Fundamentals of

Electrical Engineering
Electronic & Communication Engineering
Danang University of Technology

Lecture 3
Simple Resistive Circuits
(chapter 3)

Preview
To recognize resistors connected in series and in
parallel
To know how to design simple voltage divider and
current divider.
To use voltage division and current
appropriately to solve simple circuits

division

To use a Wheatstone bridge

To use delta-to-wye equivalent circuits to solve simple
circuits

Series/Parallel Circuits
There are two types of electrical circuits:

SERIES CIRCUITS

PARALLEL CIRCUITS

Series Circuit

The components are connected end-to-end, one

after the other

Parallel Circuit

The components are connected side by side

Resistors In Series
R1

R2

KVL around loop (a):

i1
Vs

+
-

i2
i3

is
R5
i5

R4

R3

-vs + i1R1 i2R2 + i3R3

i4R4 + i5R5 = 0
KCL at each node:
is = i1 = -i2 = i3 = -i4 = i5

i4

vs = isR1 + isR2 + isR3 + isR4 + isR5

Or
vs = is(R1 + R2 + R3 + R4 + R5)

Combining Resistors In Series

Vs

+
-

R1

R2

i1

i2

is

Vs
i3

a
R5

R4

i5

i4

R3

+
-

vs = is(R1 + R2 + R3 + R4 + R5)
Req

is

Req

Formula For series

is

R1

is

R2

+
Vs
-

R3
Rk

+
Vs
-

Req
b

R eq = R i = R 1 + R 2 + . . . R k
i =1

Series Circuit Properties

A single closed loop through the circuit
The current is the same everywhere in the circuit
Each component provides resistance and total
resistance is the sum of the component resistance
Voltage divides among the components
Voltage dropping across each device is iRcomponent
Add component, higher total resistance

Resistors Connected In Parallel

is

Vs
+
-

KCL at node a:

R1

R2

R3

R4

is = i1 + i2 + i3 + i4
Ohms law:
i1R1 = i2R2 = i3R3 = i4R4 = vs
i1 = vs / R1; i2 = vs / R2; i3 = vs / R3 ; i4 = vs / R4

Combining Resistors In Parallel

is

a
Vs

Vs
+
-

R1
R4

R2

+
-

R3

is = i1 + i2 + i3 + i4
i1 = vs / R1; i2 = vs / R2; i3 = vs / R3 ; i4 = vs / R4
is = vs(1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5)
1/Req

is

Req

Formula For Parallel

is

is

+
Vs

R1

R2

Rk

+
Vs
-

Req
b

1
1
1
1
=
+
+ ... +
R eq R 1 R 2
Rk

Parallel Circuit Properties

Each component connects to the voltage source
Voltage is the same across each component
Current from source divides into components
Total current is the sum of component currents
Current in each component is just v/Rcomponent
Add component, lower total resistance

Example
Find is, i1 & i2
4

is
120V

+
-

i1

18

i2

Example
Find is, i1 & i2

is
120V

+
-

i1

18

i2

is = 120/10 = 12(A)
vxy = 72 (V)
i1 = 72:18 = 4(A); i2 = 72/9 = 8(A)

Problem 3.1
Find vab

7.2

is
5A

i2
i1

30

i4
i3

64

10

7.2

is
5A

i2
i1

30

i4
i3

64

10

Find vab
b

- Find equivalent resistance

is

- vab = 5 x 12 = 60(V)
Power Balance ?
- Dissipated power:
- Developed power:

+
12

5A
b

Find the power dissipated in the 10ohm resistor

a

7.2

is
5A

i2
i1

30

i4
i3

64

10

P = 10x(2.4)2 = 57.6(W)
Change directions of currents & check its power again ?

