Finding temperature coefficient of resistivity of the alloy

[Searay330]

Suppose a wire made from an unknown alloy and having a temperature of 20.0°C carries a current of 0.529 A. At 52.4°C the current is 0.378 A for the same potential difference. Find the temperature coefficient of resistivity of the alloy.

tempinital = 20C
tempfinal = 52.4C
currentInital = .529A
currentFinal = .378A
Voltage = constant but unknownhow do you go about this problem i have tried manipulating ohms law ( I= V/R ) to remove voltage but that always leaves me with IR= IR.

TCOR = temperature coefficient of resistivity;

because the resistance isn't a constant i can't factor it out of RF = RI (1 - TCOR(TF - TI))

i can't use either form because i don't have resistance to solve for resistivity.

if someone could point me in the right direction that would be great.

 

[berkeman]

leaves me with IR= IR

Since the two measurement voltages were the same, you can write that equation, but you should distinguish between them...

$$I_1 R_1 = I_2 R_2$$

So you can find the ratio of the two resistances. Then use the ratio of the two temperatures (be sure to use absolute temperature, not degrees Celsius) to help you find the coefficient of resistivity...

 

[Chestermiller]

Suppose a wire made from an unknown alloy and having a temperature of 20.0°C carries a current of 0.529 A. At 52.4°C the current is 0.378 A for the same potential difference. Find the temperature coefficient of resistivity of the alloy.

tempinital = 20C
tempfinal = 52.4C
currentInital = .529A
currentFinal = .378A
Voltage = constant but unknownhow do you go about this problem i have tried manipulating ohms law ( I= V/R ) to remove voltage but that always leaves me with IR= IR.

TCOR = temperature coefficient of resistivity;

because the resistance isn't a constant i can't factor it out of RF = RI (1 - TCOR(TF - TI))

i can't use either form because i don't have resistance to solve for resistivity.

The resistance is constant in each of the tests. I don't see why you can't use the equation. Also, resistivity is proportional to resistance for the same wire. Is it that you don't know how to solve the equation for TCOR?

Also, RF = V/I_F and RI=V/I_I

 

[Searay330]

how can the resistance be constant if the voltage is constant and the current changes?

 

[Chestermiller]

how can the resistance be constant if the voltage is constant and the current changes?

It's constant with location along the wire.

$R_1=V/I_1$ and $R_2=V/I_2$. So,
$$\frac{R_2}{R_1}=\frac{I_1}{I_2}$$

 

[Searay330]

It's constant with location along the wire.

$R_1=V/I_1$ and $R_2=V/I_2$. So,
$$\frac{R_2}{R_1}=\frac{I_1}{I_2}$$

right so you can changed the formula to be
I₂/I₁ = (1 - TCOR(TF - TI))

 

[Chestermiller]

right so you can changed the formula to be
I₂/I₁ = (1 - TCOR(TF - TI))

Excellent!

 

[Searay330]

Excellent!

so then plugging everything in you would get

.378/.529 = (1- x(52.4-20))
meaning x = .0088100072

 

[berkeman]

so then plugging everything in you would get

.378/.529 = (1- x(52.4-20))
meaning x = .0088100072

Always remember to include units in your calculations and answers...

 

[Searay330]

Always remember to include units in your calculations and answers...

well that's my problem it says my answers units should be C^-1 but by using the formula wouldn't the units just be C

 

[berkeman]

well that's my problem it says my answers units should be C^-1 but by using the formula wouldn't the units just be C

so then plugging everything in you would get

.378/.529 = (1- x(52.4-20))
meaning x = .0088100072

Include units in your calculation above at each step for each quantity. What do you get now?

 

[Searay330]

Include units in your calculation above at each step for each quantity. What do you get now?

.7145557656 (A/A cancels) = (1 - (32.4xC))
- (32.4xC) = .2854442344
- xC = .2854442344/32.4
x = .0088100072 C

 

[berkeman]

No. Keep the units with the quantity that has them. You don't know x's units until you have solved the equations...

0.378[A]/0.529[A] = (1- x(52.4[C]-20[C]))

Now can you keep going with that?

 

[Searay330]

No. Keep the units with the quantity that has them. You don't know x's units until you have solved the equations...

0.378[A]/0.529[A] = (1- x(52.4[C]-20[C]))

Now can you keep going with that?

well that would mean the Amps cancel leaving that as just a number and the only unit left is C which x picks up when its distributed across those numbers. correct

 

[berkeman]

well that would mean the Amps cancel leaving that as just a number and the only unit left is C which x picks up when its distributed across those numbers. correct

Nope.

Since the lefthand side (LHS) of the equation is unitless, the RHS has to also be unitless. On the RHS, "1" is unitless, so x(52.4[C]-20[C]) has to be unitless. The temperatures have units of [C], so what units must x have?

 

[Searay330]

Nope.

Since the lefthand side (LHS) of the equation is unitless, the RHS has to also be unitless. On the RHS, "1" is unitless, so x(52.4[C]-20[C]) has to be unitless. The temperatures have units of [C], so what units must x have?

oh it must have 1/C

 

[Searay330]

It Makes perfect sense to me but the homework program told me i got the wrong answer. is there some mistake in my calculation?

 

[berkeman]

It Makes perfect sense to me but the homework program told me i got the wrong answer. is there some mistake in my calculation?

RF = RI (1 - TCOR(TF - TI))

Looking back at this equation, resistance goes up with temperature. If the final temperature is higher than the initial temperature, should you be subtracting or adding the delta resistance? I don't know if that's the error, but it jumps out at me...