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CHAPTER 6 Direct-Current Bridge.

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Presentation on theme: "CHAPTER 6 Direct-Current Bridge."— Presentation transcript:

1 CHAPTER 6 Direct-Current Bridge.
School of Computer and Communication Engineering, UniMAP Prepared By: Amir Razif b. Jamil Abdullah EMT 113: V-2008

2 6.0 Direct Current Bridge. 6.1 Introduction to Bridge.
6.2 The Wheatstone Bridge. 6.2.1 Sensitivity of the Wheatstone Bridge. 6.2.2 Unbalance Wheatstone Bridge. 6.3 Kelvin Bridge.

3 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.

4 6.2 The Wheatstone Bridge. 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. Figure 6.1: Wheatstone Bridge Circuit.

5 Cont’d… 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).

6 Cont’d… 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 the value of unknown resistor.

7 Example 6.1: Wheatstone Bridge.
Determine the value of unknown resistor, Rx in the circuit of Figure 6.2 assuming a null exist, current through the galvanometer is zero. Solution: From the circuit, the product of the resistance in opposite arms of the bridge is balance, so solving for Rx . Figure 6.2: Circuit For Example 6.1.

8 Example 6.1A(T2 2005): Wheatstone Bridge.
Calculate the value of Rx in the circuit of Figure 4 if VTh = 24 mV and Ig =13.6 uA. Solution: Calculate Rth Figure 6.2A: Circuit For Example 6.1A.

9 Calculate Rx .

10 6.2.1 Sensitivity of the Wheatstone Bridge.
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,

11 Figure 6.3: Unbalance Wheatstone Bridge.
6.2.2 Unbalanced Wheatstone Bridge. The current flows through the galvanometer can determine by using Thevenin theorem. Figure 6.3: Unbalance Wheatstone Bridge.

12 The deflection current in the galvanometer is,
Cont’d… Figure 6.4: Thevenin’s Equivalent Circuit for an Unbalanced Wheatstone Bridge. The deflection current in the galvanometer is, Rg = the internal resistance in the galvanometer

13 Example 6.2: 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: (1) Find Thevenin equivalent circuit as seen from by the galvanometer,Vth is, Figure 6.5: Circuit for Example 6.2.

14 (2) Find Thevenin’s equivalent resistance (Rth )is,
Cont’d… (2) Find Thevenin’s equivalent resistance (Rth )is, . Figure 6.6: Thevenin’s Equivalent Circuit for the Example 6.2 Unbalance Bridge.

15 Example 6.3: Slightly Unbalanced Wheatstone Bridge.
Use the approximate equation to calculate the current through the galvanometer in Figure 6.7. The galvanometer resistance, Rg is 125Ω and is center-zero uA movement. E=10V, R1=500Ω, R2=500Ω, R3 = 500 Ω and R4=525 Ω. Solution: From formula, (1) Find Thevenin equivalent voltage (Vth) is, (2) Find Thevenin equivalent resistance (Rth )is, Figure 6.7: Circuit for Example 6.3.

16 Cont’d… (3) The current through the galvanometer (Ig)is, Observation: If the deflector is a uA galvanometer, the pointer deflected full scale for a 5% change in resistance. .

17 6.3 Kelvin Bridge. 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. Figure 6.8: Basic Kelvin Bridge.

18 Cont’d… 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.

19 Example 6.4: Kelvin Bridge.
Figure 6.9 is the Kelvin Bridge, the ratio of Ra to Rb is R1 is 5 Ohm and R1 =0.5 R2. Find the value of Rx. Solution: Calculate the resistance of Rx, R1 =0.5 R2, so calculate R2 Calculate the value of Rx . Figure 6.9: For Example 6.4.

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