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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” | | | | | ….. | | | |

| | | | | | | | | | |

| | | | | | | | | | |...

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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