DR / MOSTA

Chapter Eight
Capacitance
Capacitors store energy. Not charge
A capacitor consists of two parallel conducting plates separated by an
insulator (A dielectric material).
When it is connected to a voltage, supply
charge (Q) flows onto the capacitor plates
until the potential difference across them
(V) is the same as that of the supply. It can
be seen that the charge Q is related to the
potential V by
Hence
Q=CV
Where C is a constant which depends on the size of the conductor. C
is known as the capacitance of the conductor.
Capacitance: is the ratio of charge to potential of the conductor: is the
charge Q on one plate required to cause unit potential difference, V,
across both plates of capacitor
Q
Charge (C)
C=
V

Unit of capacitance is Farad Voltage (V)

“1 Farad is the capacitance of a conductor, which has potential
difference of 1 volt when it carries a charge of 1 coulomb”.
Farad = Coulomb/Volt
• In Electronic circuits the useful range used is [10-12F-10-3F]
• Capacitors usually have their values marked in picofarads (pF=10-
12F) or microfarads (F =10-6F)

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Energy stored in a capacitor

Electrical potential energy is stored when a

W=vxQ
capacitor is charged. The energy stored is
equal to the work done to force extra charge
Δq. From the definition of potential difference
The work done is the product of potential and
charge:
W ( ¿ E P )=QV

Then with a small amount of charge ∆q

∆ E=V ∆ q
Is equal to the shaded area under the line between q and q+∆q for the
whole process.
The energy transferred from a battery when a capacitor is charged is
given by the area under the graph line when charge is plotted against
potential difference The area under the line represents a right-angled
triangle. Thus,
1
E P= QV
2
Since Q = CV
2
1 Q
Then 2
E P= CV =
2 2C
Example (1)
A 2000 F capacitor is charged to a p.d of 10 V. Calculate the
energy stored by the capacitor.
Solution
C = 2000 F V = 10v
1 2
E p = CV
2
1 2 2
¿ ×2000 ×10 × ( 10 )
−6
2
= 0.10J

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Factors affecting capacitance

The capacitance of an air-filled capacitor can be increased
by putting an insulating material, such as mica or waxed
paper, between the plates. The material between the plates is
called the dielectric. In a type of capacitor known as an
electrolytic capacitor.
The capacitance is directly proportional to the area A of the
plates, and inversely proportional to the distance d between them
A
C∝
d
For a capacitor with air or vacuum between the plates, the constant of
the proportionality is the permittivity of free space
∈0 A
C=
d
C is measured in farad A is measured in m2
d is measured in m
Then ∈0 is measured in Fm−1
If a dielectric used to increase the capacitance, then another quantity
called the relative permittivity ∈r of the dielectric increases.
∈0 ∈r A
C=
d
The relative permittivity is the capacitance of a parallel-plate
capacitor with the dielectric between the plates divided by the
capacitance of the same capacitor with vacuum between the plates
• It is a ratio and has no units
Functions of capacitor
● Smoothing. ● To store (electrical) energy.
● Timing/ (time) delay ● Smoothing/reduce ripple (on direct voltages/
● Tuning current.
● Oscillator ● To block d.c.
● Blocking d.c. ● Timing/time delay(circuits)
●Surge protection. ● In oscillator (circuits)

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● Temporary power ● In tuning (circuits)

supply ● To prevent arcing/ sparks
Capacitors in series and parallel

Circuits often contain combination of two or more capacitors. Like

resistors, they can be connected in series or parallel
(a) Capacitors in parallel
The diagram shows two capacitors connected in parallel
Here are 3 facts that you should know for this circuit
* The potential difference across each capacitor is the same
and equal to v.
* The total charge Q is the sum of the charges on each capacitor
Q=Q1+Q2
* The combined capacitance C is the sum of the two capacitances
C=C1 +C2
To prove this formula Suppose the opposite circuit We know =Q1+Q2
But Q=CV
Then Q1=C1V and Q2=C2V
By substitution
CV= C1V+ C2V
By cancelling the V’s
C=C1 +C2
Example (1)
Calculate the charge on each capacitor in
this diagram
Solution
Using Q=CV
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Q1=C 1 V =400 × 6=2400 μC

