# Finding temperature coefficient of resistivity of the alloy

• Searay330
In summary: 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...In summary, to find the temperature coefficient of resistivity of an unknown alloy, you can use the equation R2/R1 = I1/I2, where R is the resistance, I is the current, and the subscripts represent the two different temperatures and currents. By rearranging the equation, you can solve for the temperature coefficient of resistivity (TCOR) by plugging in the known values and solving
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.

Searay330 said:
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...

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Searay330 said:
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/IF and RI=V/II

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

Searay330 said:
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,
$$\frac{R_2}{R_1}=\frac{I_1}{I_2}$$

Chestermiller said:
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))

Searay330 said:
right so you can changed the formula to be
I2/I1 = (1 - TCOR(TF - TI))
Excellent!

Chestermiller said:
Excellent!
so then plugging everything in you would get

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

Searay330 said:
so then plugging everything in you would get

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

berkeman said:
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

Searay330 said:
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
Searay330 said:
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?

berkeman said:
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

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?

berkeman said:
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

Searay330 said:
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?

berkeman said:
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

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?

Searay330 said:
It Makes perfect sense to me but the homework program told me i got the wrong answer. is there some mistake in my calculation?

Searay330 said:
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...

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## 1. How is the temperature coefficient of resistivity of an alloy defined?

The temperature coefficient of resistivity is defined as the change in resistivity per degree Celsius change in temperature. It is usually denoted by the symbol α (alpha).

## 2. Why is it important to find the temperature coefficient of resistivity of an alloy?

The temperature coefficient of resistivity is an important property of an alloy as it helps in understanding how the resistivity of the material changes with temperature. This information is crucial for designing electronic circuits and devices that operate at different temperatures.

## 3. What is the method used to find the temperature coefficient of resistivity of an alloy?

The most common method is to measure the resistivity of the alloy at different temperatures and then plot a graph of resistivity against temperature. The slope of this graph gives the temperature coefficient of resistivity.

## 4. Can the temperature coefficient of resistivity be negative?

Yes, the temperature coefficient of resistivity can be negative. This means that as the temperature increases, the resistivity of the alloy decreases. This is known as a negative temperature coefficient of resistivity and is observed in some materials like semiconductors.

## 5. How does the temperature coefficient of resistivity affect the performance of electronic devices?

The temperature coefficient of resistivity plays a crucial role in the performance of electronic devices. A high temperature coefficient of resistivity can cause a significant change in the resistance of the material with temperature, which can affect the accuracy and stability of devices. Therefore, it is important to choose materials with a low temperature coefficient of resistivity for electronic applications.

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