English .   Español  .

The application of Appropriate Technology

# Part 2: Resistors And Resistances

Sections:

### 2.1 Resistance And Ohm’s Law

Electrical resistance is a little like friction and electrons flowing in wires lose energy overcoming it. All conductors have some resistance and the larger the resistance, the larger the amount of energy dissipated within it as electrons flow, and therefore, the greater the energy needed to move electrons around a circuit. Hence, if a fixed voltage supply is connected to a circuit with a low overall resistance more current (i.e. coulombs of charge per second) will flow than if the same supply is connected to a higher resistance circuit.

Some electrical components, such as water heaters, are designed to have a large resistance so that a lot of electrical energy is converted into heat as current flows through them. Wires and cables on the other hand should have a low resistance so that not much energy is lost when electricity is supplied from one place to the other.

The unit of resistance (R) is the ohm (Ω) and one ohm is the resistance that causes a drop of one volt when a current of one ampere flows. For a metallic conductor which remains at a constance temperature Ohm’s law applies:

$\mathbf{R=\dfrac{U}{I}\qquad I=\dfrac{U}{R}\qquad U=I\times R}$

Components in circuits that are designed to have a fixed resistance are usually shown as a simple box, as depicted in figure 2.1.

Figure 2.1: a simple circuit containing a battery and a resistor.

Example

An electrical heater is used on a 240V supply draws a current of 12A. Its resistance is:

$R=\dfrac {U}{I}=\dfrac {240}{12}=20\Omega$

### 2.2 Resistivity

(also known as specific resistance)

When electrons flow through a wire they experience resistance and lose energy, the furtherer along the wire they flow the more energy they lose, therefore, we can say that the total resistance of a wire is proportional to its length.

Since the electrons are evenly distributed throughout the wire and since the current is the rate at which a charge passes any point on that wire, we can see that to provide any specific current the electrons in a wider wire will have to flow a shorter distance than electrons in a narrower wire (figure 2.2). We can therefore say that resistance is inversely proportional to the conductor’s cross-sectional area. Therefore thicker wires have less resistance per meter and will cause less energy to be lost as heat.

Figure 2.2: The effect of wire diameter on current.

Putting the previous two concepts together given us:

$R\propto \dfrac{l}{a}$

where l is the length of wire, a is the cross-sectional area and α means proportional to.

The resistivity (ρ) of a material is defined as the resistance between opposite faces of a cube of that material with a given side length. ρ is very small for most conductors and is usually quoted in micro-ohms (µΩ) for a 1 meter cube expressed as µΩm. For example aluminium has a resistivity of 0.0285µΩm. Thus the resistivity of a wire is proportional to the resistivity of the material from which it is made. We can combine resistivity with the previous equation to give:

$R=\dfrac {\rho l}{a}$

Note that the resistance calculated from this equation will be given in the same units of resistance used for the resistivity (i.e. µΩ). Also l and a must be in the same length units as ρ so that if ρ is in µΩm l must be in m and a must be in m2. Table 2.1 contains the resistivities of some metals.

Table 2.1: The resistivities of some metals.
Resistivity
µΩm µΩcm µΩmm
Copper (annealed)  0.0172 1.72 17.2
Copper (hard drawn)  0.0178 1.78  17.8
Aluminium  0.0285  2.85  28.5
Tin  0.114  11.4  11.4
Silver  0.0163 1.63  16.3
Brass  0.06-0.09 6-9  100
Iron  0.1  10.0  100

Example

Calculate the resistance of 1000m of 16mm2 single annealed copper wire.
From the table: ρ = 17.2µΩmm (since the cross sectional area is given in mm2)

$l=1000m=10^6m$
$R=\dfrac{\rho l}{a}=\dfrac{17.2\times 10^6}{16}=1\ 075\ 000\mu \Omega =1.075\Omega$

All wires and cables have some resistance, therefore there will always be some energy lost and a voltage drop within. Thin wires will get very hot and may burn out if they carry too much current, they may also cause such a large voltage drop that equipment may not function properly. Thick wires will reduce these phenomena however, because they contain more copper they will be considerably more expensive than thin wires.