Q2=C 2 V =400 × 6=1200 μC

(a) Capacitors in Series

The diagram shows two capacitors connected in series

Here are 3 facts that you should know for this circuit
* The charge on each capacitor is the same and equal to Q.
* The supply potential difference V is shared between the two
capacitor
V=V1+V2
* The combined capacitance C is the sum of the two capacitances
1 1 1
= +
C C1 C 2

To prove this formula

Suppose the opposite circuit
We know V=V1+V2
Q
But V = C
Q Q
Then V 1= C 1 and V 2= C 2
By substitution
Q Q Q
= + =C 1 V +C 2 V
C C 1 C2

By cancelling the Q's

1 1 1
= +
C C1 C 2

Care that:
Series connection → (I) cons so (Q) cons
Q = It
V T =V 1 +V 2 +V 3 +. … …

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Q Q Q Q
= + + + .… … . .
CT C 1 C2 C 3
1 1 1 1
= + +
CT C 1 C2 C 3

Example (2)

A 3.0 F and a 5.0 F capacitor are connected in series with a 12 V

battery.
Solution
a. Find the equivalent capacitance.
1 1 1
= +
C S 3.0 μF 5.0 μF
1 5+3
=
C S 15 μF
15 μF
C S= →C S =1.9 μF
8

b. Find the charge on each capacitor.

Q=C S V

Q= (1.9 μF ) (12 V )

Q=23 μC

c. Find the potential drop (or voltage) across each capacitor.

q 23 μC
V 1= : V 1= ; V =7.7 V
C1 3.0 μF 1
q 23 μC
V 2= :V 2= ; V =4.6 V
C2 5.0 μF 2
charge
* Capacitor = voltege (Pd) (1)

Charge of one plate (1)

Pd bef. 2 plates (1)
Capacitor store (E) not charge

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Bec charge upper plate = (-ve) charge lower plate

So total charge = zero

Charging and discharging capacitors

A capacitor is charged or discharged when it is connected to a d.c

energy source (battery or rectified a.c input) while also connected to a
second parallel circuit for discharge and a resistor to control the
charging/discharging time and intensity as to not damage the
capacitor.

Graphs of charge and discharge:

While charging, the current I in the capacitor starts at the maximum
value and then gradually decreases as more charge is stored on the
capacitor plates.
The same thing happens while discharging as more charge Q leaves
the plates.
The potential difference V across the capacitor increases while
charging as it is directly proportional to the charge Q.
Charging:

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I Q V

t t t

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Discharging:

I Q V

t t t

Equation for the time taken to discharge:

We can derive that the time taken for a capacitor to discharge is

dependent on the resistance of the circuit and the capacitance of the
capacitor. This is shown by the equation: t=RC (very important)
where the Greek letter t (tau) is the constant for time.

Equations for the decrease in charge, p.d, current:

There are equations that describe the “decay” of charge, current and
p.d across the capacitor when it discharges. These decay equations are
very similar as p.d is directly proportional to charge and also to
current:

I =I 0 exp ( RC
−t
)
V =V 0 exp ( RC
−t
)
Q=Q 0 exp ( RC
−t
)

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What happens if we change the resistance in the circuit?

There will be no change in the initial potential difference across the

capacitor, but the initial current through the resistor will be changed.
Increased resistance will mean decreased current, so charge flows
off the capacitor plates more slowly and the capacitor will take longer
to discharge. Of course, this means decreasing the resistance will
cause the capacitor to discharge more quickly.

What happens if we increase the capacitance of the capacitor?

The initial p.d. across the capacitor is also unchanged. So, with an
unchanged resistance, the initial current will be unchanged.
However, there will be more charge on the capacitor and so it will
take longer to discharge.

Care that:

Prove that // → capacity

C t=C 1 +C2 +C 3

In // connect. I T =I 1 + I 2+ I 3

QT =Q1 +Q2+ Q3

C T V =C 1 V +C 2 V +C 3 V

C T =C 1 +C2 +C 3

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N.B

If you encounter a question like this: [Nov 20 V2, Q6(a),(b)]

(a) (i) Define the capacitance of a parallel plate capacitor

(ii) State three functions of capacitors in electrical circuits

1.

2.

