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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.
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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 ;
 𝒊𝒄 = 𝑪.
𝒅𝑽𝒄
𝒅𝒕
 Circuit representation is
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Terminologies
 Electric field strength,
 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
capacitor
C = 0 r A / d
plate
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Equivalent Capacitance
 In Series:
𝟏
𝟏
𝟏
𝟏
=
+
+ ……………+
𝑪𝒆𝒒 𝑪𝟏 𝑪𝟐
𝑪𝒏
 In Parallel:
𝑪𝒆𝒒 = 𝑪𝟏 + 𝑪𝟐 + … . . + 𝑪𝒏
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Charging of a Capacitor through a Resistor
𝐖𝐫𝐢𝐭𝐢𝐧𝐠 𝐊𝐂𝐋 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧;
𝒗𝒄 − 𝑽
𝒅𝒗𝒄
−𝑪
=𝟎
𝑹
𝒅𝒕
Initial conditionsat t=0 seconds, Vc= 0
Solving:
𝟏
−𝑹𝑪 𝒕
𝒗𝒄 = 𝑽 𝟏 − 𝒆
𝒊𝒄 =
𝑽
𝑹
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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
vc
ic
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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
t=0
𝑣𝑐
𝑑𝑣𝑐
+𝐶
=0
𝑅
𝑑𝑡
𝒗𝒄 = 𝑽 𝒆
a
V
Solving,
𝟏
−(𝑹𝑪 )𝒕
𝒊𝒄 = −𝑰𝒆
R
𝟏
−(𝑹𝑪 )𝒕
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b
i
C
+
vc
Discharging of Capacitor through a Resistor
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Energy stored in a Capacitor
 Instantaneous power
𝒅𝒗𝒄
𝒑 = 𝒗𝒄 ∗ 𝒊 = 𝑪 𝒗𝒄
𝒅𝒕
 Energy supplied during ‘dt’ time is ;
𝒅𝒘 = 𝑪 𝒗𝒄 𝒅𝒗𝒄
Energy stored when potential rises from 0 to ‘V’ volts
is,
𝑽
𝑾=
𝟎
𝟏
𝑪 𝒗𝒄 𝒅𝒗𝒄 = 𝑪. 𝑽𝟐 𝐉𝐨𝐮𝐥𝐞𝐬
𝟐
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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.
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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
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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
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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
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