Voltage drops in cables are normally recommended to be no more than 4% of the supply voltage. That means a 9.6V drop for a 240V supply or a 4V drop for a 110V supply. Note that you can only measure a voltage drop across a length of cable when a current is flowing in it.

Example

A twin 2.5mm2 MI cable feeds a heater which takes a current of 20A. If the cable is 100m long calculate the voltage drop in it and the PD across the heater if the supply voltage is 240V. What must the minimum cross sectional area of a replacement cable be if the voltage drop is not to exceed 9.6V.

MI cables have hard draw copper cores, therefore ρ = 17.8µΩm (table 2.1). Since it is a twin cable the total length of conductor is 200m.

Cable resistance:

$R=\dfrac {\rho l}{a}=\dfrac{17.8\times 200\times 1000}{2.5}\times 1\ 424\ 000\mu \Omega =1.424\Omega$

Cable voltage drop:

$U=I\times R=20\times 1.424=28.5V$

PD across the heater:

$U=240-28.5=211.5V$

Note that this is a 28.5V or 11.9% drop. For a drop of 9.6V:

Resistance of the cable:

$R=\dfrac{U}{I}=\dfrac{9.6}{20}=0.48\Omega$

Minimum cross sectional area:

$a=\dfrac{\rho l}{a}=\dfrac{17.8\times 200\times 1000}{1\: 000\: 000\times 0.48}=17.4mm^2$

This is not a standard size so a 10mm2 would be used. Such a cable is rated at 77A although we only require it to carry 20A. Therefore voltage drop and current rating must be considered when selecting cables.

### 2.3 Temperature And Resistance

When a 2V supply is connected to a 60W, 240V bulb it draws a current of 25mA, its resistance is thus:

$R=\dfrac{U}{I}=\dfrac{2}{0.025}=80\Omega$

When the same bulb is connected to the correct 240V supply it glows white hot and draws a current of 250mA and its resistance is:

$R=\dfrac{U}{I}=\dfrac{240}{0.250}=9600\Omega$

Thus we can see that a larger current has caused the bulb to get very hot and its resistance has increased by 12 times.

To take account of the change in resistance with temperature the temperature coefficient of resistance (α) is used. α for a material at 0°C is the change in resistance of a 1Ω sample of that material when the temperature increases from 0°C to 1°C. Matters are complicated furtherer because it is not easy to measure the resistance of a conductor at 0°C hence, a value for α is often quoted for a temperature increase of 20°C to 21°C. Table 2.2 contains the temperature coefficient of some metals.

Table 2.2: The temperature coefficient of some metals.
* Carbon has a negative temperature coefficient of resistance meaning that unlike most metals its resistance decreases as the temperature increases.
Temperature Coefficient of Resistance
(/°C at 0°C) (/°C at 20°C)
Copper +0.0043 +0.00396
Aluminium +0.0040 +0.00370
Brass +0.0010 +0.00098
Iron +0.0066 +0.00583
Nickel-Chrome +0.00017 +0.000169
Carbon* -0.0005 -0.00047
Silver +0.0041 +0.00379

We can see from the above table that when the temperature increases from 20°C to 21°C the resistance of a 1Ω copper resistance will increase to 1.00396Ω. Thus:

$\mathbf {R_t=R_0(1+\boldsymbol{\alpha} T)}$

where:
Rt = total conductor resistance at T (Ω)
R0 = resistance of a conductor at 0°C (Ω)
α = temperature coefficient of resistance
T = temperature (°C)

And:

$\mathbf {R_t=R_{20}(1+\boldsymbol{\alpha} T)}$

where:
R20 = resistance of a conductor at 20°C (Ω)

If the resistance of a conductor is not known at the temperature for which α is known the following method can be used:

R1 = the conductor resistance at temperature T1
R2 = the conductor resistance at temperature T2

$R_1=R_0(1+\alpha T_1)\qquad \text{and}\qquad R_2=R_0(1+\alpha T_2)$

Dividing:

$\dfrac {R_1}{R_2}=\dfrac {R_0\left ( 1+\alpha T_1 \right )}{R_0\left ( 1+\alpha T_2 \right )}=\dfrac{1+\alpha T_1}{1+\alpha T_2}$

Hence:

$\mathbf{R_2=\dfrac {R_1\left (1+\boldsymbol{\alpha} T_2 \right )}{1+\boldsymbol{\alpha} T_1}}$

Therefore the value of R0 has been eliminated from the equation.