No-load Voltage-Divider Circuit

i
Vs

+
-

R1
R2

R1
v1 = iR 1 = v s
R1 + R 2
R2
v 2 = iR 2 = v s
R1 + R 2

vs
i=
R1 + R 2

v1 and v2 are fraction

of vs
v1 and v2 are always
less than vs

Voltage-Divider Circuit with Load

i
Vs

+
-

v 0 = vs

R1
R2

+
V0
-

RL

R eq
R 1 + R eq

R 2R L
R eq =
R2 + RL

R 2R L
v 0 = vs
R 1R L + R 1R 2 + R 2 R L

R2
v0 = vs
R 1[1 + (R 2 / R L )] + R 2

what happen if
changing value of RL:
Measuring effect

Example 3.2
The resistors have a tolerance of 10%.
Find the maximum and minimum value of v0

100V

+
-

25k

R1

100k

R2

+
V0
-

Example 3.2
The resistors have a tolerance of 10%.
Find the maximum and minimum value of v0

100V

+
-

25k

R1

100k

R2

R2
v0 = vs
R1 + R 2
+
V0
-

1
= vs
R1 / R 2 + 1

Maximum value of v0 occurs when R1 is min and R2 is max

Minimum value of v0 occurs when R1 is max and R2 is min

Example 3.2
R 1min = 25 2.5 = 22.5k
R 1max = 25 + 2.5 = 27.5k
R 2 min = 100 10 = 90k
R 2 max = 100 + 10 = 110

110
v 0 max = 100
= 83.02(V)
22.5 + 110
90
v 0 min = 100
= 76.60(V)
27.5 + 90

Current-Divider Circuit
+
is

v
-

i1

R1

i2

R2

R 1R 2
v = i s R eq = i s
R1 + R 2

v
R2
i1 =
= is
R1
R1 + R 2
v
R1
i2 =
= is
R2
R1 + R 2

Example 3.3
Find the power dissipated in the 6 resistor
a

1.6

is
10A

i2
i1

16

i3

P = 6 x (i4 )2

i4

find i4

P= 61.44w

Ass. Problem 3.3

Find R that will cause i1 = 4A
60
is
20A

40
i1

i2
R

80

Ass. Problem 3.3

Find R that will cause i1 = 4A
60

60
is
20A

40
i1

i2

is

i2
R

20A

80

R
R (20)
i1 =
is =
=4
R + 120
R + 120
R = 480 / 16 = 30()

i1

120

Ass. Problem 3.3

Dissipated power in the R resistor ?
60
i2

is
20A

i1

120

Ass. Problem 3.3

Dissipated power in the R resistor ?
60
i2

is
20A

i1

120

120
120(20)
i2 =
is =
= 16(A)
R + 120
30 + 120
2
2
P = R (i) = 30(16) = 7680( W )

Ass. Problem 3.3

How much power the current source generate for the circuit?
60
i2

is
20A

i1

120

30

Ass. Problem 3.3

How much power the current source generate for the circuit?
60

60
i2

is
i1

20A

120

is
20A

is
30

20A

24

P = (20)2 x 84 = 33600(W)
84

Other solutions ? Use KVL

Voltage/ Current Division

Voltage division and current division are very useful

circuit analysis tools
Voltage division is used to find the voltage drop across
a single resistor from a set of series-connected resistors
when we know the voltage drop across the set
Current division is used to find the current through a
single resistor from a set of parallel-connected resistors
when we know the current into the set

Voltage Division Equation

How to find vj in terms of v ?
R1

R2

Circuit

+
i

V
-

Rj

vj
-

Rn

v
v
i=
=
R 1 + R 2 + ... + R n R eq

Rn-1

v j = iR j =

Rj
R eq

Voltage division equation

Current Division Equation

How to find ij in terms of i ?
+

ij

Circuit

R1

R2

Rj

Rn

V
-

v = iR eq =

i
1
1
1
+
+ ... +
R1 R 2
Rn

v R eq
ij =
=
i
Rj Rj
Current division equation

Example
Find i0 and v0
i0

40
36
10

8A
44

10

+
30 v0
-

24

+
v
-

Example
Find i0 and v0
i0

40
36
10

8A
44

+
30 v0
-

Req = (36 + 44) // 10 //(40 + 10 + 30) // 24 = 6

v = 24x 2 = 48(V)

24

10

i0 =

+
v
-

R eq
24

(8) = 2(A)

30
v0 =
(48) = 18(V)
40 + 10 + 30

Problem 3.4
i0

40

Find i0 and i1 ?

