ObjectMeasurement of resistance by kelvins bridge.
To determine unknown low value resistance (range up to 1 ohm)
S.No | Name | Type | Quantity |
1 | Kelvins Bridge training kit | EE-123 | 1 |
2 | Patch cords | - | 5-8 |
3 | Multimeter | Digital | 1 |
4 | Audio oscillator | 1 |
Kelvin Bridge is the modification of the Wheatstone’s bridge & provides greatly increased accuracy in measurement of low value resistance. An understanding of the kelvin bridge arrangement may be obtained by a study of difficulties that arise in a Wheatstone’s bridge on account of the resistance of the leads & the contact resistance while measuring low valued resistance.
Consider the bridge circuit shown in fig. * Where r represents the resistance of the lead that connects the unknown resistance R to the standard resistance S. Two galvanometer connections indicated by dotted lines, are possible .The connection may be either to m or to point n.
When the galvanometer is connected to point m, the resistance are of the connecting leads is added to the standard resistance “S”, resulting in too low an indication for unknown resistance “R”. When the connection is made to point n, the resistances are added to the unknown resistance resulting in too high a value for R.
Suppose that instead of using m , which gives a low result or n which makes the result high , we make the galvanometer connection to any intermediate point c as shown by full line in fig.* . If at point c the resistance r is divided into two parts r 1& r 2 such that :
r1/r2=P/Q ................................................(1)
Then the presence of r1 the resistance of connecting leads causes no error in the result we have,
R+r1= P (S+r2)/Q ................................................(2)
But r1/r2=P/Q from equation (1)
Or r1/(r1+ r2)= P/(P+Q)
r1/r = P/(P+Q)
When r1 + r2 =r
r1 = P.r/(P+Q)................................................(3)
From equation (1) again (r1 + r2) / r2 = (P+Q)/Q
r1 / r2 = (P+Q)/Q
r2 = Q.r/ (P+Q)................................................(4)
Using equation (3) & (4) We can write equation (2) as
R + [P.r/(P+Q)] = P[S + - [Q.r/(P+Q)]]/Q
Q.R + [Q.P.r/(P+Q)] = P.S + [P.Q.r/(P+Q)]
Or Q.R = P.S
R= P.S/Q..........................................(5)
Therefore we conclude that making the galvanometer as at c, the resistance of the leads doesn’t affect the result . The process describing the above is not a practical way of achieving the desired result as there would certainly be a trouble in determining the correct point for galvanometer connection .It does however suggest that simple modification , that two actual resistance units of correct ratio be connected between points m & n , the galvanometer be connected to the junction of resistors .this is the actual kelvin bridge arrangement which is showing fig.
The kelvins double bridge incorporates the idea of a second set of ratio arms, hence the name double bridge & the use of 4 terminals resistors for low resistance arms .fig shows the schematic diagram of the kelvin bridge. the first ratio arm is P1 & Q1 The second set of ratio arms PQ is used to connect the galvanometer to a point c, at the appropriate potential between points m & n to eliminate the effect of connecting leads of resistance r between the known resistance R & standard S.
The ratio P/Q is made equal to PQ. Under balance condition there is no current through the galvanometer which means the voltage drop E ab between a & d is equal to the voltage drop Eac between a & c
Now
Ead/P1 = Eab/(P1+ Q1)
Ead= P1Eab/(P1+ Q1)
Eab= I [ R + S + (P2 + Q2).r /P2 + Q2 +r]
Eamc= I [ R + P1 (P1 + Q1).r /(P1 + Q1)(P1 + Q1+ r)]
For zero galvanometer deflection , Ead= Eamc
P1 .I/ (P1+ Q1) [R+S+(P2 + Q2).r/(P2 + Q2)+r] = I[R+ P2.(P2 + Q2).r/(P2 + Q2).(P2 + Q2+r)]
After solving it
R=P1.S/Q1
This equation is the usual working equation for the kelvin’s bridge . It indicates that the resistance of the connection lead r has no effect on the measurement. Provided that the two sets of ratio arms have equal ratios.
Circuit Diagram:
I 2= Current in the resistor R 4of other arm
Observation Table:
Given value of P1= 1ohm………..
given value of Q1= 100ohm…….
S.No | S in ( ohm) | Theoretical value of R in ohm Rx(in Ohms) R = P 1S/Q 1 | Practical value of R |
1 | |||
2 | |||
3 | |||
4 | |||
5 |
Conclusion:-The value of unknown capacitor C2has been calculated and when compared with standard values were found close to each other.
Viva-Voice Question:-
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