ELE 1001 Basic Electrical Technology Lecture 6 Capacitors

ELE 1001: Basic Electrical Technology Lecture 6 Capacitors

Overview of Topics Ø What is a Capacitor? Ø Transient behavior of a Capacitive circuit. Ø Capacitor as an energy storage element. Department of Electrical & Electronics Engineering

Capacitors Ø Department of Electrical & Electronics Engineering

Terminologies q Electric field strength, ü E = V / d volts / m q Electric flux density, ü D = Q / A coulombs / m 2 q Permittivity of free space, q 0 = 8. 85 x 10 -12 F /m q Relative permittivity, r q Capacitance of parallel plate capacitor C = 0 r A / d Department of Electrical & Electronics Engineering

Equivalent Capacitance Ø Department of Electrical & Electronics Engineering

Charging of a Capacitor through a Resistor Initial conditionsat t=0 seconds, Vc= 0 Solving: Department of Electrical & Electronics Engineering

Charging of a Capacitor through a Resistor Ø Time Constant, = RC Ø Time taken by the voltage of the capacitor to reach its final steady state value, had the initial rate of rise been maintained constant vc ic Department of Electrical & Electronics Engineering

Discharging of Capacitor through a Resistor Ø Solving, Department of Electrical & Electronics Engineering

Discharging of Capacitor through a Resistor Department of Electrical & Electronics Engineering

Energy stored in a Capacitor Ø Department of Electrical & Electronics Engineering

Summary Ø Department of Electrical & Electronics Engineering

Illustration 1 A Capacitor consists of 2 metal plates each 400 by 400 mm spaced 6 mm apart. The space between the metal plate is filled with a glass plate 5 mm thick and a layer of paper 1 mm thick. The relative permittivity of glass and paper are 8 and 2 respectively. Calculate the equivalent capacitance. Also find the electric field strength in each dielectric due to a potential difference of 10 k. V between the metal plate. Ans: Cg = 2. 265 x 10 -9 F; Cp = 2. 832 x 10 -9 F; Ceq = 1. 258 x 10 -9 F Q = 1. 258 x 10 -5 C Vg = 5. 554 k. V; Vp = 4. 446 k. V Eg = 1. 11 k. V / mm ; Ep = 4. 446 k. V / mm Department of Electrical & Electronics Engineering

Illustration 2 A 10 F capacitor is initially charged to 100 V dc. It is then discharged through a resistance R Ω and the p. d. across the capacitor after 20 seconds is 50 V. Calculate the value of R. Ans: 2. 886 MΩ Department of Electrical & Electronics Engineering

Illustration 3 An 8 u. F capacitor is connected in series with a 0. 5 MΩ resistor, across a 200 V d. c supply through a switch. At t=0 seconds, the switched is turned on, Calculate Ø Time constant of the circuit Ø Initial charging current. Ø Time taken for the potential difference across the capacitor to grow to 160 V. Ø Current & potential difference across the capacitor 4. 0 seconds after the switch is turned on. Derive the expressions used. Ans: (i) 4 seconds, (ii) 400 μA , (iii) 6. 44 seconds (iv) 126. 424 V & 147. 15 μA Department of Electrical & Electronics Engineering