Studdy AI Logo
Math  /  Algebra

Question14. 17. Find the temperature coefficient of resistance of iron at 20∘C20^{\circ} \mathrm{C}, if iron has an inferred zero resistance temperature −162∘C-162^{\circ} \mathrm{C} ? Ans 0,00551/∘C−0,00551 /{ }^{\circ} \mathrm{C}-

Studdy Solution

STEP 1

1. The temperature coefficient of resistance (α\alpha) is a measure of how much the resistance of a material changes with temperature.
2. The formula for the temperature coefficient of resistance is given by: $ \alpha = \frac{1}{R_0} \cdot \frac{R - R_0}{T - T_0} \] where \(R_0\) is the resistance at the reference temperature \(T_0\), and \(R\) is the resistance at temperature \(T\).
3. The inferred zero resistance temperature is the temperature at which the resistance is theoretically zero.

STEP 2

1. Identify the known values and the formula to use.
2. Substitute the known values into the formula.
3. Solve for the temperature coefficient of resistance (α\alpha).

STEP 3

Identify the known values and the formula to use.
- The reference temperature T0=20∘CT_0 = 20^{\circ} \mathrm{C}. - The inferred zero resistance temperature T=−162∘CT = -162^{\circ} \mathrm{C}. - The formula to use is: $ \alpha = \frac{1}{T_0 - T} \]

STEP 4

Substitute the known values into the formula.
Substitute T0=20∘CT_0 = 20^{\circ} \mathrm{C} and T=−162∘CT = -162^{\circ} \mathrm{C} into the formula: α=120−(−162)\alpha = \frac{1}{20 - (-162)}

STEP 5

Solve for the temperature coefficient of resistance (α\alpha).
Simplify the expression: α=120+162\alpha = \frac{1}{20 + 162} α=1182\alpha = \frac{1}{182}
Calculate the value: α≈0.00549/∘C\alpha \approx 0.00549 /{ }^{\circ} \mathrm{C}
The temperature coefficient of resistance of iron at 20∘C20^{\circ} \mathrm{C} is approximately 0.00549/∘C0.00549 /{ }^{\circ} \mathrm{C}.
Note: The provided answer 0.00551/∘C0.00551 /{ }^{\circ} \mathrm{C} might be due to rounding differences.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord