Question

An engineering intern is studying the effect of temperature on the resistance of a current carrying wire. She applies a voltage to a aluminum wire at a temperature of 61.0°C and notes that it produces a current of 1.40 A. If she then applies the same voltage to the same wire at −88.0°C, what current should she expect (in A)? The temperature coefficient of resistivity for aluminum is 3.90 ✕ 10−3 (°C)−1. (Assume that the reference temperature is 20°C.)

Answer 1

Answer:

1.280 A

Explanation:

Thinking process:

The resistiviy is related to temperature by the following equation:

R  = R₀ (1 + α ΔT)

   = R₀ (1 + 3.9 × 10⁻³ × (61- 20)⁰C)

   =R₀(1.1559)

The current is given by:

I = $\frac{V}{R}$

 =  $\frac{V}{R_{0}(1.1599) }$

1.4 = $\frac{V}{R_{0}(1.1599) }$

calculating the current when the temperature is 88⁰ gives:

R = R₀ (1+ αΔT)

  = V/R₀ (0.5788)

combining the equations gives:

$\frac{I}{1.4} = \frac{1.1599}{0.5788}$

I   = 1.280 A

Answer 2

Answer:

I=2.80 A

Explanation:

We Know that    R =R₀(1+∝ ΔT)

                           R=R₀ (1+3.9*10⁻³ *(61-20))

                           R=R₀ (1.1599)

                           I=V/R=V/(R₀ (1.1599)

                           1.4 = V/(R₀ (1.1599)                       ∵ equation 1

                     We have to calculate I when T=-88°

                              R =R₀(1+∝ ΔT)

                             R=R₀ (1+3.9*10⁻³ *(-88-20))

                             R=R₀ (0.5788)

                             I=V/(R₀ (0.5788)                          ∵equation 2

Dividing equation 2 by equation 1

                             $\frac{I}{1.4} =\frac{1.1599}{0.5788}$

                            I = 2.80 A