e e 2315 lecture 03 - simple resistive circuits and applications

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E E 2315 Lecture 03 - Simple Resistive Circuits and Applications

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E E 2315

Lecture 03 - Simple Resistive Circuits and

Applications

Calculating ResistanceWhen conductor has uniformcross-section

Area,A

l

RA

l

6

6

1.67 10

2.70 10

cu

al

cm

cm

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

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

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

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

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

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

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

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

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

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

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

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

Example: 10V F.S. Voltmeter

Scale chosen for 10 volts full deflection.

9950

+50 mV

-

+ 9.95 V -+10 V

-

1 mA

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

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

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

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

Example: Voltage Measurement

True Voltage:

(If voltmeter removed)

45 V

400

100 +Vo-

10 kvolt-

meter

10045 9

500oV V V

Example: Voltage Measurement

Measured Voltage:

10045 8.9286

100400 1 100

10

oV V

k

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

Another Voltage Measurement (2/2)

Measured Voltage:

5.0% 1 100% 44.44%

9.0

VError

V

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

Example: Current Measurement (2/2)

Measured Current:

255 0.9996

125.05oI A A

0.9996% 1 100% .04%

1.0

AError

A

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

Another Current Measurement (2/2)

Measured Current:

255 0.7143

175o

mI A A

m

Measuring Resistance

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

• d’Arsonval Ohmmeter– Very Simple– Inaccurate

• Wheatstone Bridge (Most Accurate)

D’Arsonval Ohmmeter

Need to adjust Radj and zero setting each scale change.

Vb

Rb

Radj

Rx

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

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

Example: Wheatstone Bridge

1 kV

100 150 300

450 900

a b

c

d

RqI

150 300

450 900