Basic Circuit of Wheatstone Bridge
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Figure 1 : Wheatstone Bridge Circuit
Wheatstone can be used to accurately measure unknown resistance values by the use of a variable resistance and a simple mathematical formula.
The Wheatstone Bridge can be used to determine very low values of resistances down in the milli-Ohms (mΩ) range to be measured.
The Wheatstone bridge circuit can be used with modern operational amplifiers to interface various transducers and sensors.
A Wheatstone bridge circuit has two input terminals and two output terminals consisting of four resistors configured in a diamond-like arrangement as shown in the figure.
Using The Wheatstone Bridge
Figure 2 : Basic Wheatstone Bridge with one transducer
Figure 3 : Basic Wheatstone Bridge with one resistor
PURPOSE : -
PURPOSE : -
It is used to measure an unknown electrical resistance by balancing two legs of a bridge circuit. One of its legs include the unknown component.
It is used with transducer to measure physical quantities like temperature, pressure, strain etc.
Two voltage dividers are the fundamental concept. Both were fed by the same input.
The circuit output is taken from both voltage divider outputs. A galvanometer (DC current meter) is connected between the output terminals. It is used to monitor the current flows from one voltage divider to the other.
The bridge is said to be balanced if the two voltage dividers have the same ratio and no current flows in either direction through the galvanometer.
Wheatstone Bridge At Balanced Condition
Figure 4 : Wheatstone Bridge Circuit
Using voltage divider rule : -
VR3 = VDC = (R3/R1+R3)(VS)
VR4 = VBC = (R4/R2+R4)(VS)
Now, for balanced condition, the voltage across the R1 and R2 is equal. Then : -
V1 = V2
Similarly, the voltage across R3 and R4 are also equal. So,
V3 = V4
The ratios of the voltage can be written as : -
V1/V3 = V2/V4
Therefore, from Ohm’s Law we get : -
I1 R1 / I3 R3 = I2 R2 / I4 R4
Since I1 = I3 and I2 = I4, So : -
R1 / R3 = R2 / R4
Find The Unknown Resistance Using The Balanced Wheatstone Bridge
Figure 5 : Wheatstone Bridge Circuit (Linear)
In the above circuit, let us assume that R1 is an unknown resistor. So, let us call it RX. The resistors R2 and R4 have a fixed value. Which means, the ratio R2 / R4 is also fixed. Now, from the above calculation, to create a balanced condition, the ratio of resistors must be equal i.e.,
Rx / R3 = R2 / R4
Since the ratio R2 / R4 is fixed, we can easily adjust the other known resistor (R3) to achieve the above condition. Hence, it is important that R3 is a variable resistor, which we call RV.
But how do we detect the Balanced Condition? This is where a Galvanometer (an old school Ammeter) can be used. By placing the Galvanometer between the points A and B, we can detect the Balanced Condition.
With RX placed in the circuit, adjust the RV until the Galvanometer points to 0. At this point, note down the value of RV. By using the following formula, we can calculate the unknown resistor RX.
Rx = Rv (R2 / R4)
Example with Solution : Balanced Wheatstone Bridge
Question : What is the value of x when the Wheatstone’s network is balanced?
Given : P = 500 Ω, Q = 800 Ω, R = x + 400 Ω, S = 1000 Ω
Figure 6 : Wheatstone Bridge Circuit
Solution : -
P/Q = R/S
x + 400 = 0.625 × 1000
x + 400 = 625
x = 625 – 400
x = 225 Ω
Learning Video on Youtube : Balanced Wheatstone Bridge
For those who are visual or kinesthetic learners, kindly check out this video which may assist you on how to solve balanced wheatstone bridge step-by-step with some explanation regarding to how strain gauge works on it.
Video Link : https://youtu.be/GjmM6KKhgp0
Sensitivity of The Wheatstone Bridge
When bridge is unbalanced, current flows through the galvanometer causing the deflection.
The amount of deflection depends on the sensitivity of the galvanometer.
This sensitivity can be expressed as total amount of deflection per unit current :
S = Total deflection / unit current
S = mm/μA, radians/μA or degrees/μA
As the current is in μA and deflection can be measured in mm, radians or degrees. (linear or angular motion)
Higher the sensitivity of the galvanometer, greater the deflection of the pointer with the same amount of current.
