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E E 2315
Lecture 03 - Simple
Resistive Circuits and
Applications
Calculating Resistance
R
 l
A
When conductor has uniform
cross-section
cu  1.67 106   cm
 al  2.70 10   cm
6
Area,
A
l
Temperature Coefficient of Resistance
Metallic conductors have a linear increase of resistance with
increased temperature.
R(T )  Ro 1  a T  To 
To is the reference temperature (usually 20oC) and Ro is the
resistance at the reference temperature. a is the temperature
coefficient of resistance for the material. At 20oC, some
values for a are:
Material
Aluminum
Copper
Alpha @ 20oC
0.004308
0.004041
I1
Is
Vs
Resistors in Series
R1
+ V1 -
I2
Is
+
R2 Vs
V2
-
+
Vs
-
Req
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
Resistors in Parallel (1/2)
Is
+ +
Vs V1
- -
I1
R1
+
V2 R2 Is
-
By KVL: Vs = V1 = V2
By Ohm’s Law:
Combine:
I2
V1
I1 
R1
+
Vs
-
Req
By KCL: Is = I1 + I2
and
V2
I2 
R2
Resistors in Parallel (2/2)
For two resistors:
For many resistors:
1
1
1
 

Req R1 R2
1

Rn
In terms of conductance:
Geq  G1  G2 
 Gn
Voltage Divider Circuit
+ V1 I
Vs
R1
R2
+
V2
-
Measure
V2
Vs
I
R1  R2
Vs
R2
V2  I  R2 
R2 
Vs
R1  R2
R1  R2
Loaded Voltage Divider
Req
R1
R2 RL
R2  RL
Vs
+
R2 Vo
-
R2 RL
Vo  Vs
R1  R2  RL   R2 RL
RL
Voltage Divider Equations
Unloaded:
Loaded:
If RL >> R2:
R2
Vo  Vs
R1  R2
Vo  Vs
R2
 R2

R1 
 1  R2
 RL

Current Divider Circuit (1/2)
Is
+
vo
-
i1
i2
G1
G2
Is
i1
i2
vo 


G1 G2 G1  G2
G2
i2  I s
 Is
1
G1  G2
1
R1
R2
 1
R2
Current Divider Circuit (2/2)
+
vo
-
Is
i1
i2
G1
G2
1
R2
R1 R2
If there are
i

I

2
s
only
1  1 R1 R2
R1
R2
two paths:
In general:
Gn
in  I s
G1  G2 
 Gn
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
Id'A
+ + VRv Vx
-
+
Vd'A
-
Example: 1 Volt F.S. Voltmeter
950  1 mA
+ + 0.95 V 1.0 V
-
+
50 mV
-
Note: d’Arsonval movement has resistance of 50 
Scale chosen for 1.0 volt full deflection.
Example: 10V F.S. Voltmeter
9950  1 mA
+ + 9.95 V 10 V
-
+
50 mV
-
Scale chosen for 10 volts full deflection.
D’Arsonval Ammeter
• Small voltage rating on movement (~50 mV)
• Small current rating on movement (~1 mA)
• Must use current bypass conductor, Ga
Ix
IGa
Ga
+
Vd'A
-
Id'A
Example: 1 Amp F.S. Ammeter
1.0 A
19.98 S
999 mA
+
50 mV
-
1 mA
Note: d’Arsonval movement has conductance
of 0.02 S
Ga = 19.98 S has ~50.050 m resistance.
Scale chosen for 1.0 amp full deflection.
Example: 10 Amp F.S. Ammeter
10 A
199.98 S
9.999 A
+
50 mV
-
1 mA
Ga = 199.98 S has ~5.0005 m resistance.
Scale chosen for 10 amp full deflection.
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
400 
45 V
True Voltage:
100 
+
Vo
-
10 k
voltmeter
100 
Vo  45V
 9V
500 
(If voltmeter removed)
Example: Voltage Measurement
Measured Voltage:
100 
Vo  45V
 8.9286
 100  
400  1 
 100 

 10 k  
Another Voltage Measurement (1/2)
40 k
45 V
True Voltage:
10 k
+
Vo
-
10 k
voltmeter
10 k 
Vo  45V
 9V
50 k 
(If voltmeter removed)
Another Voltage Measurement (2/2)
Measured Voltage:
 5.0V

% Error  
 1100%  44.44%
 9.0V

Example: Current Measurement (1/2)
100 
5A
25 
True Current:
Io
50 m
Ammeter
25 
Io  5 A
 1.0 A
125 
(If ammeter replaced by short circuit)
Example: Current Measurement (2/2)
Measured Current:
25 
Io  5 A
 0.9996 A
125.05 
 0.9996 A 
% Error  
 1100%  .04%
 1.0 A

Another Current Measurement (1/2)
100 m
5A
25 m
True Current:
Io
50 m
Ammeter
25 m
Io  5 A
 1.0 A
125 m
(If ammeter replaced by short circuit)
Another Current Measurement (2/2)
Measured Current:
25 m
Io  5 A
 0.7143 A
175 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
Rb
Vb
Rx
Radj
Need to adjust Radj and zero setting each scale
change.
Ohmmeter Example
5
45
0
2.
5
10 mA Full Scale (Outer Numbers)
Vb=1.5 V
Rb+Radj+Rd’A=150 
Inner (Nonlinear) Scale in Ohms
10
0
0
50
8
5
7.
150
Wheatstone Bridge
Vab= 0 and Iab= 0
c
I1
Rg
R1
R2
Vad = Vbd
+ Vab -
a
Vg
I2
b
Iab
R3
I3
d
R1I1=R2I2
Rx
Ix
R3I3=RxIx
Example: Wheatstone Bridge
150  300 

450  900 
c
100 
150 
300 
I
a
1 kV
450 
Rq
b
900 
d