1. ELE 1001: Basic Electrical Technology Lecture 6 Capacitors

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

3. 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

4. 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 = x F /m
Relative permittivity, r
Capacitance of parallel plate capacitor
C = 0 r A / d

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

6. 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

7. 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

8. 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

9. Discharging of Capacitor through a Resistor

10. 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

11. 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.

12. 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 = x 10-9 F; Cp = x 10-9 F; Ceq = x 10-9 F
Q = x 10-5 C
Vg = kV; Vp = kV
Eg = 1.11 kV / mm ; Ep = kV / mm

13. 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

14. 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) V & μA