The total amount of deflection also can be denote as :
D = S*I
Where,
S = Sensitivity
I = Current in μA
Unbalanced Wheatstone Bridge
Figure 7 : Unbalanced Wheatstone Bridge
Steps to determine whether wheatstone bridge is in unbalanced condition
Steps to determine whether wheatstone bridge is in unbalanced condition
Using R₁/R₃ = R₂/R₄ to determine both sides of equation are equivalent or not.
If both sides are not equivalent, where R₁/R₃ ≠ R₂/R₄ , the wheatstone bridge circuit can be said as unbalanced.
or
Using Va= [R₁/(R₁+R₃)] x V and Vb= [R₂/(R₂+R₄)] x V to find the voltage difference between point a and point b or Vab
Vab can be calculated using VOUT= l Va-Vb l
If VOUT in the above circuit is not equal to 0 (VOUT ≠ 0), the Wheatstone is said to be an Unbalanced Wheatstone Bridge.
Load Resistance or Galvanometer Resistance
Figure 8 : Wheatstone Bridge with Load Resistance, RL
Figure 9 : Wheatstone Bridge with Galvanometer Resistance, RG
The load resistance and galvanometer resistance are the same as they are due to the unbalanced wheatstone bridge circuit.
When the wheatstone bridge is unbalanced, the current and voltage will flow through the galvanometer and cause a deflection of its pointer.
Since there is the flow of voltage and current in between the terminal A and B for left circuit or terminal C and D for right circuit, therefore the resistance is existed.
Thevenin's Theorem : Thevenin Equivalent Circuit
Figure 10 : Thevenin Equivalent Circuit
Figure 11 : Thevenin Circuit without Load Resistor
Thevenin's Theorem consists of Kirchhoff’s Circuit Laws and Ohm’s Law.
This theorem states that any linear circuit which may contain several voltages and resistances, can simplify the circuit into one voltage source and series resistance connected to a load.
Rather than having to recalculate the current and voltage of entire circuit each time when load changes, we can simplify this process with Thevenin’s Theorem.
Application of Thevenin's Theorem on Unbalanced Wheatstone Bridge
Steps to analyse unbalanced wheatstone bridge by using Thevenin's theorem
Steps to analyse unbalanced wheatstone bridge by using Thevenin's theorem
Open the load resistor, RL or remove the galvanometer, RG.
Calculate and measure the open circuit voltage. This is the Thevenin Voltage (VTH).
Open current sources and short voltage sources.
Calculate and measure the open circuit resistance. This is the Thevenin Resistance (RTH).
Redraw the circuit with measured open circuit Voltage (VTH) in Step (2) as voltage source and measured open circuit resistance (RTH) in step (4) as a series resistance and connect the load resistor, RL or galvanometer, RG , which we had removed in Step (1). This is the simplified equivalent Thevenin circuit from that linear circuit and analysed by Thevenin’s Theorem.
Now find the total current flowing through load resistor, RL or galvanometer, RG by using the Ohm’s Law : IT = VTH / (RTH + RL). (Note that : RL can be replaced by RG)
Example with Solution : Unbalanced Wheatstone Bridge
Question : An unbalanced wheatstone bridge circuit is given as the figure below. Find the Thevenin voltage (VTH), Thevenin resistance (RTH) and current flow through the galvanometer, IG.
Let R1 = 12 Ω R4 = 12 Ω
R2 = 18 Ω RG = 6 Ω
R3 = 18 Ω VS = 10 V
Firstly, remove the galvanometer and let the terminal A and B in open circuit condition.
We can either redraw the circuit into the linear form and apply the voltage divider rule.
