Transcript chapter 4 - WordPress.com

Chapter 04
DC BRIDGES
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Summary
4.1 Fundamental concept of bridge
circuit
4.2 Principle of DC bridge
4.2.1Wheatstone Bridge and application
4.2.2Kelvin Bridge
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Fundamental Concept of
Bridges Circuit
Bridge - to measure unknown values of resistance
- the bridge circuit works as a pair
of two-component voltage dividers
connected across the same source voltage, with a
null-detector meter movement
connected between them to
indicate a condition of "balance" at zero volts
- In a condition of balance;
R1 R3
or R1R4  R3 R2
R2

R4
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Principle of DC Bridges
Type of DC Bridges
1. Wheatstone Bridges - used for the
measurement of dc resistance
2. Kelvin Bridges - used for the
measurement of low resistance
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Wheatstone bridge
Consists of two parallel resistance
branches
- A voltage source is connected across
the resistance network
- A null detector, usually a galvanometer
is connected between the parallel branches
to detect a balanced condition
R1 & R2 = precision resistors
(standard) ± known
R3 = adjustable resistor (precisely
calibrated) - measured
R4 / RX = unknown ± to be calculated
-
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Wheatstone Bridge

Initially, the bridge is in unbalanced condition;
- galvanometer will give a reading
(current flows through the galvanometer) - V2 ≠ •
Vx
 R3 is carefully adjusted so that no current will flow
through the galvanometer - IG =0 & V2 =Vx
 the bridge is now in balanced condition
Rx
R2
 From the figure,
V2 
 VS  Vx 
 VS
R2  R1
Rx  R3
 In balanced condition, V2 = VS,
 hence, R1Rx = R2R3
 So,
R2  R3
Rx 
R1
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Wheatstone bridge
The bridge is balance when no current through the Galvanometer (Ig =0)
VAB  VAC
or
VBD  VCD
RX
R1
E 
E
RX  R3
R2  R1
RX ( R1  R2 )  R1 ( RX  R3 )
R1 RX  RX R2  R1 RX  R1 R3
RX R2  R1 R3
RX  ( R1 R3 ) / R2
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Example 1
Given value R1 = 1.5KΩ, R2 = 1KΩ, R3 = 3KΩ &
Rx = 2KΩ. Prove the bridges in balance condition.
Solution:
The bridges in balance condition when:
R3 R2  Rx R1
(3k )(1k )  (2k )(1.5k )
3M  3M ( The bridges is in balance condition)
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EXAMPLE 2
Refer to figure below, calculate the Rx when the
bridges balance.
Solution :
Rx  R1
200Ω
R3
R2
200  800
750
Rx  213.33
Rx 
800Ω
750Ω
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Example 3
Calculate the value of unknown resistance at the
Wheatstone bridge in Figure 3, assuming the
bridge to be in balanced condition.
Solution:
R1 = 2kΩ, R2= 10kΩ, R3 = 5kΩ,
and R4 = Rx
R1 R4  R2 R3
R2
Rx  R3
R1
10k  5k
Rx 
2k
Rx  25k
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Example 4
A Wheatstone bridge has R2 = 3.5 kΩ, R3 = 5.51kΩ
and R1 = 7 kΩ.
(a) Calculate for Rx
(b) Determine the measurement range if R3 is
adjusted from 1 kΩ - 8kΩ
Ans:
(a) 2.755 kΩ (b) 500Ω - 4 kΩ
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Example 5
1.
2.
Given the Wheatstone bridge with R1 =
15 kΩ, R2 = 10 kΩ, and R3 = 4.5 kΩ.
Find RX.
Calculate the current through the
Galvanometer in the circuit. Given R1 =
1 kΩ, R2 = 1.6 kΩ, R3 = 3.5 kΩ, R4 =
7.5 kΩ, RG = 200Ω and V = 6V.
Ans: 1. 3KΩ, 2.
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Kelvin Bridge
a modification of Wheatstone’s bridge
 also called a Kelvin double bridge and some
countries Thomson bridge
 used to measure values of resistance below
1Ω
 In low resistance measurement, the resistance
of the leads connecting the unknown
resistance to the terminal of the bridge circuit
may affect the measurement.

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Kelvin bridge schematic diagram
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Bridge Balance Equation for
Kelvin Bridge





Ry - resistance of the connecting leads from R3 to Rx.
Galvanometer can be connected either to point c or to
point a.
connected to point a, the resistance Ry, of the
connecting lead is added to the indication for Rx.
connection is made to point c, Ry is added to the bridge
arm R3 - resulting measurement of Rx is lower than the
actual value. (Actual value of R3 is higher than its
nominal value by the resistance Ry)
If the galvanometer is connected to point b, in between
points c and a, in such a way that the ratio of the
resistance from c to b and that from a to b equals the
ratio of resistances R1 and R2, then
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Bridge Balance Equation for
Kelvin Bridge
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Bridge Balance Equation for
Kelvin Bridge
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Example 5
If in Figure 5 the ratio of Ra to Rb is 1000Ω, R1 is
5Ω and R1 = 0.5R2. What is the value of Rx.
Solution:
R1
Rx  R3
R2
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