+ v0 60V

+
-

20
70

50
i1

30

V
-

10

Problem 3.4
i0

40
+ v0 60V

+
-

20
70

40
v0 =
(60)
40 + R eq + 70

50
i1

30

10

40
v0 =
(60) = 20(V)
40 + 10 + 70

Req = 20 // 30 //(50 + 10)

1
=
= 10()
1 / 20 + 1 / 30 + 1 / 60

v0
i0 =
= 0.5(A)
40

Problem 3.4
i0

40
+ v0 60V

+
-

20
70

i1 =

R eq
30

(i 0 )

Req = 20 // 30 //(50 + 10) = 10()

50
i1

30

10

10
i1 = (0.5) = 166.67(mA)
30

Problem 3.4
How much power is absorbed by the 50 resistor?
40

50

+ v0 - i0
60V

+
-

20

i1
30

i2

10

70

R eq

10
1
i2 =
(i 0 ) = (0.5) = (A)
50 + 10
60
12

P = (1 / 12) 2 (50) = 347.22(mW )

Measuring Current
Ammeter: instrument to measure current
o

o
o

In series with circuit

components to be measured
Ideally, RA = 0.
Practically, make RA as small as
possible
Add parallel resistor RP
to increase measured VS
range. How ?

RP
R1
dArsonval meter
movement

+
-

R2

d'Arsonval meter

Example
Ammeter with range limit 1mA at 50mV
However, we want a full-scale reading of 100mA
(means expanding range 100x)

iP
RP

Internal resistance of ammeter

RA = 50mV / 1mA = 50
We want 50mV at 100mA total current
iA = 1mA through ammeter & iP = 99mA through RP
RP = 50mV / 99mA = 0.505

iA

Measuring Voltage
Volmeter: instrument to measure voltage
o

o
o

In parallel with circuit

components to be measured
Ideally, RV = .
Practically, make RV as large as
possible
Add serial resistor RS to
increase measured
VS
range. How ?

+
-

R1

R2
RS

The Wheatstone Bridge

The Wheatstone bridge is a circuit that is used to:
- Measure resistance
- Precisely measure resistances of medium values, that is,
in the range of 1 to 1M
The Wheatstone bridge consists of:
-

4 resistors

a dc voltage source

a detector (often ammeter)

The Wheatstone Bridge Circuit

R2

R1

A
R3

ig
Rx

?
DC voltage source
Indicator: a dArsoval meter movement - galvanometer
R1, R2, R3: known resistors (R3: variable)
Rx: unknown resistor

Using Bridge to Measure Resistance

i0

We adjust R3 until

R2

R1

ig = 0
v

A
R3

ig

(balanced bridge)

Rx

v ab = 0 i 3 R 3 = i x R x
i3
i1
Rx = R3 = R3
ix
i2
We also have:

i1R 1 = i 2 R 2

R2
Rx =
R3
R1

Ass. Problem 3.7

The bridge is balanced when R1=100, R2=1000, R3=150
Find the value of Rx.
i0

R2

R1

5V
a

A
R3

ig
Rx

Ass. Problem 3.7

The bridge is balanced when R1=100, R2=1000, R3=150
Find the value of Rx.
i0

R2

R1

5V
a

A
R3

ig
Rx

R2
1000
Rx =
R3 =
150 = 1500()
R1
100

Ass. Problem 3.7

Suppose each resistor is capable of dissipating 250mW. Can
the described bridge be left in the balanced state without
exceeding the power-dissipating capacity of the resistors?
i0

100

i1 i2

1000

5V
A
150

ig = 0
1500

Ass. Problem 3.7

5
i1 =
= 0.02(A)
100 + 150
5
i2 =
= 0.002(A)
1000 + 1500

100

i1 i2

1000

5V
A
150

P1 = 100(0.02)2 = 40mW < 250mW

P2 = 1000(0.002)2 = 4mW < 250mW
P3 = 150(0.02)2 = 60mW < 250mW
P4 = 1500(0.002)2 = 6mW < 250mW

ig = 0
1500

 to Y Transform
Delta (Pi) circuit: a circuit with three resistors connected
in a shape ( shape)
Wye (Tee) circuit: a circuit with three resistors
connected in a Y shape (T shape)
circuit can be transformed into equivalent Y circuit
-to-Y transformation is a very helpful circuit analysis
tool

