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E E 2315 Lecture 03 - Simple Resistive Circuits and Applications
32

E E 2315

Jan 03, 2016

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E E 2315. Lecture 03 - Simple Resistive Circuits and Applications. Calculating Resistance. When conductor has uniform cross-section. Temperature Coefficient of Resistance. Metallic conductors have a linear increase of resistance with increased temperature. - PowerPoint PPT Presentation
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Page 1: E E 2315

E E 2315

Lecture 03 - Simple Resistive Circuits and

Applications

Page 2: E E 2315

Calculating ResistanceWhen conductor has uniformcross-section

Area,A

l

RA

l

6

6

1.67 10

2.70 10

cu

al

cm

cm

Page 3: E E 2315

Temperature Coefficient of Resistance

Metallic conductors have a linear increase of resistance with increased temperature.

To is the reference temperature (usually 20oC) and Ro is the resistance at the reference temperature. is the temperaturecoefficient of resistance for the material. At 20oC, some values for are:

Material Alpha @ 20oC

Aluminum 0.004308

Copper 0.004041

( ) 1o oR T R T T

Page 4: E E 2315

Resistors in Series

By KCL: Is = I1= I2

By Ohm’s Law: V1 = R1·I1 and V2 = R2·I2

Combine: Vs = R1I1 + R2I2 = (R1 + R2) Is = ReqIs

In General: Req = R1 + R2 +···+ Rn

Vs

R1

R2

+ V1 -Is

I1

I2+V2-

+Vs-

Vs Req

Is

Page 5: E E 2315

Resistors in Parallel (1/2)

By KVL: Vs = V1 = V2 By KCL: Is = I1 + I2

By Ohm’s Law: and

Combine:

Is R1 R2

+V1-

I1I2+

V2-

+Vs-

+Vs-

Is Req

11

1

VI

R 2

22

VI

R

Page 6: E E 2315

Resistors in Parallel (2/2)

For two resistors:

For many resistors:

In terms of conductance:

1 2

1 1 1 1

eq nR R R R

1 2eq nG G G G

Page 7: E E 2315

Voltage Divider Circuit

Vs

R1

R2

+ V1 -

+V2-

MeasureV2

I

1 2

sVIR R

22 2 2

1 2 1 2

ss

V RV I R R V

R R R R

Page 8: E E 2315

Loaded Voltage Divider

Vs

R1

R2 RL

+Vo-

2

2

Leq

L

R RR

R R

2

1 2 2

Lo s

L L

R RV V

R R R R R

Page 9: E E 2315

Voltage Divider Equations

Unloaded:

Loaded:

If RL >> R2:

2

1 2o s

RV V

R R

2

21 21

o s

L

RV V

RR R

R

Page 10: E E 2315

Current Divider Circuit (1/2)

Is G1 G2

+vo-

i1 i2

2 22

1 21 2

1

1 1s s

G Ri I I

G GR R

1 2

1 2 1 2

so

Ii iv

G G G G

Page 11: E E 2315

Current Divider Circuit (2/2)

In general:

If there are onlytwo paths:

Is G1 G2

+vo-

i1 i2

2 1 22

1 21 2

1

1 1s

R R Ri I

R RR R

1 2

nn s

n

Gi I

G G G

Page 12: E E 2315

D’Arsonval Meter Movement

• Permanent Magnet Frame• Torque on rotor proportional to coil

current• Restraint spring opposes electric torque• Angular deflection of indicator

proportional to rotor coil current

S N

Page 13: E E 2315

D’Arsonval Voltmeter• Small voltage rating on movement (~50 mV)• Small current rating on movement (~1 mA)

• Must use voltage dropping resistor, Rv

Rv

+Vd'A

-

+ VRv -+Vx-

Id'A

Page 14: E E 2315

Example: 1 Volt F.S. Voltmeter

Note: d’Arsonval movement has resistance of 50

Scale chosen for 1.0 volt full deflection.

