Part 103-108: Resistors and Resistance Worksheet 22A - Answers with Commentary

Introduction

A table of resistivity values and temperature coefficients is provided.

NOTE: The resistivity values are provided at 20°C. The temperature coefficient is provided at 0°C.

The resistivity unit is given in the table as µΩm. This unit is handy, as the most common resistance formula is \( R = \frac{\rho L}{A} \). If \( \rho \) is in µΩm, and \( A \) is in mm², no numbers need to be "converted" to Ωm or m². This is because 1 µΩm = 10^-6 Ωm, and 1 mm² = 10^-6 m². The two 10^-6 multiplications cancel each other out. Doing this also avoids cumbersome scientific notation workings.

The resistance at a given temperature is given by the formula below.

\( R_1 = R_0 \cdot (1 + \alpha_0 T_1) \)

where \( R_1 \) is the resistance at temperature \( T_1 \), \( T_1 \) is the temperature in °C, and \( \alpha_0 \) is the temperature coefficient of resistance at 0°C.

Resistances at different temperatures can be calculated by first converting the resistance to 0°C, then calculating the temperature at the second temperature. The formulas are shown below.

\( R_1 = R_0 \cdot (1 + \alpha_0 T_1) \)

\( R_2 = R_0 \cdot (1 + \alpha_0 T_2) \)

It is possible to define \( R_2 \) in terms of \( R_1 \) only, by substituting the equation \( R_0 = \frac{R_1}{1 + \alpha_0 T_1} \) into the equation for \( R_2 \). The worked equation is given below.

\( R_2 = R_1 \cdot \frac{1 + \alpha_0 T_2}{1 + \alpha_0 T_1} \)

Question 1

A cable is made of copper and measures 45m long, the conductor CSA is 10mm². Calculate the resistance of the cable.

NOTE: CSA stands for "conductor surface area".

Solution

The temperature is not stated, this solution assumes that the temperature is 20°C.

The formula to use is \( R = \frac{\rho L}{A} \).

The material is copper → \( \rho = \) 0.017 µΩm (at 20°C).

The resistance may be calculated directly by plugging into the standard resistance formula.

\( R = \frac{\rho L}{A} = \frac{0.017 \cdot 45}{10} = \frac{0.765}{10} = \) 0.0765 Ω #.

The resistance may also be calculated using "prefix-free" quantities. Remember that 0.017 µΩm = 0.017 · 10^-6 Ωm, and 10 mm² = 10 · 10^-6 m².

\( R = \frac{\rho L}{A} = \frac{0.017 \cdot 10^{-6} \cdot 45}{10 \cdot 10^{-6}} = \frac{0.765}{10} = \) 0.0765 Ω #.

The 10^-6 factors on the area and resistivity cancel out.

Question 2

Find the difference in Ohms between a bar 15 m long and measuring 3 mm × 10 mm if it was made from Copper or Brass.

Solution - Method 1 - Difference of Resistivity

The bar sizes are the same, so the difference in the resistance is proportional to the difference in the resistivities.

The resistivity of brass is 0.066 µΩm, and the resistivity of copper is 0.017 µΩm. The difference in resistivity \( \Delta \rho \) is therefore 0.049 µΩm.

The formula to use is \( \Delta R = \frac{\Delta \rho L}{A} \).

The resistance may be calculated directly by plugging into the standard resistance formula.

\( \Delta R = \frac{\Delta \rho L}{A} = \frac{0.049 \cdot 15}{3 \cdot 10} = \frac{0.735}{30} = \) 0.0245 Ω #.

So the brass bar has a resistance 0.0245 Ω higher than the copper bar.

Solution - Method 2 - Difference of Resistance

The bar sizes are the same, so the length and CSA are the same. The length is 15 m, the CSA is \( 3 \cdot 10 \) = 30 mm².

The resistivity of brass is 0.066 µΩm, and the resistivity of copper is 0.017 µΩm. The respective resistances are calculated below.

Quantities calculated for copper will have he subscript "Cu", quantities for brass will have the subscript "Br" (I am aware that Br is the chemical symbol for bromine, but that will not cause confusion here).

\( R_{\mathrm{Cu}} = \frac{ \rho_{\mathrm{Cu}} L}{A} = \frac{0.017 \cdot 15}{30} = \frac{0.255}{30} = \) 0.0085 Ω.

\( R_{\mathrm{Br}} = \frac{ \rho_{\mathrm{Br}} L}{A} = \frac{0.066 \cdot 15}{30} = \frac{0.99}{30} = \) 0.033 Ω.

The difference of resistance is \( 0.033 - 0.0085 = \) 0.0245 Ω. So the brass bar has a resistance 0.0245 Ω higher than the copper bar.

Question 3

Find the Voltage drop that occurs on a 120 mm² Aluminium cable, 2 km long and carrying 50 A.

Solution

This question is a straightforward application of the resistance formula and Ohm's Law. The resistivity of aluminium is 0.028 µΩm, assuming a conductor temperature of 20°C.

First, the resistance is calculated. For the resistance formula to give an output in ohms, the length unit must be in metres.

\( R = \frac{\rho L}{A} = \frac{0.028 \cdot 2 \cdot 10^3}{120} = \frac{56}{120} = \) 0.466667 Ω (6 sf).

The voltage drop is calculated using Ohm's Law.

\( V = I R = 50 \cdot 0.466667 = \) 23.33 V (4 sf) #.

Question 4

Calculate the length of 0.75 mm² nichrome wire required to make an element of 12 Ω resistance.

