ENERGY-STORAGE

ELEMENTS
BEV10403 SEM II 2023/2024
S1 - Mr. Suhaimi bin Saiman
S2 - Ms. Rohaiza Binti Hamdan
S3 - Dr. Mimi Faisyalini Binti Ramli
S4 - Assoc. Prof. Ts. Dr. Mohd Noor Bin Abdullah
S5 - Dr. Khairul Anuar Bin Mohamad
ROHAIZA HAMDAN | JKEK | FKEE
ENERGY STORAGE ELEMENT
[1.1] Capacitors and energy storage

[1.2] Series and parallel capacitors

[1.3] Inductors and energy storage in inductors

[1.4] Series and parallel inductors

[1.5] Practical capacitors and inductors

▪ In contrast to resistors,
which dissipate energy,
capacitors and inductors
do not dissipate but store
energy, which can be
retrieved at a later time.

▪ For this reason, capacitors

and inductors are called
storage elements.
[1.1] Capacitors and energy storage
▪ A capacitor is a passive element designed to store energy in its electric field.
▪ Capacitors are used extensively in electronics, communications, computers, and
power systems. For example, they are used in the tuning circuits of radio
receivers and as dynamic memory elements in computer systems.
▪ A capacitor consists of two conducting plates separated by an insulator (or
dielectric).
▪ Typical construction of capacitor is shown in Fig. 1.1.

Figure 1.1
A typical capacitor construction.
▪ When a voltage source (v) is connected to the capacitor, as in Fig. 1.2, the
source deposits a positive charge (q) on one plate and a negative charge (–q) on
the other (i.e electrons from the left plate are removed by the source and
deposited onto the right plate).
▪ The left plate becomes positively charged and the right plate becomes negatively
charged. The capacitor is said to store electric charges.

Figure 1.2
A capacitor with applied voltage v.
▪ The amount of charge stored by the capacitor is represented by the positive charge q
stored on the left plate.
▪ The amount of charge stored q is a function of the applied voltage v.
▪ Fig. 1.3 shows that for a linear capacitor, the amount of charge stored, q, is directly
proportional to the applied voltage v. Meanwhile, For a nonlinear capacitor, q is a
nonlinear function of the applied voltage v.

Linear capacitor
q(t)

Nonlinear capacitor
v(t)

Figure 1.3
Typical capacitor charge vs voltage
characteristics
▪ For a linear capacitor, the charge stored, q(t), can be related to the applied voltage v(t) by the
straight-line equation

q(t) = Cv(t) (1.1)

where C is a constant of proportionality and is equal to the gradient of the straight-line characteristic.
▪ The constant C is called the capacitance of the capacitor.
▪ The MKS unit for C is coulomb/volt.
▪ The SI unit for C is the farad (symbol: F).

Note: For convenience, the time t is sometimes dropped, and Eq. (1.1) is alternatively written as

q = Cv (1.2)
▪ The capacitance C depends on the physical dimensions of the capacitor and is a measure of the
capacitor’s ability to store charge.
▪ For a parallel-plate capacitor, the capacitance is given by:
A
C= (1.3)
d
where A = surface area of each plate (m2),
d = distance between the plates (m)
ε = permittivity of dielectric material (F/m)

Figure 1.4
▪ Fig. 1.5 below illustrates the circuit symbols for fixed and variable capacitors.

Figure 1.5 Circuit symbols for:

(a) fixed capacitor
(b) variable capacitor.
 [1.1.1] Current-Voltage Relationship for a Linear Capacitor

▪ Consider the situation where a voltage v(t) is applied to a capacitor C (refer to Fig. 1.6(a)).
▪ Let current i(t) flowing into the capacitor causes the amount of positive charge stored on
the upper plate to increase from q to (q + dq) in the time dt, and the voltage across the
capacitor to increase from v to (v + dv) (refer to Fig. 1.6(b)).

i(t) q + dq
i

V + dv
v(t) q(t)

(a) (b)

Figure 1.6
▪ For a linear capacitor, we have from Eq. (1.2)
q = Cv
▪ Hence, if the voltage increases from v to (v + dv) units, then q increases from q to (q + dq); that is
q + dq = C(v + dv) (1.4)
▪ Thus, subtracting Eq. (1.2) from Eq. (1.4) we obtain
dq = Cdv (1.5)
▪ The charging current i(t) flowing through the capacitor can be found by differentiating Eq. (1.5) with
respect to time. Thus,
dq dv
i(t ) = =C (1.6)
dt dt
▪ Eq.(1.9) is the fundamental equation relating the charging current to the voltage across the
capacitor. Thus, Fundamental current-voltage relationship for a linear capacitor:
dv
i(t ) = C (1.7)
dt
• The exact form of Eq. (1.7) to use in circuit analysis will depend on the reference
directions chosen for both the terminal voltage and the terminal current of the capacitor.

