ELE 1001: Basic Electrical Technology Lecture 6 Capacitors

Overview of Topics What is a Capacitor? Transient behavior of a Capacitive circuit. Capacitor as an energy storage element.

Capacitors A capacitor is a passive electric device that can store energy in the electric field between a pair of closely spaced conductors Capacitance (C): Property which opposes the rate of change of voltage; Unit – Farad (F) The capacitive current is proportional to the rate of change of voltage across it ; 𝒊 𝒄 =𝑪. 𝒅 𝑽 𝒄 𝒅𝒕 Circuit representation is Picture Courtesy: Google

Terminologies Electric field strength, Electric flux density, E = V / d volts / m Electric flux density, D = Q / A coulombs / m2 Permittivity of free space, 0 = 8.85 x 10-12 F /m Relative permittivity, r Capacitance of parallel plate capacitor C = 0 r A / d

Equivalent Capacitance In Series: 𝟏 𝑪 𝒆𝒒 = 𝟏 𝑪𝟏 + 𝟏 𝑪𝟐 + ……………+ 𝟏 𝑪𝒏 In Parallel: 𝑪 𝒆𝒒 = 𝑪 𝟏 + 𝑪 𝟐 + ….. + 𝑪 𝒏

Charging of a Capacitor through a Resistor 𝐖𝐫𝐢𝐭𝐢𝐧𝐠 𝐊𝐂𝐋 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧; 𝒗 𝒄 −𝑽 𝑹 −𝑪 𝒅 𝒗 𝒄 𝒅𝒕 =𝟎 Initial conditions- at t=0 seconds, Vc= 0 Applying KCL, 𝑉𝑐 −𝑉 𝑅 −𝐶 𝑑𝑉𝑐 𝑑𝑡 =0 𝑑𝑉𝑐 𝑑𝑡 + 1 𝑅𝐶 𝑉𝑐= 𝑉 𝑅𝐶 Complimentary function = 𝐾 𝑒 − 1 𝑅𝐶 𝑡 Particular integral = V Total solution: 𝑉𝑐=𝑉+𝐾 𝑒 − 1 𝑅𝐶 𝑡 Applying initial conditions, 𝑡=0, 𝑉𝑐=0;𝐾=−𝑉; Total Solution: 𝑣 𝑐 =𝑉 1 − 𝑒 − 1 𝑅𝐶 𝑡 Current 𝑖 𝑐 = 𝑉 𝑅 𝑒 − 1 𝑅𝐶 𝑡 Solving: 𝒗 𝒄 =𝑽 𝟏 − 𝒆 − 𝟏 𝑹𝑪 𝒕 𝒊 𝒄 = 𝑽 𝑹 𝒆 − 𝟏 𝑹𝑪 𝒕

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

Discharging of Capacitor through a Resistor Capacitor is initially charged to a voltage V. At t= 0 sec, switch is moved from position a to b 𝑣 𝑐 𝑅 +𝐶 𝑑𝑣 𝑐 𝑑𝑡 =0 Solving, 𝒗 𝒄 =𝑽 𝒆 −( 𝟏 𝑹𝑪 )𝒕 𝒊 𝒄 =−𝑰 𝒆 −( 𝟏 𝑹𝑪 )𝒕

Discharging of Capacitor through a Resistor

Energy stored in a Capacitor Instantaneous power 𝒑= 𝒗 𝒄 ∗𝒊=𝑪 𝒗 𝒄 𝒅𝒗 𝒄 𝒅𝒕 Energy supplied during ‘dt’ time is ; 𝒅𝒘=𝑪 𝒗 𝒄 𝒅𝒗 𝒄 Energy stored when potential rises from 0 to ‘V’ volts is, 𝑾= 𝟎 𝑽 𝑪 𝒗 𝒄 𝒅𝒗 𝒄 = 𝟏 𝟐 𝑪. 𝑽 𝟐 𝐉𝐨𝐮𝐥𝐞𝐬

Summary Capacitor stores energy in its electrical field. When a series RC circuit is connected to a d.c voltage source, there is an exponential growth of voltage across the capacitor and it decays exponentially when the voltage source is removed. Time constant of a series RC circuit is 𝑅𝐶 𝑠𝑒𝑐𝑜𝑛𝑑𝑠.

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 kV 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 kV; Vp = 4.446 kV Eg = 1.11 kV / mm ; Ep = 4.446 kV / mm

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

Illustration 3 An 8uF capacitor is connected in series with a 0.5MΩ resistor, across a 200 V d.c supply through a switch.At t=0 seconds, the switched is turned on, Calculate Time constant of the circuit Initial charging current. Time taken for the potential difference across the capacitor to grow to 160V. 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