2. Contents
• Capacitor Symbol and Application
• Capacitance of Capacitor
• Grouping of Capacitors
• Electrical Energy Stored in Capacitor
• Charging and Discharging of Capacitor
3. Capacitor (Condenser)
1. Capacitor : Is an electric device made up by two parallel
metallic poles separated by insulator called dielectric.
1. Electrical Symbol : Or
Capacitor Variable
Capacitor
4. Capacitors are used in many applications
• Storage of electrical charges
• Convertor circuits from AC DC
• Telecommunication circuits such as TV, Radio …
• Startup circuits for motors.
• Protection circuits.
5. Capacitance of a Capacitor
• A capacitor is initially uncharged, If a battery is connected across its plates,
the positive (+) terminal of the battery will attract free electrons from plate
A, poles becomes positive (+) charge .
• Plate B receiving these electrons becomes negative (-) charged . An equal
number of (+) and (-) charges will exists on both plates of the capacitor .
• The Total positive Storage = Q
• C =
𝑄
𝑉
or Q = C x V Q : Charge measured in coulombs (C) .
V : Potential difference or voltage across the
capacitor , in volts (V) .
C : Capacitance of the capacitor , in Farad (F) .
6. In general we use the sub-units of Farad.
• 1µF = 10-6 F
• 1nF = 10-9 F
• 1pF = 10-12 F
7. Example 1
1. Determine the potential difference across a 4µF when charged with 5 mC.
2. Find the charge on a 50 pF capacitor when the voltage applied is 2KV.
Solution
1) Q = C x V V =
𝑄
𝐶
=
5 𝑥 10_3
5 𝑥 10_6
= 1.25 x 103 V .
2) Q = C x V = 50 x 10-12 x 2 x 103 =10-7 C = 0.1 µF .
8. Example 2
• A direct current of 4A flows into a 20 µF capacitor for a time of 3ms.
• Determine the potential difference across the capacitor.
Solution:
I =
𝑄
𝑡
Q = I x t = 4 x 3 x 10-3 = 12 x 10-3 C .
V =
𝑄
𝐶
=
12 𝑥 10_2
20 𝑥 10_6
= 600 v.
9. • The capacitance of a capacitor depends on the geometry of the capacitor
and on the insulator.
• C =
Ɛ𝑟
.Ɛ0
.𝐴
𝑑
• A : Area of the plate (m2)
• d : distance between plates
• Ɛ0 : 8.85 x 10-12 (F/m) = constant of electric
Permittivity in vacuum.
• Ɛr : without unit , constant of relative
Permittivity of the insulator.
10. Example 3
1. A capacitor ha an effective area of 4 cm2 separated by 0.1 mm of ceramic
of relative permittivity (Ɛr =100). Calculate the Capacitance .
2. If the capacitor in part (1) Is given a change of 0.1 µC ,what will be the
voltage between the plates . (Ɛ0 = 8.85 x 10-12 (F/m) )
11. Solution
1 . C =
Ɛ𝑟
.Ɛ0
.𝐴
𝑑
=
8 . 85 x 1 0−12x 100 x 4 x 1 0−4
1 0−4 = 3 .5 4 x
1 0−4 F
2 . V =
𝑄
𝐶
=
0.1 x 10 −1 2
3 .54 x 1 0−4
= 2 8 .2 4 8 v.
12. Example 4
• A capacitor of effective area 800 cm2. If the capacitance is 4435 pF ; Determine the
distance between plates if its relative permittivity is 2.5.
• S o l u t i o n :
C =
Ɛ 𝑟
. Ɛ 0
. 𝐴
𝑑
d =
Ɛ 𝑟 . Ɛ 0 . 𝐴
C
=
8 . 8 5 x 1 0 − 1 2 x 2 . 5 x 8 x 1 0 − 2
4 4 3 5 x 1 0 − 1 2
= 0 . 4 m m .
13. Grouping of Capacitors
a) Grouping in parallel :
KCL at node A:
I =I1 + I2
I.t = I1.t + I2 .t
Q = Q1 + Q2
C.V =C1.V +C2.V
C =C1 + C2
14. b) Grouping in Series
VAC =VAB + VBC
𝑞
𝑐
=
𝑞
𝐶1
+
𝑞
𝐶2
1
𝑐
=
1
𝐶1
+
1
𝐶2
+ … +
1
𝐶𝑛
15. Example 5
• What capacitance must be connected in series with a 30 µF for the
equivalence capacitance to be 12 µF.
• Solution:
1
𝑐
=
1
𝐶1
+
1
𝐶2
1
12
=
1
30
+
1
𝐶2
𝐶2 = 20 µF
16. Example 6
• 3 Capacitors of 3 µF, 6 µF and 12 µF are connected in series across a 350 v
supply
a) Calculate the equivalent capacitance of C1,C2 and C3 .
b) The charge stored in each capacitor
c) The voltage across each capacitor
19. Electric Energy Stored In Capacitor
• Electric Energy = W
• W = ½ Q V
• W = ½ C V2
• W = ½ Q2/C
20. Example 7
1. Determine the energy stored in a 3 µF capacitor when charged to 400 V.
2. Find the average power if the energy is dissipated in a time of 10 µs.
• Solution:
1. W = ½ Q V = ½ [(3 x 400 x 10-6) 400] = 0.24 J
2. W = P x t P =W / t = 0.24 / 10 x10-6 = 24000 w =24kw