(a) Passive sign convention

i(t)

dv
i-v relationship: i (t ) = C
v(t) q(t)

dt

(b) Active sign convention

i(t)

dv
v(t) q(t) i-v relationship:i (t ) = −C
dt

Figure 1.7 Capacitor i-v relationships according to voltage and current reference directions
 Example
Voltage waveform across a capacitor C = 5
F is as given by Fig. 1.8. Obtain a time plot v(t) (V)

for the charging current, i(t).

24

NOTE: This example illustrates the 0 6 8

(t) (ms)

application of Eq. (1.7).

Figure 1.8
Solution
We start by using the passive sign convention to define the reference directions for the
terminal voltage and terminal current of the capacitor.

i(t)

v(t) C = 5 F

Figure 1.9

Next, we need to find a mathematical expression for the voltage waveform given in Figure
1.8. From the v(t) graph given in Figure 1.8, we can write the voltage as a piecewise
function

24v / 6ms = 4 10 t ,
3
0  t  6ms
v(t ) = 

(−24v /(8 − 6)ms) = −12 10 t + 96, 6ms  t  8ms
3
(1.8)
Solution (Cont.)
For 0  t  6 ms,

i(t ) = C
dv
dt
= 5 10−6
d
dt
( )
4 103 t = 20 mA
(1.9)

For 6ms  t  8ms,

i(t ) = C
dv
dt
= 5 10−6
d
dt
( )
− 12 103 t + 96 = −60 mA
(1.10)

Therefore the graph of i(t) is given by Fig. 1.10 i(t)

20 mA
(t) (ms)
0 6 8

– 60 mA
Figure 1.10
 Example
(a) Calculate the charge stored on a 3 pF capacitor with 20 V across it.
(b) Find the energy stored in the capacitor.

Solution
 [1.1.2] Voltage-Current Relationship for a Linear Capacitor

To find an expression relating the terminal voltage v(t) to the charging current i(t) we can
start by rewriting Eq.(1.7) as

i(t ) dt
1
dv = (1.11)
C
To find the terminal voltage at any time t, we first assume that the current i(t) starts from 0
at time t = - ∞ and gradually increases in value as time progresses, as depicted in Figure
1.11.

i(t) 1
Area dv = i(t )dt
C
-∞ t
0
dt
Figure 1.11
We note from Figure 1.11 that the value of the terminal voltage at time t is given by the area
under the curve from -∞ to t. Hence, to find the value of the terminal voltage at time t we
need to integrate both sides of Eq. (1.11) with respect to time, from - ∞ to t. Mathematically,
we write
t t

−C i( ) d
1
−
 dv = where  is a dummy variable
t
Therefore vt− =  i( )d
1
C −
v(t) - v(-) =  i( )d
1 t
Or (1.12)
C − 

Since physically v(-∞) = 0, we then have

i( )d
1 t (1.13)
v(t) =
C − 

Eq.(1.13) is the fundamental voltage-current relationship for a linear capacitor, relating the
charging current to the terminal voltage of the capacitor.
We can split the integration limits in Eq.(1.13) into two-time intervals, from -∞ to 0 and from 0
to t. Thus, we can rewrite Eq.(1.13) as
0 t
v(t) =  i( )d =  i( )d +  i( )d
1 t 1 1
(1.14)
C − C − C0
Term represents capacitor voltage generated by the current flow from time - until time t = 0. It
is thus the initial value of the capacitor voltage at time t = 0. Hence, it is called the initial voltage
and given the label v(0); that is we write
0
v(0) =  i( ) d
1
(1.15)
C −
Eq. (1.14) can thus be .rewritten as
t
v(t ) = v(0) +  i( ) d
1 (1.16)
C0
For the special case where i(t) = 0 for t  0, then v(0) = 0 and the Eq.(1.14) reduces to the form
t
v(t ) =  i ( ) d
1
(1.17)
C0
The exact form of Eq.(1.17) to use in circuit analysis will depend on the reference directions
chosen for both the terminal voltage and the terminal current of the capacitor.

(a) Passive sign convention

i(t)

t
v-i relationship: v(t ) = i ( ) d
1

v(t) q(t)

C0

(b) Active sign convention

i(t)
t
v-i relationship: v(t ) = − i ( ) d
1
v(t) q(t)
C0

Figure 1.12 Capacitor v-i relationships according to voltage and current reference directions
 Example
The waveform of charging current i(t) that
flows through a 5 μF capacitor is shown in i(t)

Fig. 1.13. Obtain waveform of v(t). Assume

capacitor C is without charge at time, t = 0.
20 mA
(t) (ms)
0 6 8

– 60 mA

Figure 1.13
Solution
For 0  t  6ms
The terminal voltage of the capacitor is related to its terminal current via Eq. (1.18).
t t
v(t ) =  i( ) d = v(0) +  i( ) d
1 1
C − C0
where v(0) = 0 (given that the capacitor is initially uncharged) and i(t) = 20 mA for 0 < t < 6
ms. Therefore

