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.

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=U/I = 2/0.025 =80ohms

• 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=U/I = 240/0.250 = 9600ohms

• 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.

•  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: 

                                                          Rt=Ro(1+αT)
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: Rt=R20(1+αT)
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
R1=Ro(1+αT1) and R2=Ro(1+αT2)
Dividing: R1/R2 = (Ro(1+αT1))/(Ro(1+αT2)) = (1+αT1)/(1+αT2)

Hence:
                                   R2=(R11+αT2)/(1+αT1)
Therefore the value of R0 has been eliminated from the equation.

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.

• Putting the previous two concepts together given us:

                                           R~1/a (*proportional)

• 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=ρl / 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.

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=ρl / a = (17.2*10^6)/16 =1075000μΩ=1.075Ω

• 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.