Solution - Method 1

This question requires solving for conductor length. The resistivity of nichrome is 1.122 µΩm, assuming a conductor temperature of 20°C.

First, the length is calculated. The length formula is obtained by rearranging \( R = \frac{\rho L}{A} \).

\( L = \frac{R A}{\rho} = \frac{12 \cdot 0.75}{1.122} = \frac{9}{1.122} = \) 8.021 m (4 sf) #.

Solution - Method 2

This question requires solving for conductor length. The resistivity of nichrome is 1.122 µΩm, assuming a conductor temperature of 20°C.

First, the resistance per unit length \( r \) is calculated.

\( r = \frac{\rho}{A} = \frac{1.122}{0.75} = \) 1.496 Ω.m^-1.

The length is calculated by dividing the target resistance by the resistance per length.

\( L = \frac{R}{r} = \frac{12}{1.496} = \) 8.021 m (4 sf) #.

Question 5

Fill in the blanks. The answers are in bold.

For any conductor, an increase in LENGTH, increases resistance and a decrease in AREA increases resistance.

For a conductor with a positive Tempco, resistance increases as Temperature increases.

For a conductor with a negative Tempco, resistance decreases as Temperature increases.

Question 6

The working temperature of a heater element using nichrome wire is 600°C. Find the resistance at this temperature if the resistance at 0°C is 5 Ohms.

Solution

The tempco of nichrome is 0.00017°C^-1.

The formula for resistance \( R_1 \) at a temperature \( T_1 \) in terms of the resistance at 0°C is given below.

\( R_1 = R_0 \cdot (1 + \alpha_0 T_1) \)

Plugging the quantities into the formula gives the result below.

\( R_1 = R_0 \cdot (1 + \alpha_0 T_1) = 5 \cdot (1 + 0.00017 \cdot 600) = 5 \cdot (1 + 0.102) = 5 \cdot 1.102 = \) 5.51 Ω at 600°C #.

Question 7

A copper cable has a resistance of 0.2 Ohms at 0°C. When a short circuit occurs, 3 kV are dropped across the cable as 8 kA flows. Find the temperature the cable reaches during the short circuit - is this good for the insulation? - Hint: Find the R during the short circuit using Ohms Law then apply the Resistivity formulae.

Solution

The resistance during the short circuit is calculated.

\( R = \frac{V}{I} = \frac{3000}{8000} = \) 0.375 Ω.

The resistance of the cable is 0.2 Ω at 0°C, so the problem involves finding the temperature that caused the resistance to increase to 0.375 Ω.

The formula for resistance \( R_1 \) at a temperature \( T_1 \) in terms of the resistance at 0°C is given below.

\( R_1 = R_0 \cdot (1 + \alpha_0 T_1) \)

This formula must be rearranged to solve for \( T_1 \).

\( R_1 = R_0 \cdot (1 + \alpha_0 T_1) \rightarrow \frac{R_1}{R_0} = 1 + \alpha_0 T_1 \rightarrow T_1 = \frac{\frac{R_1}{R_0} - 1}{\alpha_0} \)

The formula for \( T_1 \) is shown below.

\( T_1 = \frac{\frac{R_1}{R_0} - 1}{\alpha_0} \)

Plugging in the information we have, we can calculate \( T_1 \).

\( T_1 = \frac{\frac{0.375}{0.2} - 1}{0.00427} = \frac{1.875 - 1}{0.00427} = \frac{0.875}{0.00427} = \) 204.9°C (4 sf) #.

This temperature is right at the upper limit for non-mineral insulation. It is more likely than not that the cable insulation is permanently damaged.

Question 8

A special motor with silver conductors for windings, has a resistance of 2.7 Ohms at room temperature of 18°C and when run fully loaded for 30 minutes has a resistance of 3.37 Ohms. Find the average temperature of the conductors at full load.

NOTE: This is quite an effective method of finding the average temperature rise in a conductor. It is good because it is completely non-invasive and non-destructive.

Solution - Method 1 - Normalised Resistance Method

The "cold" resistance (resistance at 18°C) is first "normalised" to 0°C. The tempco of silver at 0°C is 0.004°C^-1.

\( R_0 = \frac{R_1}{1 + \alpha_0 T_1} = \frac{2.7}{1 + 0.004 \cdot 18} = \frac{2.7}{1.072} = \) 2.51866 Ω (6 sf).

This "normalised" resistance is then used to calculate the temperature rise required to get the "hot" temperature \( T_2 \).

\( T_2 = \frac{\frac{R_2}{R_0} - 1}{\alpha_0} = \frac{\frac{3.37}{2.51866} - 1}{0.004} = \frac{1.33801 - 1}{0.004} = \frac{0.33801}{0.004} = \) 84.5°C #.

Solution - Method 2 - Direct Calculation Method

This method uses the "direct" equation below.

The objective is to solve for \( T_2 \).

\( R_2 = R_1 \cdot \frac{1 + \alpha_0 T_2}{1 + \alpha_0 T_1} \)

Rather than rearrange the entire equation, all known quantities will be plugged in, and the equation solved from there.

\( 3.37 = 2.7 \cdot \frac{1 + 0.004 T_2}{1 + 0.004 \cdot 18} \)

All calculable quantities are "collapsed".

\( 1.24815 = \frac{1 + 0.004 T_2}{1.072} \)

Round 2.

\( 1.33801 = 1 + 0.004 T_2 \)

The equation can be easily solved from here.

\( 1.33801 = 1 + 0.004 T_2 \rightarrow 0.004 T_2 = 0.33801 \rightarrow T_2 = \frac{0.33801}{0.004} = \) 84.5°C #.