# Specific Resistance Of The Material Of A Wire

## SPECIFIC RESISTANCE OF THE MATERIAL OF A WIRE USING A METER BRIDGE

 A known length (L) of a wire is connected in one of the gaps (P) of a meter bridge, while a Resistance Box is inserted into the other gap (Q). The circuit is completed by using a battery (B), a rheostat (Rh), a key (K) and a galvanometer (G). The balance length (l) is found by closing key K and momentarily connecting the galvanometer until it gives zero deflection (null point).  Then,

(using the expression for the meter bridge at balance.)

Here, P represents the resistance of the wire while Q represents the resistance in the resistance box. The key K is kept open when the circuit is not in use.

The resistance of the wire, P = ⇒ ρ =

where r is the radius of wire and L is the length of the wire, r is measured using a screw gauge while L is measured with a scale.

### Errors

The major systematic errors in this experiment are due to the (i) heating effect, (ii) end corrections introduced due to shift of the zero of the scale at A and B, (iii) stray resistances in P and Q, (iv) errors due to non−uniformity of the meter bridge wire.

### Error analysis:

End corrections can be estimated by including known resistances P1 and Q1 in the two ends and finding the null point:

(where α and β are the end corrections.)

When the resistance Q1 is placed in the left gap and P1 in the right gap,

which gives two linear equations for finding α and β.

In order that α and β be measured accurately P1 and Q1 should be as different from each other as possible.

For the actual balance point,

Errors due to non−uniformity of the meter bridge wire can be minimised by interchanging the resistances in the gaps P and Q.

,  where δl′1 and δl′2 are of the order of the least count of the scale.

The error is, therefore, minimum if l′1 = l′2 i.e. when the balance point is in the middle of the bridge. The error in ρ is .

Illustration 13.With two resistances R1 and R2(> R1) in the two gaps of a metre bridge, the balanced point was found to be 1/3m from the zero end.  When a 6 Ω resistance is connected in series with the smaller of the two resistances, the point is shifted to 2/3 m from the same end.  Calculate R1 and R2.

Solution:, where l is in metre.

or,R2 = 2R1 . . . (i)

Again,

or,  R1 + 6 = 2R2 . . . (ii)

From (i) and (ii)

R1 = 2Ω  and R2 = 4 Ω.