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CHAPTER CHAPTER 66

EMT 113: V-2008EMT 113: V-2008

School of Computer and School of Computer and Communication Engineering, UniMAPCommunication Engineering, UniMAP

Prepared By: Prepared By: Amir Razif b. Jamil AbdullahAmir Razif b. Jamil Abdullah

Direct-Direct-Current Current Bridge.Bridge.

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6.1 Introduction to Bridge.6.1 Introduction to Bridge.6.2 The Wheatstone Bridge.6.2 The Wheatstone Bridge.

6.2.1 Sensitivity of the Wheatstone 6.2.1 Sensitivity of the Wheatstone Bridge.Bridge.6.2.2 Unbalance Wheatstone Bridge.6.2.2 Unbalance Wheatstone Bridge.

6.3 Kelvin Bridge.6.3 Kelvin Bridge.

6.0 Direct Current Bridge.6.0 Direct Current Bridge.

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6.1 Introduction to Bridge.6.1 Introduction to Bridge. Bridge circuits are the instruments for making

comparison measurements, are widely used to measure resistance, inductance, capacitance and impedance.

Bridge circuits operate on a null-indication principle, the indication is independent of the calibration of the indicating device or any characteristics of it. It is very accurate.

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The Wheatstone bridge consists of two parallel resistance branches with each branch containing two series resistor elements, Figure 6.1.

A DC voltage source is connected across the resistance network to provide a source of current through the resistance network.

A nul detector is the galvanometer which is connected between the parallel branches to detect the balance condition.

The Wheatstone bridge is an accurate and reliable instrument and heavily used in the industries.

6.2 The Wheatstone 6.2 The Wheatstone Bridge.Bridge.

Figure 6.1: Wheatstone Bridge Figure 6.1: Wheatstone Bridge Circuit.Circuit.

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Operation:Operation: We want to know the value of R4, vary one of the

remaining resistor until the current through the null detector decreases to zero.

The bridge is in balance condition, the voltage across resistor R3 is equal to the voltage drop across R4,(R3 = R4).

Cont’d…Cont’d…

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At balance the voltage drop at R1 and R2 must be equal to.

No current go through the galvanometer G, the bridge is in balance so,

This equation, R1R4 = R2R3 , states the condition for a balance Wheatstone bridge and can be used to compute thevalue of unknown resistor.

31 II

2211 RIRI

4433 RIRI

42 II

3241

4

2

3

1

4231

RRRR

or

R

R

R

R

RIRI

Cont’d…Cont’d…

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Example 6.1:Example 6.1: Wheatstone Bridge. Wheatstone Bridge.Determine the value of unknown resistor, Determine the value of unknown resistor, RRxx in the circuit of in the circuit of Figure 6.2Figure 6.2 assuming a null exist, current through the galvanometer is zero.assuming a null exist, current through the galvanometer is zero.

Solution:Solution:From the circuit, the product of the resistance in opposite arms of the bridge is balance, so solving for Rx

.321 RRRRx

Figure 6.2: Circuit For Figure 6.2: Circuit For Example 6.1.Example 6.1.

KK

KK

R

RRRx

4012

32*151

32

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Example 6.1A(T2 2005):Example 6.1A(T2 2005): Wheatstone Wheatstone Bridge.Bridge.Calculate the value of RCalculate the value of Rxx in the circuit of in the circuit of Figure 4Figure 4 if if VVThTh = 24 mV and Ig =13.6 uA. = 24 mV and Ig =13.6 uA.

Solution:Solution:Calculate RCalculate Rthth

Figure 6.2A: Circuit For Figure 6.2A: Circuit For Example 6.1A.Example 6.1A.

KR

A

mVR

RI

VR

RR

VI

Th

Th

gg

thTh

gTh

thg

665.1

1006.13

24

9

Calculate RCalculate Rxx

.

KR

KK

KKKK

KK

KKKK

R

RR

RRRR

RR

RRRR

R

RR

RRR

R

R

R

RR

RRR

R

R

RR

RRRR

RRRRR

RR

RR

RR

RRR

RR

RR

RR

RRR

RRRRRR

x

x

Th

Th

x

Thx

Thx

x

xThx

x

xTh

x

xTh

xabTh

941.4

51

5*1665.11

51

5*1665.11

1

)(

*)(

////

31

312

31

312

31

31

22

31

31

22

231

312

2

2

31

31

2

2

31

31

231

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When the bridge is in unbalance condition, current flows through the galvanometer causing a deflection of its pointer.

The amount of deflection is a function of the sensitivity of the galvanometer.

Sensitivity is the deflection per unit current. The more sensitive the galvanometer will deflect

more with the same amount of current.

Total deflection D is,

A

radian

A

rees

A

etersmiS

deglim

6.2.1 Sensitivity of the 6.2.1 Sensitivity of the Wheatstone Bridge.Wheatstone Bridge.

ISD *

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6.2.2 Unbalanced 6.2.2 Unbalanced Wheatstone Bridge.Wheatstone Bridge.

