# DETERMINATION  OF  THE  VALUE  OF  AN  UNKNOWN  RESISTANCE  BY  CAREY  FOSTER’S  METHOD

THEORY :      When  two  equal  resistance are placed  in two  arms  of  the  Wheatstone  bridge , a  fractional  resistance  (X)  in the  left  gap  and  a  copper  strip  in  the  right  gap , the  null  point  is  obtained  at  a  length  l1 , if the  position  of  the  copper  strip  are interchanged  the  null  point  is  obtained  at  a  length  l2 . Then  the  resistance  per  unit  length  of  the  wire  is  given  by ,

$$λ=X/(l_1~l_2 )$$   …………………………(1)

If  now an  unknown  resistance  (Y)  is  placed  in  the  right  gap  and a known resistance in the right gap, the null point is obtained at  . If now there positions be interchanged then the null point is obtained at  . Hence the unknown resistance is given by:

Y =  X- λ( $$l’_1~l’_2$$ )………………………..(2)

Circuit diagram for Carey Foster’s Bridge

APPARATUS:

1) Carey Foster Bridge.                                        5) Null detector.

2) Two Equal Resistance.                                      6) A Commutator.

3) A Fractional Resistance Box.

4) A constant voltage power supply( 0~2 V,1A)

Procedure:

1) To determine the resistance per unit length of the bridge wire: i) First the circuit is connected as in the fig. for which decimal resistance box X is connected in the left gap ab and copper strip Y is connected in the right gap gh of the bridge. Now both the lower fixed ends of the rheostat are connected to terminals A and C respectively and its variable end is connected to terminal B. Thereafter the leclanche cell E and the plug key K are joined in series in between the terminals B, its other end is connected to the jockey D. ii) The variable end of the rheostat is adjusted in middle such that both the resistances P and Q are nearly equal. iii) Now inserting some resistance X through the resistance box, the jockey D is pressed on the bridge wire and it is slided on it until zero deflection is obtained in the galvanometer. In this position, the distance l1 of jockey from left end on wire is noted. iv) Thereafter the positions of resistance box X and copper strip Y are interchanged and then without changing the resistance box, again the position of jockey is adjusted on the bridge wire in order to obtain zero deflection in the galvanometer. In this position, the length l2 of the jockey on the wire from the left end is noted. v) Now the experiment is repeated three – four times by changing the resistance X from the resistance box and each time the values of l1 and l2 are noted corresponding to the value of X. vi) Then using the relationship ρ = X/ (l2-l1), the value of ρ is calculated for each observation and its mean value is calculated.

2) To determine the resistance of a given wire: i) To determine the resistance of a given wire, from the electric circuit as in the fig. The copper strip connected in the left is withdrawn and in its place the given wire is connected. ii) The above steps 2, 3, 4 and 5 in part (i) of the experiment are repeated.iii) Now using the relation Y = X- ρ (l2-l1), the value of Y is calculated from each observation and its mean value is obtained.

EXPERIMENTAL RESULTS:

1.Position of the null point when two metal strips are connected at left and right gap of the bridge

= ……………………….(cm)

2.Table for determining per unit length of the wire:

Table-I

 Sl.no. Resistance in ohm.(Ω) Position of the null points (cm) in cm λ in Ω/cm Mean λ in Ω/cm Left Gap Right Gap Direct Current Reverse Current Mean

3.Table for determining the unknown resistance :

Table-II

 Sl.no. Resistance in ohm(Ω) Position of the null points (cm) in cm λ in Ω/cm from Table-I Y in Ω Mean Y  in Ω Left Gap Right Gap Direct Current Reverse Current Mean

Results:

1.Resistance per unit length of the wire(λ)……………..Ω/cm.

2.The unknown resistance is(Y)………………………………Ω.

Calculation of percentage error:

As   $$λ=X/(l_1~l_2 )$$

So,

$$δλ/λ x 100=( δX/X+δ(l_1~l_2 )/(l_1~l_2 ) )x100$$ ,   Where $$dX=0$$ (X is given)  and

$$δ(l_1~l_2 )$$= 2 x smallest scale division of the meter scale.

$$l_1~l_2$$ = Put any value of from Table-I.

Again, Y =  X- λ(

$$l’_1~l’_2$$).

So   $$dY/Y*100=(dX/x+dY/y+d(l_1’~l_2′)/l_1′-l_2′)*100$$ where $$dX=0$$ (X is given), $$δ(l_1~l_2 )$$= 2 x smallest scale division of the meter scale.

$$l_1’~l_2′$$ = put any value of  from Table-II

$$δλ/λ$$ = put the value from the error calculation of$$δλ/λ$$

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