lec3_voltagecurrent

8/10/2019 lec3_VoltageCurrent

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Fundamentals of

Electronic & Communication Engineering

Danang University of Technology


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Lecture 3

(chapter 3)


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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 divisionappropriately to solve simple circuits

To use a Wheatstone bridge

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


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Series/Parallel Circuits

There are two types of electrical circuits:

SERIES CIRCUITS PARALLEL CIRCUITS


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Series Circuit

The components are connected end-to-end, oneafter the other


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Parallel Circuit

The components are connected side by side


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Resistors In Series

KVL around loop (a):

-vs + i1R1 i2R2 + i3R3i4R4 + i5R5 = 0

R3Vs

i1

+

i2

R1 R2

i3is a

KCL at each node:

is = i1 = -i2 = i3 = -i4 = i5

vs = isR1 + isR2 + isR3 + isR4 + isR5

Or

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

R5 R4

i5 i4


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Combining Resistors In Series

RVs

i1

+

i2

R1 R2

i3is

Vs

+ isReq

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

Req

-

R5 R4

i5 i4

-


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Formula For series

ReqVs

a

+

is

-R3Vs

R1

R2

+

is

-

b

k21

k

1i

ieq R...RRRR ++===

Rk


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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


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Resistors Connected In Parallel

Vs

is a

+-

R1 R2 R3 R4

Ohms law:

i1R1 = i2R2 = i3R3 = i4R4 = vs

KCL at node a:

is = i1 + i2 + i3 + i4

i1 = vs / R1; i2 = vs / R2; i3 = vs / R3 ; i4 = vs / R4

b


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Combining Resistors In Parallel

Vs

is a

+-

R1 R2 R3R4

Vs

+-

is Req

is = i1 + i2 + i3 + i4

i1 = vs / R1; i2 = vs / R2; i3 = vs / R3 ; i4 = vs / R4

b

is = vs(1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5)

1/Req


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Formula For Parallel

ReqVs

a

+

is

-Vs

is a

R1 R2 Rk

+

b

k21eq

R

1...

R

1

R

1

R

1+++=

-

b


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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


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Example

Find is, i1 & i2

is

4 x 3

120V i1 18+-

i2 6

y


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Example

Find is, i1 & i2

is

4 x 3

1

-2

y

is = 120/10 = 12(A)

vxy = 72 (V)

i1 = 72:18 = 4(A); i2 = 72/9 = 8(A)


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Problem 3.1

Find vab

is

a 7.2 x 6

i2 i4

i3 645A i1 30

b y

10


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Find vab

i3

645A

is

i1 30

a 7.2 x 6

i2 i4

10

- Find equivalent resistance

- vab = 5 x 12 = 60(V)

5A

is

12

a

+

-

b

b y

Power Balance ?

- Dissipated power:

- Developed power:


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Find the power dissipated in the 10ohm resistor

is

a 7.2

x 6

i2 i4

P = 10x(2.4)2 = 57.6(W)

i3 645A i1 30

b y

10

Change directions of currents & check its power again ?


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No-load Voltage-Divider Circuit

Vs +-

iR

1

R2 21

s

RRvi+

=

21

2s22

21

1s11

RR

RviRv

RR

RviRv

+

==

+

== v1 and v2 are fraction

of vs

v1 and v2 are alwaysless than vs


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Voltage-Divider Circuit with Load

L2

eq1

eq

s0

RR

RR

Rvv

+

=

Vs +-

iR1

R2 R+V

L221L1

L2s0

RRRRRR

RRvv

++

=

L2

eq

RR +-

2L21

2s0

R)]R/R(1[R

Rvv

++

=

what happen ifchanging value of RL:

Measuring effect


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Example 3.2

25k R

The resistors have a tolerance of10%.

Find the maximum and minimum value of v0

100V +- +

-V0

100k R2


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Example 3.2

Rvv 2=25k R

The resistors have a tolerance of10%.