The current flows through the galvanometer can determine by using Thevenin theorem.

42

42

31

31

4231 ////

RR

RR

RR

RR

RRRRRR abTh

42

4

31

3

RR

RE

RR

RE

VVV baTh

Figure 6.3: Unbalance Figure 6.3: Unbalance Wheatstone Bridge.Wheatstone Bridge.

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The deflection current in the galvanometer is,

Rg = the internal resistance in the galvanometer

gth

thg RR

VI

Figure 6.4: Thevenin’s Equivalent Circuit for an Unbalanced Figure 6.4: Thevenin’s Equivalent Circuit for an Unbalanced Wheatstone Bridge.Wheatstone Bridge.

Cont’d…Cont’d…

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Example 6.2:Example 6.2: Unbalance Wheatstone Bridge. Unbalance Wheatstone Bridge.Calculate the current through the galvanometer in the circuit Figure 6.5. Given that E=6V, R1= 1kΩ, R2= 1.6kΩ, R3 = 3.5kΩ, R4= 7.5kΩ and Rg=200Ω.

Solution:Solution:(1) Find Thevenin equivalent circuit as seen from by the galvanometer,Vth is,

Figure 6.5: Circuit for Example 6.2.

VV

KK

K

KK

KV

RR

R

RR

REVTh

276.0824.0778.06

6.15.7

5.7

15.3

5.36

24

4

13

3

14

(2) Find Thevenin’s equivalent resistance (Rth )is,

.

.

K

KK

KK

KK

KK

RR

RR

RR

RRRTh

097.2

5.76.1

5.7*6.1

5.31

5.3*16

42

42

31

31

Figure 6.6: Thevenin’s Equivalent Circuit for the Figure 6.6: Thevenin’s Equivalent Circuit for the Example 6.2 Unbalance Bridge.Example 6.2 Unbalance Bridge.

Cont’d…Cont’d…

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Example 6.3:Example 6.3: Slightly Unbalanced Wheatstone Bridge.Slightly Unbalanced Wheatstone Bridge.Use the approximate equation to calculate the Use the approximate equation to calculate the current current through the through the galvanometer in galvanometer in Figure 6.7Figure 6.7. The galvanometer resistance, R. The galvanometer resistance, Rgg is is 125Ω and is center-zero 200-0-200-uA movement.125Ω and is center-zero 200-0-200-uA movement.E=10V, RE=10V, R11=500Ω, R=500Ω, R22=500Ω, R=500Ω, R33 = 500 Ω and R = 500 Ω and R44=525 Ω.=525 Ω.

Solution:Solution:From formula,

(1) Find Thevenin equivalent voltage (Vth) is,

(2) Find Thevenin equivalent resistance (Rth )is,

VVR

rVTh 125.0

2000

25*10

4

Figure 6.7: Circuit for Example 6.3.

500RRTh

gTh

Thg RR

VI

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(3) The current through the galvanometer (Ig)is,

Observation: If the deflector is a 200-0-200-uA galvanometer, the pointer deflected full scale for a 5% change in resistance.

.

.

A

V

RR

VI

gTh

Thg

200

125500

125.0

Cont’d…Cont’d…

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The Kelvin Bridge is the modified version of the Wheatstone Bridge.

The modification is done to eliminate the effect of contact and lead resistance when measuring unknown low resistance.

By using Kelvin bridge, resistor within the range of 1 Ω to approximately 1uΩ can be measured with high degree of accuracy.

Figure 6.8 is the basic Kelvin bridge. The resistor Ric represent the lead and contact resistance present in the Wheatstone bridge.

6.3 Kelvin Bridge.6.3 Kelvin Bridge.

Figure 6.8: Basic Kelvin Bridge.

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The second set of Ra and Rb compensates for this relatively low lead contact resistance .

At balance the ratio of Ra and Rb must be equal to the ratio of R1 to R3.

a

bx

x

x

R

R

R

R

R

R

R

R

R

R

R

RRR

1

3

2

1

3

2

1

32

Cont’d…Cont’d…

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Example 6.4:Example 6.4: Kelvin Bridge. Kelvin Bridge.Figure 6.9Figure 6.9 is the Kelvin Bridge, the ratio of R is the Kelvin Bridge, the ratio of Raa to R to Rbb is 1000. R is 1000. R11 is 5 Ohm and R is 5 Ohm and R11 =0.5 =0.5 RR22. . Find the value of Find the value of RRxx..

Solution:Solution:Calculate the resistance of Rx,

R1 =0.5 R2, so calculate R2

Calculate the value of Rx

.

01.01000

110

1000

12RRx

1000

1

52

ax R

R

R

105.0

5

5.01

2

RR

Figure 6.9: For Example 6.4.