Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "1 CHAPTER 6 EMT 113: V-2008 School of Computer and Communication Engineering, UniMAP Prepared By: Prepared By: Amir Razif b. Jamil Abdullah Direct-Current."— Presentation transcript:

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

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

3 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 4  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 Bridge. Figure 6.1: Wheatstone Bridge Circuit.

5 5 Operation:  We want to know the value of R 4, 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 R 3 is equal to the voltage drop across R 4,(R 3 = R 4 ).Cont’d…

6 6  At balance the voltage drop at R 1 and R 2 must be equal to.  No current go through the galvanometer G, the bridge is in balance so,  This equation, R 1 R 4 = R 2 R 3, states the condition for a balance Wheatstone bridge and can be used to compute the value of unknown resistor.Cont’d…

7 7 Example 6.1: Wheatstone Bridge. Determine the value of unknown resistor, R x 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 R x . Figure 6.2: Circuit For Example 6.1.

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

9 9 Calculate R x .

10 10  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, Sensitivity of the Wheatstone Bridge.

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

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

13 13 Example 6.2: Unbalance Wheatstone Bridge. Calculate the current through the galvanometer in the circuit Figure 6.5. Given that E=6V, R 1 = 1kΩ, R 2 = 1.6kΩ, R 3 = 3.5kΩ, R 4 = 7.5kΩ and R g =200Ω.Solution: (1) Find Thevenin equivalent circuit as seen from by the galvanometer,V th is, Figure 6.5: Circuit for Example 6.2.

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

15 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, R g is 125Ω and is center-zero uA movement. E=10V, R 1 =500Ω, R 2 =500Ω, R 3 = 500 Ω and R 4 =525 Ω. Solution: From formula, (1) Find Thevenin equivalent voltage (V th ) is, (2) Find Thevenin equivalent resistance (R th )is, Figure 6.7: Circuit for Example 6.3.

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

17 17  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 R ic represent the lead and contact resistance present in the Wheatstone bridge. 6.3 Kelvin Bridge. Figure 6.8: Basic Kelvin Bridge.

18 18  The second set of R a and R b compensates for this relatively low lead contact resistance.  At balance the ratio of R a and R b must be equal to the ratio of R 1 to R 3.Cont’d…

19 19 Example 6.4: Kelvin Bridge. Figure 6.9 is the Kelvin Bridge, the ratio of R a to R b is R 1 is 5 Ohm and R 1 =0.5 R 2. Find the value of R x. Solution: Calculate the resistance of R x, R 1 =0.5 R 2, so calculate R 2 Calculate the value of R x . Figure 6.9: For Example 6.4.


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

Similar presentations


Ads by Google