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Practical

  • Date Submitted: 09/14/2015 05:36 AM
  • Flesch-Kincaid Score: 88.5 
  • Words: 511
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Theory:
If connections are made (as in Fig.) with two equal resistances P and Q (each equal to one ohm) in the two middle gaps G2 and G3,a resistance X in the resistance box(X) placed in the extreme left gap G1 and another resistance Y in the extreme right gap G4,the null point will be obtained at J1,at a distance l from the left end of the bridge wire of average resistance p per unit length. By interchanging the resistances X and Y , if null point is obtained at J2 at a distance l’ from the end of the bridge wire then ,     X-Y=p(l’-l) …………………………(1)
  1. To find p, the average resistance per unit length of the bridge wire the procedure is to make , Y=0;   X=R1;   l=l1 . Hence we   get,     p=R1l1'-l1 ………………….(2)
  2. To find a length from the total length L’ of the sample wire required for one ohm , let X be replaced by sample wire of length L’ and resistance R2, and Y by r (a Known Resistance ) then Eq. (1).
R2= r + p (l’2-l2) …………………………(3)
Where we put   l=l2 and l’=l2’
Therefore, L =   L'R2 ………………………………(4)
  3. To compare prepared ohm (p-ohm) with a standard ohm(s-ohm), let us put , Y = s-ohm;   X=p-ohm. Hence   p- ohm = s-ohm + p(l’-l) …………………………………(5)

Experimental Result:
  A. To find p of bridge wire:
TABLE-1
P=1Ώ;   Q=1Ώ
No. of Obs. | Res.(R1)in ohm in the extreme | Null Points in cm. with current | Mean of (m) and (n) in cm | (l’1-l1)in cm |     p=R1l1'-l1in ohm per cm | Mean p ohm per cm |
| Left gap | Right Gap | Direct(m) | Mean(m) | Reversed(n) | Mean(n) | | | | |
      a)1.     b) | …R1’ | 0 | | | | | ...l1 | | | |
| | | | | | | | | | |
| | | | | | | | | | |
| 0 | …R1’ | | | | | ...l1’ | | | |
| | | | | | | | | | |
| | | | | | | | | | |
    a)2     b) | …R1” | 0 | | | | | ….. | | | |
| | | | | | | | | | |
| | | | | | | | | | |
| 0 | …R1” | | | | | ….. | | | |
| | | | | | | | | | |
| | | | | | | | | | |...

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