Rankers Physics
Topic: Current Electricity
Subtopic: Variation of Resistance with Temperature

The ratio of the resistance of conductor at temperature 15°C to its resistance at temperature 37.5°C is 4 : 5. the temperature coefficient of resistance of the conductor is : (reference is taken as 0°C)
\[1/25 ^{o}C^{-1}\]
\[1/50 ^{o}C^{-1}\]
\[1/80 ^{o}C^{-1}\]
\[1/75 ^{o}C^{-1}\]

Solution:

To find the temperature coefficient of resistance α\alpha of the conductor, we use the formula for the change in resistance with temperature:

RT=R0(1+αT)R_T = R_0 \left( 1 + \alpha T \right)

Where:

  • RTR_T is the resistance at temperature TT,
  • R0R_0 is the resistance at the reference temperature (0°C),
  • α\alpha is the temperature coefficient of resistance,
  • TT is the temperature change in °C.

Step 1: Given

  • The ratio of resistances at 15°C and 37.5°C is given as R15R37.5=45\frac{R_{15}}{R_{37.5}} = \frac{4}{5}.
  • The resistance at temperature 15°C, R15=R0(1+α×15)R_{15} = R_0 (1 + \alpha \times 15).
  • The resistance at temperature 37.5°C, R37.5=R0(1+α×37.5)R_{37.5} = R_0 (1 + \alpha \times 37.5).

Step 2: Set up the equation based on the given ratio

R15R37.5=45\frac{R_{15}}{R_{37.5}} = \frac{4}{5}

Substituting the expressions for R15R_{15} and R37.5R_{37.5}:

R0(1+α×15)R0(1+α×37.5)=45\frac{R_0 (1 + \alpha \times 15)}{R_0 (1 + \alpha \times 37.5)} = \frac{4}{5}

Canceling R0R_0 from both the numerator and denominator:

1+15α1+37.5α=45\frac{1 + 15\alpha}{1 + 37.5\alpha} = \frac{4}{5}

Step 3: Solve for α\alpha

Cross-multiply to solve for α\alpha:

5(1+15α)=4(1+37.5α)5(1 + 15\alpha) = 4(1 + 37.5\alpha)

Expanding both sides:

5+75α=4+150α5 + 75\alpha = 4 + 150\alpha

Simplify:

54=150α75α5 - 4 = 150\alpha - 75\alpha 1=75α1 = 75\alpha α=175\alpha = \frac{1}{75}

Final Answer:

The temperature coefficient of resistance α\alpha is 175per °C\boxed{\frac{1}{75}} \, \text{per °C}.

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