Using VA= [R₃/(R₁+R₃)] x VS
VA= [18 Ω /(12 Ω+18 Ω)] x 10 V
VA= 6 V
Using VB= [R₄/(R₂+R₄)] x VS
VB= [12 Ω /(18 Ω+12 Ω)] x 10 V
VB= 4 V
By applying the Kirchhoff's voltage law (2nd Law), where ΣV = 0 in a closed loop. The following equation is obtained : VTH + VR₄ - VR₃ = 0
VTH = l VR₃ - VR₄ l
Note that : VR₃ = VA
VR₄ = VB
Therefore, VTH = l VA - VB l
or
VTH = l VS { [R₃/(R₁+R₃)] - [R₄/(R₂+R₄)] } l
VTH = l 6V - 4V l
VTH = 2 V
Open current sources and short voltage sources and the circuit is redrawn in linear form. Hence, the Thevenin Resistance (RTH) can be calculated by using parallel resistor equation, where
RTH = R₁//R₃ + R₂//R₄
or
RTH = [(R₁R₃)/(R₁+R₃)] + [(R₂R₄)/(R₂+R₄)]
RTH = { [(12 Ω)(18 Ω)] / [(12 Ω+18 Ω)] + [(18 Ω)(12 Ω)] / [(18 Ω+12 Ω)] }
RTH = 14.4 Ω
Lastly, Connect the RTH in series with Voltage Source VTH and replug the galvanometer that we have already removed in step (1). This the Thevenin’s equivalent circuit.
From here, using Ohm’s Law to calculate the total current flowing across the galvanometer like this :
IG = VTH / (RTH+RG)
IG = 2 V / (14.4 Ω + 6 Ω)
IG = 0.09804 A
IG = 98.04 mA
Learning Video on Youtube : Unbalanced Wheatstone Bridge
For those who are visual or kinesthetic learners, kindly check out this video which may assist you on how to determine unbalanced wheatstone bridge using voltmeter.
Video Link : https://youtu.be/hp2oAgTiadc
Other Application of Wheatstone Bridge
The Wheatstone Bridge can used for measuring the resistor that has very low resistance precisely.
Wheatstone bridge installed along with transducer is used to measure the physical quantity like temperature, strain, light, etc.
We can also measure the quantities of capacitance, inductance, impedance and frequency by varying device on the Wheatstone Bridge.
Figure 12 : Wheatstone Bridge with Thermistor
Wheatstone Bridge for Temperature Measurement
- Thermistor
The unknown resistor was replaced by a transducer called thermistor which is a temperature dependent resistor.
Changes in temperature will either increase or decrease the resistance of the thermistor and it acts like a temperature sensor.
Therefore, the output voltage of the wheatstone bridge, VOUT will become a positive value.
The value of output voltage, VOUT is proportional to the temperature.
The display will show temperature in terms of the output voltage by calibrating voltmeter.
Figure 13 : Wheatstone Bridge with Strain Gauge
Wheatstone Bridge for Strain Measurement
- Strain Gauge
Strain gauge is a device where electrical resistance varies in proportional to the mechanical factors like strain, pressure or force.
In general, the range of strain gauge resistance is from 30 Ω to 3000 Ω.
To accurately measure the fractional changes of resistance, a wheatstone bridge circuit is used.
The circuit beside shows a wheatstone bridge where the unknown resistor is replaced with a strain gauge.
Example of Strain Gauge
Figure 14 : Load Cell in Weight Scale
Uses of Strain Gauge
- Weight Scale
When there is external force, the resistance of the strain gauge changes and the bridge becomes unbalanced.
The output voltage can be calibrated to display the changes in strain.
For example, one popular application of strain gauge and wheatstone bridge is in weight scales.
For this case, the strain gauges are carefully mounted as a single unit called as load cells, which is a transducer that converts mechanical force to electrical signal.
Basically, weight scales consist of four load cells, where two strain gauges expand or stretch (tension type) when external force is acting and two strain gauges compress (compression type) when load is placed.
Strain gauge will either tensed or compressed, hence the resistance also will increase or decrease. Therefore, this causes unbalancing of the bridge. This produces a voltage deflection on voltmeter according to the strain change.
If the strain applied on a strain gauge is increase, then the voltage difference across the meter terminals is also increase. If there is no strain, then the bridge is balanced and the reading of meter display zero.
Due to the fractional measurement of resistance in wheatstone bridge, it can acts as a measurement instrument accurately without influences by any other external or enviroment factor.
Thus, wheatstone bridges are mostly used in strain gauge and thermometer measurements.
Murray Loop Test
This test is used to find the fault location in an underground cable by making one Wheatstone Bridge in it.
The fault location can be found when comparing the resistance.
The length of the cables should be known for this test.