 Structure
Rc
Rc

Ra

Rb

b
Rb

Ra

structure is referred to as a structure without

disturbing the electrical equivalence of the two structures

Y Structure
a

R1

R2

R2

R1

R3

R3

c
c
Y structure is referred to as a T structure without
disturbing the electrical equivalence of the two structures

The -to-Y Transformation

a
Rc

R2

R1
Ra

Rb

R3

R ab

R c (R a + R b )
=
= R1 + R 2
Ra + Rb + Rc

R bc

R a (R b + R c )
=
= R2 + R3
Ra + Rb + Rc

R ca

R b (R c + R a )
=
= R1 + R3
Ra + Rb + Rc

The -to-Y Transformation

a
Rc

R2

R1
Ra

Rb

R bR c
R1 =
Ra + Rb + Rc
R2

R cR a
=
Ra + Rb + Rc

R aR b
R3 =
Ra + Rb + Rc

R3

The Y-to- Transformation

a

b
R2

R1

Rc

Ra

Rb

c
R3

Ra

R 1R 2 + R 2 R 3 + R 3 R 1
=
R1

Rb

R 1R 2 + R 2 R 3 + R 3 R 1
=
R2

R 1R 2 + R 2 R 3 + R 3 R 1
Rc =
R3

Example of -to-Y Application

Find the current and power supplied by the 40V source:
5
100

40V

125
25

40

37.5

100

125
R1

25

40

R2

R3

37.5

100 x 125
R1 =
= 50
250
100 x 25
R2 =
= 10
250
125 x 25
R3 =
= 12 . 5
250

40

37.5

Req = 5 + 50 + (12.5 + 37.5) //(10 + 40)

5
50

40V

= 80()

10

12.5

40

37.5

i = 40 / 80 = 0.5(A)
P = 80(0.5) 2 = 20( W )

Ass. Problem 3.8

Find v ?
28

10

20

2A

+
v

105

Can I start with KCL & KVL methods ?

Ass. Problem 3.8

28

10

20

2A

28

20

+
v
-

2A
5

105

+
v
-

R1

R2
R3

5x10 5
10x105
5x105
R1 =
= ;R2 =
= 8.75; R 3 =
= 4.375
120 12
120
120

Ass. Problem 3.8

28

20

+
2A

+
2A
v
-

v
R1

R2

R3

Req = R3 + (20 + R1 ) //( 28 + R2 ) = 17.5

v = 2 x 17.5 = 35V

Req

Study Guide Section 3.1

a. What two methods can you use to determine whether two resistors
are in series?
b. The equivalent resistance of a collection of series-connected
resistors is
smaller than / larger than / the same as (circle one)
the value of the largest single resistor.
c. Define the term black box.
d. If 100 V is applied to the black box containing the seven resistors in
Fig. 3.4, the current into the box is 25 A. What resistor can be placed in
another black box so that it is impossible two tell the two black boxes
apart?
e. Solve Chapter Problem 3.1

Study Guide Section 3.2

a. What characteristics of parallel-connected resistors are missing in
Fig. 3.6?
b. Four parallel-connected resistors have the values 1.5 k, 3 k, 4 k,
and 6 k. A friend tells you that the equivalent resistance of these four
resistors is 2 k. Without doing any computations, you tell the friend
that 2 k cannot possibly be the correct answer. How did you know
that?
c. Determine the equivalent resistance of the four resistors discussed in
part (b).
d. State in words the equivalent resistance of two resistors in parallel.
e. Show that the solution in Example 3.1 satisfies
i) KCL at each node;
ii) KVL around each loop (there are three loops);
iii) The requirement for power balance.
f. Solve Assessment Problem 3.1 and Chapter Problems 3.2 and 3.6.

Study Guide Section 3.3

a. Describe the type of circuit that is best analyzed using voltage division.
b. In words, what is the relationship between the voltage drop across a
single resistor in a collection of series-connected resistors to the source
voltage?
c. In the voltage divider circuit of Fig. 3.12, if v2 > v1 is R2 > R1 or is R1 >
R2? Why?
d. In the current divider circuit of Fig. 3.15, if i2 > i1 is R2 > R1 or is R1 > R2?
Why?
e. Define the term load.
f. What is the relationship, in words, of the output voltage of an unloaded
voltage divider to the output voltage of a loaded voltage divider?
g. Suppose the tolerance on the resistors in Fig. 3.14 of Example 3.2 is
decreased to 5%. Now what are the maximum and minimum values of vo?
h. Suppose you want vo in Fig. 3.14 of Example 3.2 to vary no more than
1% from its nominal value. What is the largest tolerance allowed for the 25
k and 100 k resistors?
i. Calculate the current in the rest of the resistors in Fig. 3.17 of Example
3.3. Solve Assessment Problems 3.2 and 3.3.