Find the maximum and minimum value of v0

1R/R

1v

21

s

21

+=

+100V +

- +

-V0

100k R2

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


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Example 3.2

==

=+=

==

k9010100R

k5.275.225R

k5.225.225R

min2

max1

min1

)V(60.76905.27

90100v

)V(02.831105.22

110100v

min0

max0

=+

=

=

+

=

=+= 11010100R max2


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Current-Divider Circuit

is i1 R1 i2 R2

+v

-

21

21

seqs RRiRiv +==

21

1s

2

2

21

2

s1

1

RR

Ri

R

vi

RR

Ri

R

vi

+

==

+

==


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Example 3.3

Find the power dissipated in the 6 resistor

is

a 1.6 x

i2

P = 6 x (i4 )2 find i4 P= 61.44w

10A i1 16 i3 4 6

b y

i4


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Ass. Problem 3.3

Find R that will cause i1 = 4A

is

60

i240

20A i1

80

R


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Ass. Problem 3.3

Find R that will cause i1 = 4A

is

60

i240is

60

i2

20A i1

80

R 20A i1 120 R

)(3016/480R

4120R

)20(Ri

120R

Ri s1

==

=+

=+

=


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Dissipated power in the R resistor ?

is

60

i2

Ass. Problem 3.3

20A i1 120 R


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Dissipated power in the R resistor ?

is

60

i2

Ass. Problem 3.3

20A i1 120 R

)W(7680)16(30)i(RP

)A(1612030

)20(120i

120R

120i

22

s2

===

=+

=+

=


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How much power the current source generate for the circuit?

is

60

i2

Ass. Problem 3.3


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How much power the current source generate for the circuit?

is

60

i2 is

60

Ass. Problem 3.3

20A

is

84

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

Other solutions ? Use KVL


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Voltage/ Current Division

Voltage division and current division are very usefulcircuit analysis tools

Voltage division is used to find the voltage drop across

a single resistor from a set of series-connected resistorswhen we know the voltage drop across the set

Current division is used to find the current through asingle resistor from a set of parallel-connected resistors

when we know the current into the set


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Voltage Division EquationHow to find vj in terms of v ?

vRV

R1 R2

+

iir uit

+

eqn21 Rv

R...RRvi =

+++=

-

Rn Rn-1

-

vRRiRveq

j

jj ==

Voltage division equation


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Current Division EquationHow to find ij in terms of i ?

V

ij+

Circuit R1 R2 Rj Rn

n21

eq

R

1...

R

1

R

1 iiRv+++

== iRR

Rvi

j

eq

j

j ==

Current division equation

-


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ExampleFind i0 and v0

8A

+

v

i0

36

44

10

40

10 24

-+30 v

0-


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ExampleFind i0 and v0

8A

+

v

i0

36

44

10

40

10 24

-+

)A(2)8(24

R

i

eq

0 ==

=+++= 624//)301040//(10//)4436(eqR

)V(482x24v ==

30 v0-

)V(18)48(

301040

30v0 =

++

=


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Problem 3.4Find i0 and i1 ?

60V

+ v0 -

40

20+- 30

50

10

i0

+

V

i1

70 -


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Problem 3.4

60V

+ v0 -

40

20+- 30

50

10

i0

+

V

i1

)V(20)60(

701040

40v0 =

++

=

)(1060/130/120/1

1

)1050//(30//20

=

++

=

+=eqR

)60(70R40

40v

eq

0++

=

70 -

)A(5.0

40

vi 00 ==


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Problem 3.4

60V

+ v0 -

40

20+- 30

50

10

i0

+

V

i1

)mA(67.166)5.0(3010i1 ==

)(10)1050//(30//20 =+=eqR

)i(

30

Ri 0

eq

1 =

70 -


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Problem 3.4

How much power is absorbed by the 50 resistor?

60V

+ v0 -

40

+

50

i0 i1i2

)mW(22.347)50()12/1(P

2==

)A(121)5.0(

6010)i(

1050Ri 0eq

2 ==+

=

70

-


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Measuring Current

Ammeter: instrument to measure current

o In series with circuitcomponents to be measured

VS

R1

R2+

-

P

dArsonval meter

movement

, .

o Practically, make RA as small aspossible

Add parallel resistor RPto increase measuredrange. How ?


