The constant resistance of a resistor is 30 ohms, independent of the temperature. The resistor is made up of an aluminium resistor with resistance R1 at 0°C in series with a carbon resistor with resistance R2 at 0°C. Evaluate R1 and R2, temperature coefficient for aluminum and carbon are 3.9×10^-3°C^-1 and -0.50×10^-3°C^-1
To evaluate the resistances R1 and R2, we need to consider the temperature coefficient of resistivity for each material and their resistance values at 0°C.
Let's denote the resistance of the aluminum resistor at 0°C as R1 and the resistance of the carbon resistor at 0°C as R2.
Given:
- Resistance of the constant resistor: 30 ohms
- Temperature coefficient of aluminum (α1): 3.9×10^-3°C^-1
- Temperature coefficient of carbon (α2): -0.50×10^-3°C^-1
The total resistance (R_total) of resistors in series is given by the sum of their individual resistances:
R_total = R1 + R2
Since the constant resistance is 30 ohms and independent of temperature, we can equate it to the total resistance when the resistors are at 0°C:
30 = R_total | at 0°C
30 = R1 + R2 | at 0°C
Now, we need to account for the effect of temperature on each resistor. The change in resistance (ΔR) for a material with temperature change (ΔT) is given by:
ΔR = R * α * ΔT
For the aluminum resistor:
ΔR1 = R1 * α1 * ΔT | at 0°C
For the carbon resistor:
ΔR2 = R2 * α2 * ΔT | at 0°C
Since the resistors are in series, the total change in resistance is simply the sum of the changes in resistances of each resistor:
ΔR_total = ΔR1 + ΔR2
At 0°C, the total change in resistance is zero, as stated in the problem. Therefore, we can write:
0 = ΔR_total | at 0°C
0 = ΔR1 + ΔR2 | at 0°C
Now, we have two equations to solve simultaneously:
1) 30 = R1 + R2
2) 0 = R1 * α1 * ΔT + R2 * α2 * ΔT
We can rearrange equation 1) as:
R1 = 30 - R2
Substituting this into equation 2):
0 = (30 - R2) * α1 * ΔT + R2 * α2 * ΔT
Simplifying:
0 = 30 * α1 * ΔT - R2 * α1 * ΔT + R2 * α2 * ΔT
Combining like terms:
0 = (30 * α1 - R2 * α1 + R2 * α2) * ΔT
Since ΔT can be any value (other than zero) for the equation to hold true, then the coefficient in front of ΔT must be zero to satisfy the equation:
30 * α1 - R2 * α1 + R2 * α2 = 0
Now, solving the above equation for R2:
R2 * α2 = R2 * α1 - 30 * α1
R2 * (α2 - α1) = -30 * α1
R2 = (-30 * α1) / (α2 - α1)
Finally, substituting the given values of α1 = 3.9×10^-3°C^-1 and α2 = -0.50×10^-3°C^-1:
R2 = (-30 * 3.9×10^-3) / (-0.50×10^-3 - 3.9×10^-3)
Thus, by substituting the values, we can calculate the value of R2. To find R1, we can use R1 = 30 - R2.
To evaluate the values of R1 and R2, we can make use of the formula for the temperature-dependent resistance of a material:
R = R0 * (1 + α * ΔT)
where R is the resistance at a given temperature T, R0 is the resistance at the reference temperature T0, α is the temperature coefficient, and ΔT is the change in temperature.
Let's consider the reference temperature T0 = 0°C.
For the aluminum resistor, we can write:
R1 = R0_1 * (1 + α_1 * ΔT)
For the carbon resistor, we can write:
R2 = R0_2 * (1 + α_2 * ΔT)
We know that the total resistance R is equal to 30 ohms, so we have:
R = R1 + R2
Substituting the equations for R1 and R2, we get:
30 = R0_1 * (1 + α_1 * ΔT) + R0_2 * (1 + α_2 * ΔT)
Now we can solve for R1 and R2 by substituting the given values of α_1, α_2, and ΔT:
α_1 = 3.9×10^-3 °C^-1
α_2 = -0.50×10^-3 °C^-1
ΔT = T - T0 = T
Substituting these values, we have:
30 = R0_1 * (1 + 3.9×10^-3 * T) + R0_2 * (1 + -0.50×10^-3 * T)
Since we have two unknowns (R0_1 and R0_2) and only one equation, we need additional information to solve for R1 and R2.
changeinR1+changeinR2=0
3.8E-3(dT*X-.5E-3*dT*(30-x)=0
dT divides out, as E-3
3.8x-.5(-x)=.5*30 check that
x(3.8+.5)=.5(30)
x= 3.448 ohms R2=30-3.448 ohms