Download Engineering Science EAB_S_127_Ch3

Engineering Science EAB_S_127
Electricity Chapter 3
Introduction




Potentiometers
Voltage dividers
Measurement of electrical resistance
The Wheatstone bridge
Potentiometers


When a resistive material is connected at both ends to a
voltage source and a sliding bar is moved along its length,
a variable voltage is output depending on the resistance
These devices are commonly called “Potentiometers” and
are typically used in volume controls on audio equipment
0V
Vout
Vin
Potentiometers continued

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It can be shown that the voltage across the bar is a
fraction of the input voltage depending on the ratio of the
input and output resistances.
Hence as I is the same for the entire resistive strip,
Ohm’s Law shows us that
-
+
Vin Vout
I

Rin Rout
I
Vin
Rin
Vout
Rout
Vout
Sliding bar
Rout
 Vin
Rin
Potentiometers: Example


Calculate Rout when we require an output voltage of 10V
from a voltage divider, which has the total resistance of
100 Ω and can supply the maximum voltage 50 V.
Answer:
Vout

Hence
Rout
Rout
 Vin
Rin
Vout
10
 Rin
 100  20
Vin
50
Voltage Dividers

Consider the circuit shown in Figure 3.2 below containing
two discrete resistors. We can develop an equation that
describes the voltage across each resistor R1 and R2 in
terms of the input voltage and a ratio of resistors
V
Vin
+
IT 
R1  R2
in
IT
- V1 +
- V2 +
R1
R2
Figure 3.2 Potential Divider
V1  IT R1
V2  IT R2
Vin
V1 V2


R1 R2 R1  R2
 R1 

V1  Vin 
 R1  R2 
 R2 

V2  Vin 
 R1  R2 
Voltage Dividers: Example

Calculate V1 and V2 when Vin = 24V, R1 = 8Ω and R2 = 40Ω
 R1 

V1  Vin 
 R1  R2 

 R2 

V2  Vin 
 R1  R2 
Answer:
 R1 
 8  24  8


V1  Vin 
 24
 4V


48
 8  40 
 R1  R2 
 R2 
 40  24  40
  24
V2  Vin 
 20V

48
 8  40 
 R1  R2 
The Wheatstone Bridge
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We use an “Ohmmeter” to measure an unknown resistance
The heart of the simplest Ohmmeter is a so-called
“Wheatstone Bridge” circuit
If R1 was a variable resistor, we can adjust it until Vab = 0
The Balanced Wheatstone Bridge

When Vab = 0, a special condition occurs: the bridge is said
to be “balanced”, i.e. Va = Vb

This implies that ig = 0, hence from KCL, i4 = i3 and i2 = i1
Further, from Ohm’s Law; i4R4 = i2R2 and i3R3 = i1R1

The Wheatstone Bridge continued

Hence
i1 R1 i3 R3

i2 R2 i4 R4
R3
R1 
R2
R4
The Wheatstone Bridge: Example

Calculate R1 in a Wheatstone bridge when it is balanced and
when R2 = 300Ω, R3 = 200Ω, R4 = 100Ω .
R3
R1 
R2
R4

Answer:
R3
200
R1 
R2 
300  600
R4
100
Summary
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Learning Outcomes:
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Potentiometers
Voltage dividers
Wheatstone Bridge
Balanced Condition