Methods For Measuring Low resistance

This ppt is about different Methods for measuring low resistance.
1) Voltmeter- Ammeter Methods
2) Kelvin's Bridge
3) Kelvin's Double Bridge

1. The figure shows the construction of low resistance For the protection of low resistance, it is constructed with four terminals CC’ = Current Terminals PP’ = Potential Terminals The value of low resistance is measured between the potential terminal P and P’.

2. R ecall our classification done above, as we move from top to bottom the value of resistance decreases hence, we require more accurate and precise device to measure the low value of resistance W e need some modification in Wheatstone bridge itself, and the modified bridge so obtained is Kelvin bridge , which is not only suitable for measuring low value of resistance but has wide range of applications in the industrial world

3. There are two methods of connecting voltmeter and ammeter for measurement of resistance as shown in figure. In both cases measured value of unknown resistance is equal to the reading of voltmeter divided by reading of ammeter. From fig.(1) R = R m Ideally R = R m only when R a = 0 From fig.(2) R = Ideally R = R m when resistance of voltmeter is ‘∞’  

4. + = + R mp ) R x + = + R x =   R y = represents resistance of connecting lead from R 3 to R x . Figure (2)

5. From Fig.(2) The balance condition of bridge is + = + R mp ) Substituting from eq.(1) R x + = + R x = T he problems with the above method are :- the method is not practical difficult to find correct galvanometer null point  

6. To Overcome the problem Of Kelvin Bridge , The New Bridge is Introduced ,which is used for precise measurement of low resistance called Kelvin’s Double Bridge. Construction Consist of 2 ratio arms Connected resistances are P, Q, p,q,r,S,R . r is the resistance of slide wire R is the unknown resistance R g is regulating resistance Galvanometer (G) is connected between point ‘F’ and ‘H’. Working By adjusting the balanced condition, we can find the unknown resistance

7. At Balance condition I g = 0 . Hence voltage across ‘P’ = voltage across ‘R’ + voltage across ‘p’ V P = V R + Vp I 1 P = IR + I 2 p ……(3) Similarly voltage across ‘Q’ = Voltage across ‘S’ + voltage across ‘q’ V Q = V S + V q I 1 Q = IS + I 2 q …….(4) Voltage across ‘r’ = voltage across ( p+q ) V r = V ( p+q ) So, I 2 = So from eq.(1) I 1 P = IR + ……….(5) I 1 Q = IS + ……….. (6)  

8. Divide eq. (5) & (6) Finally we get the below equation R = …….(7) Usually, ratio is adjusted is equal to . So eq.(7) can be expressed as R = For accurate measurement of ‘R’ two readings are taken by reversing the direction of current, the average value of these two values is taken as magnitude of unknown resistance  

9. Thank You