Study Guide Section 3.4

a. In Fig. 3.18, if the voltage drop across Rj is larger than the voltage drop
across any other resistor, is Rj larger or smaller than all the other
resistors? Use Eq. 3.30 to prove your answer.
b. In Fig. 3.19, if the current through Rj is larger than the current through
any other resistor, is Rj larger or smaller than all the other resistors? Use
Eq. 3.32 to prove your answer.
c. Use voltage division, current division, and Ohms law to find the current
and voltage for all of the resistors in Fig. 3.20 of Example 3.4.
d. Solve Assessment Problem 3.4
e. Describe the type of circuit that is best analyzed using voltage division.
f. In words, what is the relationship between the voltage drop across a
single resistor in a collection of series-connected resistors to the source
voltage?
g. Define the term load.
h. What is the relationship, in words, of the output voltage of an unloaded
voltage divider to the output voltage of a loaded voltage divider?
i. Recalculate the maximum and minimum values of the output voltage in
Example 3.2 if the resistors have a tolerance of 2%.

Study Guide Section 3.5 & 3.6

a. Fill in the blanks:
An ammeter is used to measure _________________. An
ammeter is placed in ____________ with the component it is measuring.
An ideal ammeter behaves like a(n) __________________. A real analog
ammeter consists of a meter movement in _______________ with a
resistor. The purpose of the resistor is to
_____________________________________.
A voltmeter is used to measure _________________. A voltmeter
is placed in ____________ with the component it is measuring. An ideal
voltmeter behaves like a(n) __________________. A real analog voltmeter
consists of a meter movement in _______________ with a resistor. The
purpose of the resistor is to _____________________________________.
b. From Example 3.5, what are the two methods that can be used to
determine the effective resistance of an ammeter, Rm?

Study Guide Section 3.5 & 3.6

c. From Example 3.6, what are the two methods that can be used to
determine the effective resistance of a voltmeter, Rm?
d. Solve Assessment Problems 3.5 and 3.6.
e. What is a galvanometer?
f. When Eq. 3.33 is satisfied, we say that the bridge is balanced. The
easiest way to remember the condition for a balanced bridge is to note
that if Eq. 3.33 is satisfied, the product of one set of opposite resistors
equals the product of the other set. In Fig. 3.25, resistors R1 and Rx are
opposite, as are resistors R2 and R3, so if the bridge is balanced,
R 1R x = R 2R 2
Show that this condition is equivalent to Eq. 3.33.
g. Solve Assessment Problem 3.7 and Chapter Problem 3.49.

Study Guide Section 3.7

a. Redraw the delta-connected resistors and the wye-connected
resistors in Fig. 3.31 by superimposing them. The result should look
like Fig. 3.33. Now use the superimposed resistors to find the pattern
used in transforming from one type of interconnection to the other. For
example, from Eqs. 3.44 3.46 we see that a wye-connected resistor
equals the product of the two delta-connected resistors on either side,
divided by the sum of the three delta-connected resistors. What is the
pattern used to calculate a delta-connected resistor from the wyeconnected resistors? (Use Eqs. 3.47 3.49.)
b. If R1 = R2 = R3 in the wye-connection of Fig. 3.31, what are the
values of the delta-connected resistors? If Ra = Rb = Rc in the deltaconnection of Fig. 3.31, what are the values of the wye-connected
resistors?

Study Guide Section 3.7

c. Notice that in Fig. 3.32 of Example 3.7, there are two wye-connected
sets of resistors: the 100 25 40 resistors on the left and the
125 25 37.5 resistors on the right. Repeat the problem
stated in Example 3.7, but now replace the wye-connected resistor on
the left with an equivalent set of delta-connected resistors, and then
simplify with series and parallel combinations of resistors.
d. Solve Assessment Problem 3.8 and Chapter Problem 3.53.