Pearson Edexcel Physics

Unit 2: Waves and Electricity

2.4 Electric Circuits

Pearson Edexcel Physics Unit 2: Waves and Electricity 2.4 Electric Circuits Candidates will be assessed on their ability to::
64.	Understand that electric current is the rate of flow of charged particles and be able to use the equation [math]I = \frac{\Delta Q}{\Delta t}[/math]
65.	Understand how to use the equation [math]V = \frac{W}{Q}[/math]
66.	Understand that resistance is defined by [math]R = \frac{V}{I}[/math] and that Ohm’s law is a special case when [math]I ∝ V[/math] for constant temperature
67.	(a) Understand how the distribution of current in a circuit is a consequence of charge conservation (b) Understand how the distribution of potential differences in a circuit is a consequence of energy conservation
68.	Be able to derive the equations for combining resistances in series and parallel using the principles of charge and energy conservation, and be able to use these equations
69.	Be able to use the equations [math] P= VI, W = VIt[/math] and be able to derive and use related equations, e.g. [math]P = I^2 R \quad \text{and} \quad P = \frac{V^2}{R}[/math]
70.	Understand how to sketch, recognise and interpret current-potential difference graphs for components, including ohmic conductors, filament bulbs, thermistors and diodes
71.	Be able to use the equation [math]R = \frac{ρl}{A}[/math]
72.	CORE PRACTICAL 7: Determine the electrical resistivity of a material
73.	Be able to use [math]I = nqvA[/math] to explain the large range of resistivities of different materials
74.	Understand how the potential along a uniform current-carrying wire varies with the distance along it
75.	Understand the principles of a potential divider circuit and understand how to calculate potential differences and resistances in such a circuit
76.	Be able to analyse potential divider circuits where one resistance is variable including thermistors and light dependent resistors (LDRs)
77.	Know the definition of electromotive force (e.m.f.) and understand what is meant by internal resistance and know how to distinguish between e.m.f. and terminal potential difference
78.	CORE PRACTICAL 8: Determine the e.m.f. and internal resistance of an electrical cell
79.	Understand how changes of resistance with temperature may be modelled in terms of lattice vibrations and number of conduction electrons and understand how to apply this model to metallic conductors and negative temperature coefficient thermistors
80.	Understand how changes of resistance with illumination may be modelled in terms of the number of conduction electrons and understand how to apply this model to LDRs.