3.

(b) A student has available three capacitors, each of capacitance

12μF.

Draw diagrams, one in each case, to show how the student connects
the capacitors to give a combined capacitance between the terminals
of:

(i) 18μF

(ii) 8μF

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You could answer with:

(a)(i) charge on one plate per unit potential difference between both
plates (2 mark)

(ii) any three from:

* smoothing

* timing/(time) delaying

* tuning

* oscillator

* blocking d.c. (direct current)

* surge protection

* temporary power supply

(each 1 mark)

(b)(i)

(ii)

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If you encounter a question like this: [Nov 20 V1, Q(b)]

A student has available four capacitors, each of capacitance 24μF.

The capacitors are connected as shown in Fig. 6.1.

24 μF
=8 μF
3

1 1 1 1
= + +
CT C 1 C2 C 3

Fig. 6.1

Calculate the combined capacitance between the terminals X and Y.

You could answer with:

1 1 1
= +
C C1 C 2
for capacitors in series,

therefore three capacitors in series have a combined capacitance of

8μF (1 mark), and since total capacitance for capacitors in parallel is
simply C = C1 + C2,

24μF + 8μF = 32μF (1 mark)

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If you encounter a question like this: [June 19 V1, Q6(b)]

Three uncharged capacitors of capacitances C1, C2 and C3 are

connected in series with a battery of electromotive force (e.m.f.) E and
a switch, as shown in Fig.6.1.

Fig. 6.1

Series

RT =R1 + R2 + R3 1 1 1 1
= + +
CT C 1 C2 C 3

Parallel

1 1 1 1 C T =C 1 +C2 +C 3
= + +
R T R1 R2 R3

When the switch is closed, there is a charge +q on plate P of the

capacitor of capacitance C1. Show that the combined capacitance C of
the three capacitors is given by the expression
1 1 1 1
= + +
C C1 C 2 C 3

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You could answer with

Q
V =V 1+V 2 and V=
C
thereore

Q Q Q Q
= + +
Total C C 1 C2 C 3

and Q is constant at all capacitors, so

1 1 1 1
= + +
Total C C 1 C2 C 3

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If you encounter a question like this:

The potential difference across the plates of a capacitor of capacitance

500 μF is 240 V. The capacitor is connected across the terminals of a
600 Ω resistor.

Find the time taken for the current to fall to 0.10 A.

You could answer with:

V 240
I 0= =
R 600

I 0=0.40 A

t=RC

−6
t=600 ×500 ×10

t=0.30 s

I =I 0 exp ( RCt )
0.10=0.40 exp ( 0.30t )
0.25=exp ( 0.30t )
Take in both sides
t
In 0.25= 0.30

t=0.41 s

N.B

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If power max. given

If you encounter a question like this:

Explain what is meant by the time constant of a circuit containing

capacitance and resistance.

You could answer with:

It’s the product of the capacitance of capacitor and the total resistance
of the circuit, it’s also equal to the time taken for the capacitor to fully
discharge

If you encounter a question like this: [Specimen 22, Q(12)(a)]

A sinusoidal alternating potential difference (p.d.) from a supply is

rectified using a single diode. The variation with time t of the rectified
potential difference V is shown in Fig. 5.1.

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Fig. 5.1

The alternating potential difference is rectified and smoothed using

the circuit in Fig. 5.2.

connect the capacitor parallel to (R) as shown

The capacitor has capacitance C of 85 μF and the resistor has

resistance R.

The effect of the capacitor and the resistor is to produce a smoothed

output potential difference VOUT. The difference between maximum
and minimum values of VOUT is 2.0 V.

(i) On Fig. 5.1, draw a line to show VOUT between times t = 1.0 ms and
t = 5.0ms.

(ii) Determine the time, in s, for which the capacitor is discharging

between times t = 1.0 ms and t = 5.0 ms.

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(iii) Use your answers in (b)(i) and (b)(ii) to calculate the resistance
R.

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You could answer with:

(i)

Fig. 5.1

(ii) t = (4.6 – 1)×10-3 = 3.6×10-3 s

(iii) τ = RC
−6
3.6=R ×85 ×10

−3
3.6 × 10
R= −6
85 ×10

R = 42

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