in parallel Apply KVL to CPs E = V 1 = V 2 = V 3 Similar format to seriesconnected resistor V = Q C Q = Q 1 = Q 2 = Q 3 C C 1 C 2 C 3 CP1 CP2 CP3 Q = Q 1 + Q 2 + Q 3 Q = CV C E = C 1 V 1 + C 2 V 2 + C 3 V 3 C = C 1 + C 2 + C 3

19 Example4: a. Find the total capacitance. b. Determine the charge on each plate. c. Find the voltage across each capacitor. b. Q Q Q Q C E (8 10 F)( 60V ) 480 C a. C C C C C F F F F 6 Q C c. V V 6 C F V V 2 3 Q C Q C C F C F and E V V V 60V V 48. 0V

20 Example5: a. Find the total capacitance. b. Determine the charge on each plate. c. Find the total charges. a. C C C C 800F 60F 1200F 2060F 6 b. Q C E ( F)( 48V ) 38. 4mC Q C E ( F)( 48V ) 2. 88mC Q C E ( F)( 48V ) 57. 6mC 3 1 c. Q Q Q Q mc

21 Example6: Find the voltage across and charge on each capacitor for the network ' C C C 4 F 2 F 6F C C C C ' C ' ( 3F)( 6F) 3F 6F 2 F Q C E ( F)( 120V ) 240C V 1 Q ' Q Q Q Q 1 240C Q C C ' F C 80V

22 V ' Q C ' ' F C 40V ' 6 Q C V ( 4 10 F)( 40V ) 160 C 2 2 ' 6 Q C V ( 2 10 F)( 40V ) 80 C

23 Example7: Find the voltage across and charge on each capacitor for the network, after each has charged up to its final value V C2 = (7Ω)(72V) 7Ω + 2Ω V C1 = (2Ω)(72V) 7Ω + 2Ω = 56 V = 16(V)

24 ransients in Capacitive Networks Charging phase he placement of charge on the plates of a capacitor does not occur instantaneously. Instead, it occurs over a period of time determined by the components of the network.

25 v C = E(1 e t/τ ) τ = RC

26 i C = E R e t/τ τ = RC

27 Example8: (a) Find the mathematical expressions for the transient behavior of v C, i C, and v R for the circuit when the switch is moved to position 1. Plot the curves of v C, i C, and v R. b. How much time must pass before it can be assumed, for all practical purposes, that i C 0 A and v C E volt?

28 b. 5τ = 5 32ms = 160ms a. τ = RC = Ω F = s = 32ms v C = E 1 e t τ = 40 1 e t V = 40V 8kΩ e =5mAe i C = E R e t τ t t 32ms v R = Ee t τ = 40e 32msV t

29 Discharging phase v C = Ee t τ τ = RC

30 i C = E R e t τ τ = RC v R = Ee t τ

31

32 Example9: After v C in Example 8 has reached its final value of 40 V, the switch is shown into position 2. Find the mathematical expressions for the transient behavior of v C, i C, and v R after the closing of the switch. Plot the curves for v C, i C, and v R using the defined directions and polarities in example 8. Assume that t = 0 when the switch is moved to position 2.

33 τ = RC = 32ms v C = Ee t τ = 40e t 32msV i C = E R e t τ =5mAe t 32ms v R = Ee t τ = 40e 32msV t

34 Example10: Find the mathematical expressions for the transient behavior of v C for the circuit when the switch is moved to position 1 when C=4µF, 40µF, and 100µF. Plot the curve v C. a. τ = RC 1 = Ω F = s = 32ms τ = RC 2 = Ω F = s = 320ms τ = RC 3 = Ω F = 0.8s

35 Example11: Find the mathematical expressions for the transient behavior of v C and i C for the circuit when the switch is closed position 1 at t = 0s. Find the hevenin equivalent circuit without the capacitor.

36 R h = R 1 //R 2 + R 3 = 30(kΩ) = (60kΩ)(30kΩ) 60kΩ+30kΩ E h = ER 2 R 1 + R 2 = (21V)(30kΩ) 60kΩ + 30kΩ = 7(V) τ = R h C = 30kΩ F = s = 6ms

37 τ = R h C = 30kΩ F = s = 6ms 7.23 τ = 6ms v C = 7 1 e t τ = 7 1 e t 6ms V i C = E R e t τ = 7V 30kΩ e t =0.23mAe t 6ms τ = 6ms

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