DC CIRCUITS:

CHAPTER 2

DET 101/3 Basic Electrical Circuit 1

RESISTIVE CIRCUITS

Series/Parallel Equivalent Circuits Voltage Divider Rule (VDR) Current Divider Rule (CDR) Voltage and Current Measurements Wheatstone Bridge Delta (or Pi) and Wye (or Tee) Equivalent

Circuit

Series/Parallel Equivalent Circuits

Most common connection found in circuit analysis.

Circuit simplifying technique. Several resistors are combined to represent a

single equivalent resistance. Equivalent resistance depends on two (2)

factors: Type of connection Point of terminals

Series Equivalent Circuit The equivalent resistance for any number of

resistors in series connection is the sum of each individual resistor.

N

nnNeq RRRRRRRR

154321

R1 R2

RN

R3

R4

R5

Req

x

y

x

y

(2.1)

Series Equivalent Circuit (Continued…)Apparently the single equivalent resistor is

always larger than the largest resistor in the series connection.

Series resistors carry the same current thru them.

Voltage across each of the resistors obtained using voltage divider rule principle or Ohm’s law.

Parallel Equivalent Circuit

The equivalent resistance for any number of resistors in parallel connection is obtained by taking the reciprocal of the sum of the reciprocal of each single resistor in the circuit.

1

1

1

54321

111111

N

n nNeq RR

RRRRRR

x

y

R1 R4R3R2 R5 RN Req

x

y

(2.2)

Parallel Equivalent Circuit (Continued…)Apparently, the single equivalent resistor

is always smaller than the smallest resistor in the parallel connection.

Voltage across each resistor must be the same.

Currents thru each of them are divided according to the current divider rule principle.

**When just two resistors connected in parallel the equivalent resistance is simply the product of resistances divided by its sum.

Parallel Equivalent Circuit (Continued…)Special simplified formula if the number of

resistors connected in series is limited to two elements i.e. N=2.

21

21

RR

RRReq

(2.3)

Special Cases of Connections: Open Circuit (O.C)

An opening exists somewhere in the circuit.

The elements are not connected in a closed path.

R1

R2Vs

R3

i = 0 A

a

b

O.C: i = 0 AKVL: Voc = Vs

Ohm’s Law: Rab = V/I = ∞

Special Cases of Connections: Short Circuit (S.C)

Both of its terminal are joint at one single node.

The element is bypassed.

R1

R2Vs

R3

ii

0 A

a

b

S.C: Rab = 0 Ohm’s Law: i = Vs/(R1 + R3)

: Vsc = 0 V

Practice Problem 2.9

Q: By combining the resistors in Figure below, find Req.

Practice Problem 2.10

Q: Find Rab for the circuit in Figure 2.39.

Practice Problem 2.11

Q: Calculate Geq in the circuit of Figure 2.41.

Voltage Divider Rule (VDR)

Whenever voltage has to be divided among resistors in series use voltage divider rule principle.

R1

R2Vs

R3

+ V1 -

+ V3 -

+ V

2 -sVRR

RV

21

11

sVRR

RV

21

22

VDR (Continued…)

In general, to find the voltage drop across the nth resistor in the voltage divider circuit configuration we use this formula:

sN

nn V

RRRR

RV

321

Where n = 1, 2, 3,.....N

(2.4)

Practice Problem 2.12

Find V1 and V2 in the circuit shown in Figure 2.43. Also calculate i1 and i2 and the power dissipated in the 12 and 40 resistors.

Current Divider Rule (CDR)

Whenever current has to be divided among resistors in parallel, use current divider rule principle.

R1Vs R2

is

i1 i2siRR

Ri

21

21

siRR

Ri

21

12

CDR (Continued…)

Circuit with more than two branches…

R1Vs R2 R2

is

i1 i3i2 siGG

Gi

21

11

sN

nn i

GGGG

Gi

321 n = 1, 2, 3…..N

• In general, for N-conductors the formula represents:

siGG

Gi

21

22

(2.5)

Practice Problem 2.13

Find (a)V1 and V2 (b) the power dissipated in the 3 k and 20 k resistors and (c) power supplied by the current source.

1 k

3 k 5 k10 mA

20 k

+V1-

+V2-

Figure 2.45

Chapter 2, Problem 34

Determine i1, i2, v1, and v2 in the ladder network in Fig. 2.98. Calculate the power dissipated in the 2- resistor.

Chapter 2, Problem 36

Calculate Vo and Io in the circuit of Fig. 2.100.

Voltage and Current MeasurementsTo determine the actual and quantitative

behavior of the physical system. Two most frequently used measuring

devices in the laboratories: Ammeter Voltmeter

Ammeter

Must be placed in series connection with the element whose current is to be measured.

An ideal ammeter should have an equivalent resistance of 0 and considered as short circuit equivalent to the circuit where it is being inserted.

Voltmeter

Must be placed in parallel connection with the elements whose voltage is to be measured.

An ideal voltmeter should have an equivalent resistance of ∞ and considered as open circuit equivalent to the circuit where it is being inserted.

Meter Types Analog meters

Based on the d’Arsonval meter movements. Digital meters

More popular than analog meters. More precision in measurement, less

resistance and can avoid severe reading errors.

Measure the continuous voltage or current at discrete instants of time called sampling times.

Configuration of Voltmeter and Ammeter In A Circuit

R1

R2Vs

A

V

RA = 0

RV = Inf

Wheatstone Bridge : Practical Application of Resistance Measurement

Invented by a British professor, Charles Wheatstone in 1847.

