Fig. Shows a meter bridge consisting of two resistances X and Y together in parallel with a meter-long constantan wire AC of uniform cross-section. D is a movable contact that can slide along the wire AC. The resistors X, Y and resistances of segments AD and DC of the wire constitute the four arms of the bridge. The length of wire AC is 100 cm. X is a standard 4.00 Ω resistor and Y is a coil of wire. With Y immersed in melting ice the null point is found to be at a distance of 40.0 m from point A. When the coil Y is heated to 100º C, a 100 Ω resistor has to be connected in parallel with Y in order to keep the bridge balanced at the same point. Calculate the temperature coefficient of resistance of the coil.
Text Solution
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Sol. The resistances of the two segments of the wire AD and DC are in the ratio of their lengths. If R 0 is the resistance of Y in melting ice (0ºC), the balance condition of Whetstone bridge gives
= 
= 
where k is the resistance per centimeter of wire AC. Now, l = 40.0 cm and X = 4.00 Ω . Substituting these values, we get R 0 = 6.00 Ω . Let R t be the resistance of Y when heated to a temperature t = 100ºC. When it is connected in parallel with 100 Ω resistor as shown in Fig. , the net resistance becomes
R' = 
Since the null point remains unchanged, we have
= 
⇒ R' = 6.00 Ω
Thus 6.00 = 
⇒ R t = 6.38 Ω
Temperature coefficient of resistance of the coil Y is
α = 
= 
= 6.3 × 10 –4 K –1 .
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