 (20 10 )d

t
v(t ) = 0 +
1 −3

5 10 −6 0

20 10−3 t
= −6
 0
5 10
= 4 103 t

(1.21)
Solution (Cont.)
Between 0 < t < 6 ms, voltage v(t) increases
linearly from 0 at a rate of 4  103 V/s. v(t)

At t = 6 ms, 24 4  103 V/s

( )
v(6 ms ) = 4  10 3 6  10 −3 = 24 V
(t) (ms)
0 6
Graph of v(t) between 0 and 6 ms is as shown
in Fig. 1.14.

Figure 1.14
Solution (Cont.)
For 6ms  t  8ms

Since v(0) = 0, then

t t
v(t ) = v(0) +  i( ) d =  i( ) d
1 1
C0 C0
Since the current varies at different rates for 0< t< 6 ms and for 6 ms < t < 8 ms, we need to
split the integration into two time intervals as follows:
t
v(t ) =  i ( ) d
1
C0
6 ms t
=
1
0 i ( ) d +
1
 i ( ) d
C C 6 ms
Solution (Cont.)
Furthermore, since
6 ms

 i( ) d = 24 V
1
C 0

 (− 60 10 )d
t
Then, 1 −3
v(t ) = 24 +
5  10 −6 6 ms
t
= 24 − 12  103 
6 ms
d

= 24 − 12  103  6 ms
t

= 24 − 12  103 t + 72
= 96 − 12  103 t
Solution (Cont.)
At t = 6 ms, v(t) in Eq.(1.21) takes the value v(t) (V)

( )
v(6 ms ) = 4  10 3 6  10 −3 = 24 V
24

At t = 8 ms, v(t) in Eq.(1.23) takes the value

(t) (ms)

( )
0 6 8
v(8 ms ) = 96 − 12  10 3 8  10 −3 = 0 V

Graph of v(t) is as given by Fig. 1.15

Figure 1.15
 [1.1.3] Stored Energy and Energy Calculation

In the passive sign convention, power Energy stored in capacitor from - to t is

supplied to the capacitor at time t is given by t t
e (t ) =  p ( ) d =  i ( )  v ( ) d
p(t) = v(t)  i(t) − −

 dv 
t

i(t) =  C    v( ) d
− 
d 
t
= C  v( ) dv
−
C
v(t) 0− t
= C  v( ) dv + C  v( ) dv
− 0+

1
2

= 0 + C v 2 ( ) 0+
t

Figure 1.15
= C  v 2 (t ) − C  v 2 0 +
1
2
1
2
( )
For the special case where the capacitor is initially uncharged, then
1
2
( )
C  v 2 0+ = 0

and thus, the energy stored by a capacitor at any time t is given by

e(t ) = C  v 2 (t )
1
(1.18)
2
 Example
The waveform of the current i(t) flowing through a 0.1 μF capacitor is as given in Fig. 1.16
Obtain time plots for v(t), p(t) and e(t). Assume that the capacitor is initially uncharged.

i(t)
i(t)

4 μA v(t) C = 0.1 μF

t
0 0.5 s

Figure 1.16
Solution
From the given current waveform, we obtain a piecewise function for current i(t):
4 A 0  t  0.5 s

i (t ) = 
 t  0.5
 0
Eq. (1.16) allows an expression for the terminal voltage to be found when given the terminal
current i(t). Thus, using Eq. (1.16), we obtain the expression
t
1
v(t ) = v(0) +  i ( )d
C0

 (4  10 )d
t
1 −6
= 0+
0.1  10 −6 0
Solution (Cont.)
Thus, for 0 < t< 0.5 μs, we have
i(t)

v(t ) = 40 0 = 40t

t
(1.19)
4 μA

Eq. (1.19) shows that the voltage rises

.
linearly
with time at a rate of 40 V/s 0 0.5 s
t

At the end of the charging period (at t = 0.5 s), the

voltage across the capacitor is
v(t)

v(0.5) = 40  0.5 = 20 V 20 V

t
0 0.5 s

Figure 1.17
Solution (Cont.)
The power supplied to the capacitor at time t is given by

p(t) = v(t)i(t)

Where 40 A 0  t  0.5 s


i (t ) = 
 t  0.5
 0

and 40t V 0  t  0.5 s


v(t ) = 
 20 V t  0.5

Solution (Cont.)
Hence,
1600 10−6 t W 0  t  0.5 s i(t)


p(t ) = v(t ).i (t ) =  4 μA
 0 t  0.5
 t
0 0.5 s
Thus, during the charging interval the power supplied to
p(t)
the capacitor increases linearly with time at a rate of
1600 x10-6 W per second.
80 μW