950

+50 mV

-

+ 0.95 V -+1.0 V

-

1 mA

Page 15: E E 2315

Example: 10V F.S. Voltmeter

Scale chosen for 10 volts full deflection.

9950

+50 mV

-

+ 9.95 V -+10 V

-

1 mA

Page 16: E E 2315

D’Arsonval Ammeter• Small voltage rating on movement (~50 mV)• Small current rating on movement (~1 mA)

• Must use current bypass conductor, Ga

Ga

+Vd'A

-

IGa Id'AIx

Page 17: E E 2315

Example: 1 Amp F.S. Ammeter

Note: d’Arsonval movement has conductance of 0.02 S

Scale chosen for 1.0 amp full deflection.

Ga = 19.98 S has ~50.050 m resistance.

19.98 S

+50 mV

-

999 mA 1 mA1.0 A

Page 18: E E 2315

Example: 10 Amp F.S. Ammeter

Scale chosen for 10 amp full deflection.

Ga = 199.98 S has ~5.0005 m resistance.

199.98 S

+50 mV

-

9.999 A 1 mA10 A

Page 19: E E 2315

Measurement Errors

• Inherent Instrument Error• Poor Calibration• Improper Use of Instrument• Application of Instrument Changes

What was to be Measured– Ideal Voltmeters have Infinite

Resistance– Ideal Ammeters have Zero Resistance

Page 20: E E 2315

Example: Voltage Measurement

True Voltage:

(If voltmeter removed)

45 V

400

100 +Vo-

10 kvolt-

meter

10045 9

500oV V V

Page 21: E E 2315

Example: Voltage Measurement

Measured Voltage:

10045 8.9286

100400 1 100

10

oV V

k

Page 22: E E 2315

Another Voltage Measurement (1/2)

True Voltage:

(If voltmeter removed)

45 V

40 k

10 k+Vo-

10 kvolt-

meter

1045 9

50o

kV V V

k

Page 23: E E 2315

Another Voltage Measurement (2/2)

Measured Voltage:

5.0% 1 100% 44.44%

9.0

VError

V

Page 24: E E 2315

Example: Current Measurement (1/2)

True Current:

(If ammeter replaced by short circuit)

5A

100

25 50 mAmmeter

Io

255 1.0

125oI A A

Page 25: E E 2315

Example: Current Measurement (2/2)

Measured Current:

255 0.9996

125.05oI A A

0.9996% 1 100% .04%

1.0

AError

A

Page 26: E E 2315

Another Current Measurement (1/2)

True Current:

(If ammeter replaced by short circuit)

5A

100 m

25 m 50 mAmmeter

Io

255 1.0

125o

mI A A

m

Page 27: E E 2315

Another Current Measurement (2/2)

Measured Current:

255 0.7143

175o

mI A A

m

Page 28: E E 2315

Measuring Resistance

• Indirect– Measure Voltage across Resistor– Measure Current through Resistor– Calculate Resistance (Inaccurate)

• d’Arsonval Ohmmeter– Very Simple– Inaccurate

• Wheatstone Bridge (Most Accurate)

Page 29: E E 2315

D’Arsonval Ohmmeter

Need to adjust Radj and zero setting each scale change.

Vb

Rb

Radj

Rx

Page 30: E E 2315

Ohmmeter Example

10 mA Full Scale (Outer Numbers)

Rb+Radj+Rd’A=150 Vb=1.5 V

Inner (Nonlinear) Scale in Ohms

5

1002.5 7.5

050

150

450

8

Page 31: E E 2315

Wheatstone Bridge Vab= 0 and Iab= 0

Vad = Vbd

R1I1=R2I2

R3I3=RxIx

Vg

Rg R1R2

R3 Rx

a b

c

d

+ Vab -

I1 I2

I3 Ix

Iab

Page 32: E E 2315

Example: Wheatstone Bridge

1 kV

100 150 300

450 900

a b

c

d

RqI

150 300

450 900