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d'Arsonval meter


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Example

Ammeter with range limit 1mA at 50mV

However, we want a full-scale reading of 100mA

(means expanding range 100x) iP

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

P

iA


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Measuring Voltage

Volmeter: instrument to measure voltage

o In parallel with circuitcomponents to be measured

, .

o Practically, make RV as large aspossible

Add serial resistor RS toincrease measuredrange. How ?

VS

R1

R2+

-

RS


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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)


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The Wheatstone Bridge Circuit

R1

a b

R2

Avig

R3 Rx?

DC voltage source

Indicator: a dArsoval meter movement - galvanometer

R1, R2, R3: known resistors (R3: variable)

Rx: unknown resistor


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Using Bridge to Measure ResistanceWe adjust R3 until

ig = 0(balanced bridge)

R1

a b

R2

Avig

i0

3

2

13

x

3x

xx33ab

Ri

iR

i

iR

RiRi0v

==

==

2211 RiRi =

R3 Rx?

3

1

2x R

R

RR =

We also have:


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Ass. Problem 3.7

The bridge is balanced when R1=100, R2=1000, R3=150

Find the value of Rx.

R1R2i0

5V

a b

R3 Rx?

A ig


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Ass. Problem 3.7

The bridge is balanced when R1=100, R2=1000, R3=150

Find the value of Rx.

R1R2i0

5V

a b

R3 Rx?

A ig

)(1500150

100

1000R

R

RR 3

1

2x ===


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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?

i

Ass. Problem 3.7

100

Aig = 0

5V

1000

150 1500

i1 i2


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Ass. Problem 3.7

100

Aig = 0

5V

1000i1 i2

)A(002.015001000

5i

)A(02.0150100

5i

2

1

=+

=

=+

=

1500

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


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to Y Transform

Delta (Pi) circuit: a circuit with three resistors connectedin a shape ( shape)

Wye (Tee) circuit: a circuit with three resistorsconnecte n a s ape s ape

circuit can be transformed into equivalent Y circuit

-to-Y transformation is a very helpful circuit analysis

tool


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Structure

RaRb

Rc

a b

Rb

Rc

Ra

a b

cc

structure is referred to as a structure without

disturbing the electrical equivalence of the two structures


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Y Structure

R2R1

a b

R

R2R1a b

Y structure is referred to as a T structure without

disturbing the electrical equivalence of the two structures

R3

c

c

Th t Y T f ti


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RaRb

Rca b

R2R1

a b

The -to-Y Transformation

c R3

c

31

cba

acbca

32

cba

cbabc

21

cba

bacab

RR

RRR

)RR(RR

RRRRR)RR(RR

RRRRR

)RR(RR

+=

++

+=

+=+++=

+=++

+=

The to Y Transformation


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RaRb

Rca b

R2R1

a b

The -to-Y Transformation

c R3

c

cba

ba3

cba

ac2

cba

cb1

RRR

RRR

RRR

RR

R

RRRRRR

++

=

++=

++=

The Y to Transformation


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RaRb

Rca b

c

R2R1

a b

The Y-to-Transformation

R3

c

3

133221c

2

133221b

1

133221a

R

RRRRRRR

R

RRRRRRR

R

RRRRRRR

++=

++

=

++=

E l f Y A li i


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Example of -to-Y Application

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

100 125

5

40V

40

37.5

25


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100 125

25R1

R2 R3.

4037.5

==

==

==

5.12250

25x125R

10250

25x100R

50250

125x100R

3

2

1


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5

50 )(80

)4010//()5.375.12(505

=

++++=eq

R

40 37.5

1040V 12.5

)W(20)5.0(80P

.

2==

Ass Problem 3 8


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Ass. Problem 3.8

28

1020

Find v ?

1052A

5

-

v

Can I start with KCL & KVL methods ?