Derivation of Murray Loop Test
Figure 15 : Murray Loop Test Circuit
Using wheatstone bridge concept,
R3/R1 = RX/R2
Changing R1 and RX position,
R3/RX = R1/R2
Plus 1 at both side,
(R3/RX)+(RX/RX) = (R1/R2)+(R2/R2)
(R3+RX)/RX = (R1+R2)/R2
Turn upside down,
RX/(R3+RX) = R2/(R1+R2)
Hence,
RX = [R2/(R1+R2)]*(R3+RX)
or,
(ρXℓX)/AX = [R2/(R1+R2)]*{[(ρ3ℓ3)/A3]+[(ρxℓx)/Ax]} (Note that : R = (ρℓ)/A )
If ρX = ρ3 and AX = A3,
ℓX = [R2/(R1+R2)]*(ℓ3+ℓx) (Note that : ℓ3+ℓx = 2ℓ )
Therefore,
ℓX = [R2/(R1+R2)]*(2ℓ)
Example with Solution : Murray Loop Test
Question : A Wheatstone bridge is connected for a Murray loop test as shown in Figure below and balanced. Cable a is an aerial cable with a resistance of 0.1 Ω per 1000 ft. Cable b is an underground cable with a resistance of 0.5 Ω per 1000 ft. Neglecting temperature differences, calculate the distance LX from the ground fault to the test set if La=Lb.
Figure 16 : Murray Loop Test Circuit
By using the formula,
RX = [R1/(R2+R1)]*(Ra+Rb)
RX = [100 Ω / (500 Ω+100 Ω)]*(0.1 Ω /1000 ft + 0.5 Ω / 1000 ft)
RX = 0.1 Ω / 1000 ft
Take the fraction of RX and Rb,we know
RX/Rb = (0.1 Ω / 1000 ft)/(0.5 Ω / 1000 ft)
RX/Rb = 0.2
LX is located at 1/5 of Lb
Hence, we can also say that
Every 1000 ft of Lb, the LX is located in first 200 ft of it.
Varley Loop Test
This test is used to find the fault location in an underground cable by making one Wheatstone Bridge in it.
The fault location can be found by comparing the resistance instead of calculating it from the known lengths of the cable.
The length of the cable does not required or to be known in this test.
Derivation of Varley Loop Test
Figure 17 : Varley Loop Test Circuit
When the switch is at position A, using wheatstone bridge concept
R1/R3 = R2/(Ra+Rb)
R1(Ra+Rb) = R2R3
(Ra+Rb) = (R2R3)/R1 --------- (1)
When the switch is at position B, using wheatstone bridge concept
R1/(R3'+Rx) = R2/[Ra+(Rb-RX)]
R1[Ra+(Rb-RX)] = R2(R3'+Rx)
Expand the equation,
R1Ra+R1Rb-R1RX = R2R3'+R2Rx
R1Ra+R1Rb-R2R3' = R1RX+R2Rx
RX(R1+R2) = R1Ra+R1Rb-R2R3'
Make RX be the subject,
RX = (R1Ra+R1Rb-R2R3')/(R1+R2)
RX = [R1(Ra+Rb)-R2R3']/(R1+R2) --------- (2)
Substitute equation (1) into (2),
RX = {R1[(R2R3)/R1]-R2R3'}/(R1+R2)
Cancel out R1 at numerator,
RX = (R2R3-R2R3')/(R1+R2)
Therefore,
RX = [R2(R3-R3')]/(R1+R2)
Example with Solution : Varley Loop Test
Question : A Wheatstone bridge is connected for a Varley loop test as shown below. When switch is set to position a, the bridge is balance R1 = 1k Ω, R2 = 100 Ω and R3 = 53 Ω. When switch to position b, the bridge is balance with R3 = 52.9 Ω. If the resistance of the shorted cable is 0.015 Ω/meter, how many meters the distance from the bridge to a ground fault.
Figure 18 : Varley Loop Test Circuit
By using the formula,
RX = [R2(R3-R3')]/(R1+R2)
RX = [100 Ω (53 Ω-52.9 Ω)]/(1000 Ω+100 Ω)
RX = 9.09 mΩ
Since shorted cable is 0.015 Ω per meter,
Distance = 9.09 mΩ / 0.015 Ω per meter
= 0.61 meter
Learning Video on Youtube : Wheatstone Bridge with Strain Gauge
For those who are visual or kinesthetic learners, kindly check out this video which may assist you more understanding the application of wheatstone bridge in term of strain gauge.
Video Link : https://youtu.be/qCfEjfzxZVI
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