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

1.280 A

Step-by-step 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 =
`(V)/(R)`

=
`(V)/(R_(0)(1.1599) )`

1.4 =
`(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:

`(I)/(1.4) = (1.1599)/(0.5788)`

I = 1.280 A

Answer 2

I=2.80 A

Step-by-step 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

`(I)/(1.4) =(1.1599)/(0.5788)`

I = 2.80 A