More accurate device to measure resistance in the mid-range (1 to 1 M)

In commercial models of the Wheatstone bridge, accuracies about ± 0.1% are achievable

Wheatstone Bridge (Continued…)

The bridge circuit consists of Four resistors A dc voltage source A detector known as

galvanometer (microampere range)

R1 R2

R3 Rx

Vs GIg

a b

Figure A

Balanced Bridge

If R3 is adjusted until the current Ig in the galvanometer is zero the bridge its balance state.

No voltage drop across the detector which means point a and b are at the same potential.

Implies that V3 = Vx when Ig = 0 A.

Balanced Bridge (Continued…)

Applying the voltage divider rule (VDR):

sVRR

RV

31

33

sx

xx V

RR

RVand

2

Balanced Bridge (Continued…)

Since no current flows through the galvanometer,

sx

xsx V

RR

RV

RR

RVV

231

33

132 RRRR x

31

2 RR

RRx hence (2.6)

Example 1

The galvanometer shows a zero current through it when Rx measured as 5 k. What do you expect to be the value of the adjustable resistor, R3? Show your derivation in getting the formula.

2 k 2k5

R3 Rx

Vs GIg

a b

Exercise 1

The bridge in Figure A is energized by 6V dc source and balanced when R1 = 200, R2 = 500 and R3 = 800.

(a) What is the value of Rx?

(b) How much current (in miliamperes) does the dc source supply?

(c) Which resistor absorbs the least power and which absorbs the most? How much?

Unbalanced Bridge

To find Ig when the Wheatstone bridge is unbalanced, use Thevenin equivalent circuit concept to the galvanometer terminals.

Assuming Rm is the resistance of the galvanometer yields,

mth

thg RR

VI

(2.7)

Delta (or Pi) and Wye (or Tee) Equivalent CircuitStuck with neither series nor parallel

connection of the resistors in part of a circuit.

Simplify the resistive circuit to a single equivalent resistor by means of three-terminal equivalent circuit.

Wye/Tee Circuit

Same type of connections

R2

R1

R3

a

c

b

R3

R2R1

a b

c c

(a) (b)Wye Tee

Delta/Pi Circuit

Same type of Connections

Rc

RaRb

a

c

b

Rc

RbRa

a b

c c

(a) (b)Delta Pi

Delta-to-Wye and Wye-to-Delta TransformationRemember that before and after

transformation using either Wye-to-Delta or Delta-to-Wye, the terminal behavior of the two configurations must retain.

Then only we can say that they are equivalent to each other.

Special Case of -Y Transformation

A special case occur when R1 = R2 = R3 = RY or Ra = Rb = Rc =R under which the both networks are said to be balanced. Hence the transformation formulas will become:

RY = R/3 or R = 3RY By applying Delta/Wye transformations, we may

find that this final process leads to series/parallel connections in some parts of the circuit.

Delta to Wye Transform

To obtain the equivalent resistances in the Wye-connected circuit, we compare the equivalent resistance for each pair of terminals for both circuit configurations.

21:

)()(||:

RRRconnectedY

RRR

RRRRRRRconnected

ab

cba

bacbacab

Delta to Wye Transform(Continued…)To retain the terminal behavior of both

configurations i.e. R = RY

So that,

21

)(RR

RRR

RRRR

cba

bacab

32

)(RR

RRR

RRRR

cba

cbabc

31

)(RR

RRR

RRRR

cba

cabca

(2.8)

(2.9)

(2.10)

Delta to Wye Transform(Continued…) To obtain the resistance values for Y-connected

elements, by straightforward algebraic manipulation and comparisons of the previous three equations gives,

cba

cb

RRR

RRR

1

cba

ac

RRR

RRR

2

cba

ba

RRR

RRR

3

(2.11)

(2.12)

(2.13)

Wye to Delta Transform

By algebraic manipulation, obtain the sum of all possible products of the three Y-connected elements; R1, R2 and R3 in terms of -connected elements; Ra, Rb and Rc.(From Eq. (2.11 – 2.13)

)(

)(

)(2133221

cba

cba

cba

cbacba

RRR

RRR

RRR

RRRRRRRRRRRR

(2.14)

Wye to Delta Transform(Continued…) Then we divide Eq. (2.14) by each of Eq. (2.11) to (2.13)

to obtain each of the -connected elements as to be found variable in your left-side and its equivalent in Y-connected elements.

1

133221

)(

)(R

RRRRRR

RRR

RRRRR

RRR

cba

cb

cba

cba

(2.14) / (2.11) :

Wye to Delta Transform(Continued…)

Using the same manner,1

133221

R

RRRRRRRa

2

133221

R

RRRRRRRb

3

133221

R

RRRRRRRc

(2.14) / (2.12) :

(2.14) / (2.13) :

(2.14) / (2.11) : (2.15)

(2.16)

(2.17)

Wye-Delta Transformations

)(1cba

cb

RRR

RRR

)(2cba

ac

RRR

RRR

)(3cba

ba

RRR

RRR

1

133221

R

RRRRRRRa

2

133221

R

RRRRRRRb

3

133221

R

RRRRRRRc

Delta -> Star Star -> Delta

Superposition of Delta and Wye Resistors “Each resistor in the Y-

connected circuit is the product of the two resistors in two adjacent branches divided by the sum of the three resistors”

“Each resistor in the -connected circuit is the sum of all possible products of Y resistors taken two at a time divided by the opposite Y resistors”

Rc

Ra

Rb

a

c

R2R1

R3

b

Practice Problem 2.15

Q: For the bridge circuit in Fig. 2.54, find Rab and i.

13

100 V

50

10

i a

b Figure 2.54

Exercise 2

Use -to-Y transformation to find the voltages v1 and v2.

1 10

24 V

15

+V1-

+V2-

Exercise 3

Find the equivalent resistance Rab in the circuit below.

Exercise 4

Find Rab in the circuit below.

9 k9 k

9 k 9 k9 k

9 k 9 k

9 k 9 kba