At t = 0.5 s, t
0 0.5 s
p(0.5) = 1600 10−6  0.5
= 800 W
Figure 1.20
Solution (Cont.)
The energy stored in the capacitor i(t)
1
e(t ) = Cv(t ) 2 4 μA
2
=  0.1  10 −6  (40t )
1 2
t
2 0 0.5 s

= 80  10 -6 t 2 joules
e(t)

This result shows that the energy stored in the 20 μJ

capacitor increases as a square of the charging
time t. t
0 0.5 s

At t = 0.5 s, e(t ) = 80 10-6 t 2

= 80  10-6  (0.5)2 Figure 1.21
= 20 J
[1.2] Series and parallel capacitors
 [1.2.1] Series-Connected Capacitors

When capacitors are connected in series, the impressed voltage is divided among the capacitors and
the equivalent capacitance CT is less than that of the smallest unit. This is indicated in the analysis
that follows.

a a’ b b’ c c’

C1 C2 C3

V1 V2 V3

VT
Figure 1.22
Capacitors in series.

Consider three capacitors connected in series, as in Fig. 1.22. The capacitors are initially uncharged
so that when a potential VT is applied with the indicated polarity, plate a becomes positively charged
and plate c’ becomes negatively charged. As a result of this charging action, plate a’ takes on an
equal negative charge, while plate c takes on an equal positive charge; in a similar fashion, plates b
and b’ become charged.
Thus all plates in a series capacitor circuit acquire exactly the same charge, QT = Q1 = Q2 =
Q3, and so on.
In accordance with Kirchhoff’s voltage law, the total VT is
VT = V1 + V2 + V3
Q1
where, by Eq. (1.2) V1 =
C1
Q
V2 = 2
C2
Q3
V3 =
C3

QT QT QT QT
Then VT = = + +
CT C1 C2 C3
Dividing this last equation by QT and solving for CT, one obtains
1
CT =
1 1 1
+ +
C1 C2 C3

Rearranging terms, we can write

1 1 1 1
= + + (1.20)
CT C1 C2 C3
For two capacitors in series, the total capacitance is found by the product over the sum
relationship:

C1C2 (1.21)
CT =
C1 + C2
 Example
Three series capacitors are connected to a 120 V source as depicted in Fig. 1.22. If C1 = 10
μF, C2 = 20 μF, and C3 = 30 μF, what is
(a) the total capacitance and
(b) the voltage across each capacitor?
Solution
(a) The total capacitance is calculated (b) For the series combination
from Eq. (1.20).
Q1 = Q2 = Q3 = CT VT
1 1 1 1
= + + = 120 (5.45 10 −6 ) = 6.54 10 −4 C
CT C1 C2 C3
1 1 1 Then
= + + Q1 6.54 10−4
10F 20F 30F V1 = = −6
= 65.4V
C1 10 10
CT = 5.45 F
Q2 6.54 10−4
V2 = = −6
= 32.7V
C2 20 10
Q3 6.54 10−4
V3 = = −6
= 21.8V
C3 30 10
 [1.2.2] Parallel-Connected Capacitors

When capacitors are connected in parallel, the total charge flowing into the combination is divided
among the capacitors and the total capacitance is the sum of the individual capacitances.
Consider the parallel capacitor circuit of Fig. 1.23.

VT C1 C2 C3
Figure 1.23
Capacitors in parallel.

Each capacitor takes a charge given by Q1 = C1V1 Q2 = C2V2 Q3 = C3V3

Where VT = V1 = V2 = V3
The total charge is then

QT = CT VT = Q1 + Q2 + Q3 = C1VT + C2VT + C3VT

Dividing the last equation by VT we obtain an expression for the total parallel capacitance.
This expression is given in general by Eq. 1.5.

CT = C1 + C2 + C3 + ..... + Cn (1.22)
 Example
A 600 V is applied across the capacitor configuration of
Fig. 1.24 where C1 = 12 μF, C2 = 6 μF, and C3 = 30 μF.
Find: C1

(a) The total capacitance,

VT = 600 V
(b) The charge on each capacitor, and
(c) The voltage across each capacitor. C2 C3

Figure 1.24
Solution
(a) C2 // C3 = 6 + 30 = 36 F
The total capacitance is CT =
(12)(36)
= 9 F C1
12 + 36
VT = 600 V

(b) The total charge is

(
QT = CTVT = 9 10−6 (600) )
C2 C3

= 54 10−4 = 5.4 mC
Thus,

Q1 = QT = Q2 + Q3 = 5.4 10 −4 = 5.4 mC

Solution (Cont.)
(c) Then the capacitor voltages are
Q1 5.4 10−3
V1 = = −6
= 4.5  10 2
= 450 V
C1 12 10
V2 = V3 = 600 − 450 = 150 V
Thank You