Ass Problem 3 8


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Ass. Problem 3.8

28

1020

28

20

105

2A

5

-

vR2R1

R3

2Av

-

375.4120

105x5R;75.8

120

105x10R;

12

5

120

10x5R 321 ======

Ass Problem 3 8


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Ass. Problem 3.828

20

+2A Re

+2A

v

21

R3

v

-

V355.17x2

5.17)28//()20( 213

==

=+++=

v

RRRReq

-


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Study Guide Section 3 1


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Study Guide Section 3.1

a. What two methods can you use to determine whether two resistorsare in series?

b. The equivalent resistance of a collection of series-connectedresistors issmaller 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 inFig. 3.4, the current into the box is 25 A. What resistor can be placed inanother black box so that it is impossible two tell the two black boxesapart?

e. Solve Chapter Problem 3.1

Study Guide Section 3 2


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a. What characteristics of parallel-connected resistors are missing inFig. 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 fourresistors is 2 k. Without doing any computations, you tell the friendthat 2 k cannot possibly be the correct answer. How did you knowthat?

c. Determine the equivalent resistance of the four resistors discussed inpart (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.



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Study Guide Section 3.3a. Describe the type of circuit that is best analyzed using voltage division.b. In words, what is the relationship between the voltage drop across asingle resistor in a collection of series-connected resistors to the sourcevoltage?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 unloadedvoltage 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 isdecreased to 5%. Now what are the maximum and minimum values of v

o

?h. Suppose you want vo in Fig. 3.14 of Example 3.2 to vary no more than1% from its nominal value. What is the largest tolerance allowed for the 25k 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.



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Study Gu de Sect o 3

a. In Fig. 3.18, if the voltage drop across Rj is larger than the voltage dropacross any other resistor, is Rj larger or smaller than all the otherresistors? Use Eq. 3.30 to prove your answer.b. In Fig. 3.19, if the current through Rj is larger than the current throughany other resistor, is Rj larger or smaller than all the other resistors? UseEq. 3.32 to prove your answer.c. Use voltage division, current division, and Ohms law to find the currentand voltage for all of the resistors in Fig. 3.20 of Example 3.4.

d. Solve Assessment Problem 3.4e. Describe the type of circuit that is best analyzed using voltage division.f. In words, what is the relationship between the voltage drop across asingle resistor in a collection of series-connected resistors to the sourcevoltage?g. Define the term load.h. What is the relationship, in words, of the output voltage of an unloadedvoltage 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


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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 analogammeter consists of a meter movement in _______________ with aresistor. The purpose of the resistor is to_____________________________________.

A voltmeter is used to measure _________________. A voltmeteris placed in ____________ with the component it is measuring. An idealvoltmeter behaves like a(n) __________________. A real analog voltmeter

consists of a meter movement in _______________ with a resistor. Thepurpose of the resistor is to _____________________________________.

b. From Example 3.5, what are the two methods that can be used todetermine the effective resistance of an ammeter, Rm?



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y

c. From Example 3.6, what are the two methods that can be used todetermine 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. Theeasiest way to remember the condition for a balanced bridge is to notethat 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 areopposite, as are resistors R2 and R3, so if the bridge is balanced,

R1Rx = R2R2Show that this condition is equivalent to Eq. 3.33.

g. Solve Assessment Problem 3.7 and Chapter Problem 3.49.



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y

a. Redraw the delta-connected resistors and the wye-connectedresistors in Fig. 3.31 by superimposing them. The result should looklike Fig. 3.33. Now use the superimposed resistors to find the patternused in transforming from one type of interconnection to the other. Forexample, from Eqs. 3.44 3.46 we see that a wye-connected resistorequals 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 wye-connected 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 delta-connection of Fig. 3.31, what are the values of the wye-connectedresistors?



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y

c. Notice that in Fig. 3.32 of Example 3.7, there are two wye-connectedsets of resistors: the 100 25 40 resistors on the left and the125 25 37.5 resistors on the right. Repeat the problemstated in Example 3.7, but now replace the wye-connected resistor onthe left with an equivalent set of delta-connected resistors, and thensimplify with series and parallel combinations of resistors.

d. Solve Assessment Problem 3.8 and Chapter Problem 3.53.

lec3_voltagecurrent

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