Example

The winding of a DC motor is made from annealed copper and has a resistance of 500Ω at 15°C. What current will flow at the operating temperature of 35°C if the field PD is 300V? (α = 0.0043/°C for annealed copper at 0°C)

$R_2=\dfrac {R_1\left (1+\alpha T_2 \right )}{1+\alpha T_1}=\dfrac{500\times \left ( 1+0.0043\times 35 \right )}{1+0.0043\times 15}=540\Omega$
$I=\dfrac{U}{R}=\dfrac{300}{540}=0.556A$

Changes in resistance with temperature can have serious consequences. Consider a 60W bulb connected to a 240V supply, its cold resistance is 80Ω and the current drawn as soon as it is switched on will be:

$I=\dfrac{U}{R}=\dfrac{240}{80}=3A$

As the lamp heats up the resistance quickly increases and the current reduces to 250mA. The short lived ‘transient’ that flows current when the bulb is switched on often causes the bulb to blow when it is first switched on. Also if a string of bulbs are operated from a single switch the current may be enough to blow a sensitive fuse.

### 2.4 Series And Parallel Resistors

Electrical components can be connected together in two basic ways: parallel and series.

When resistors are connected in series (figure 2.3) a fraction of the supply voltage is dropped across each resistor and each resistor dissipates some of the total energy from each coulomb of charge. The total resistance (RT) of the circuit (ignoring the resistance of the wires) is the sum of the resistances and the same current flows through each resistor.

Figure 2.3: Three resistors in series.

The voltage drop across each resistor is given by:

$U_1 = IR_1,\qquad U_2 = IR_2\qquad \text{and}\qquad U_3 = IR_3$

Total resistance:

$R_T=R_1+R_2+R_3$

Supply voltage:

$U=U_1+U_2+U_3$

The current flowing through each resistor is given by:

$I=\dfrac{U}{R_T}$

When resistors are connect in parallel the voltage across each resistor is the same since, any two components connected to the same point in a circuit must be at the same voltage (figure 2.4). The current flowing through each resistor is a fraction of the total current (IT). The reciprocal of the total resistance is the sum of the reciprocals of the resistances.

Figure 2.4: Three resistors in parallel.

The current through each resistor:

$I_1=\dfrac{U}{R_1}\qquad I_2=\dfrac{U}{R_2}\qquad I_3=\dfrac{U}{R_3}$

The total current:

$I_T=I_1+I_2+I_3$

Therefore:

$I_T=\dfrac{U}{R_1}+\dfrac{U}{R_2}+\dfrac{U}{R_3}$

Dividing through by U gives:

$\dfrac{I_T}{U}=\dfrac{1}{R_T}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$

Figure 2.5 shows how circuits containing both parallel and series resistors can be simplified.

Figure 2.5: two blocks of resistors can be simplified by summing the resistors.

Once the total resistance has been calculated (figure 2.5) we can calculate the total current:

$I=\dfrac{U}{R_T}=\dfrac{12}{12}=1A$

We can calculate the voltage drop across each bank of resistors:

$U_1=IR_1=8V,\qquad U_2=IR_2=4V$

We can now calculate the current in each resistor:

$I_1=\dfrac{8}{10}=0.8A,\qquad I_2=\dfrac{8}{40}=0.2A$

In the second block of resistors the current will split equally since each resistor is 12Ω therefore:

$I_3=I_4=I_4=1/3A$