64) Understand that electric current is the rate of flow of charged particles and be able to use the equation
[math]I = \frac{\Delta Q}{\Delta t}[/math]
⇒ Electric charge:
Some particles have an electric charge. For example, the electron has a negative charge. In SI units, electric charge is measured in coulombs (C) and the amount of charge on a single electron in these units is [math]-1.6 × 10^{-19} С[/math]
[math]e = -1.6 × 10^{-19} С[/math]
This means that you would have one coulomb of negative charge if you collected together [math]6.25 \times 10^{18}[/math] electrons, as shown in this calculation:
[math]\begin{gather}
\text{Total charge: } Q = ne = 6.25 \times 10^{18} \times (-1.6 \times 10^{-19}) \\
Q = -1 \, \text{C}
\end{gather}[/math]
The charges on fundamental particles such as electrons are fixed properties of these particles. It is impossible to create or destroy charge – the total charge must always be conserved.
⇒ Electric Current:
“The rate of movement of charge”.
As it is usually a physical movement of billions of tiny charged particles, such charge movements are often said to flow.
The flow of charge like the current in a river can be useful, and we will see later how current splits and recombines at circuit junctions in a manner that is like water flow.
As the total amount of water in a river at a given time does not change, even if the river splits, this again shows the conservation of charge.
Electric current occurs when a charged particle, which is free to move, experiences an electric force. If it can move it will be accelerated by the force. This movement of charge forms the electric current.
Most electric circuits are made from metal wiring in which there are electrons that are free to move. These conduction electrons then form the current.
⇒ Calculating Current:
The SI unit for electric current is the ampere, A. Current can be calculated from the equation:
[math]\begin{gather}
\text{Current (A)} = \frac{\text{Charge passing a point (C)}}{\text{Time for that charge to pass (s)}} \\
I = \frac{\Delta Q}{\Delta t}
\end{gather}[/math]
One ampere (1 A) is the movement of one coulomb (1 C) of charge per second (1 s).
Example:
If the lightning takes 0.1 seconds to transfer 1150 coulombs of charge, we could calculate the current in the lightning:
[math]\begin{gather}
I = \frac{\Delta Q}{\Delta t} \\
I = \frac{1150}{0.1} \\
I = 11500 \, \text{A}
\end{gather}[/math]
65) Understand how to use the equation
[math]V = \frac{W}{Q}[/math]
⇒ Voltages:
The electrical quantity voltage is a measure of the amount of energy a component transfers per unit of charge passing through it. It can be calculated from the equation:
[math]\begin{gather}
\text{Voltage (V)} = \frac{\text{Energy transferred (J)}}{\text{Charge passing (C)}} \\
V = \frac{E}{Q}
\end{gather}[/math]
⇒ Electromotive force:
For a supply voltage – a component which is putting electrical energy into a circuit – the correct term for the voltage is electromotive force, or emf.
If a cell supplies one joule (1 J) of energy per coulomb of charge (1 C) that passes through it, it has an emf of 1 volt (1 V).
[math]\begin{gather}
\text{emf (V)} = \frac{\text{energy transferred (J)}}{\text{charge passing (C)}} \\
\varepsilon = \frac{E}{Q}
\end{gather}[/math]
For example, if the bicycle lamp in figure 1 was powered by a cell with an emf of 1.5 V, then this cell would transfer 1.5 J of electrical energy to each coulomb of charge that passed through it.

Figure 1 Electric circuits transfer energy.
⇒ Potential difference:
For a component which is using electrical energy in a circuit and transferring this energy, the correct term for the voltage is potential difference, or pd.
If a component uses one joule (1 J) of energy per coulomb of charge (1 C) that passes through it, it has a pd of 1 volt (1 V).
The energy being used by the component could be referred to as work done, W.
[math]\begin{gather}
\text{pd (V)} = \frac{\text{energy transferred (J)}}{\text{charge passing (C)}} \\
V = \frac{W}{Q}
\end{gather}[/math]
For example, if the bicycle light in figure 1 produced light from an LED with a pd of 1.5 V, then this LED would transfer 1.5 J of electrical energy from each coulomb of charge that passed through it to light energy and a little thermal energy.
Example:
A cell transfers 40J of chemical energy for every 15 C of electrons that pass through it, in order to power an electrical circuit. What is the cell’s emf?
Solution:
[math]\begin{gather}
\text{emf (V)} = \frac{\text{energy transferred (J)}}{\text{charge passing (C)}} \\
\varepsilon = \frac{E}{Q} \\
\varepsilon = \frac{40}{15} \\
\varepsilon = 2.7 \, \text{V}
\end{gather}
[/math]
Example:
For every 2000 C of charge passing through it, a loudspeaker transfers 18 500J of electrical energy to sound and thermal energy. What is the loudspeaker’s pd?
Solution:
[math]\begin{gather}
\text{pd (V)} = \frac{\text{energy transferred (J)}}{\text{charge passing (C)}} \\
V = \frac{W}{Q} \\
V = \frac{18500}{2000} \\
V = 9.25 \, \text{V}
\end{gather}[/math]
66) Understand that resistance is defined by [math]R = \frac{V}{I}[/math] and that Ohm’s law is a special case when I ∝ V for constant temperature
⇒ Resistance:
“Electrical resistance is considered to be the opposition to the flow of current within a conductor.”
It can be calculated from the equation:
[math]\begin{gather}
\text{Resistance (}\Omega\text{)} = \frac{\text{Potential difference (V)}}{\text{Current (I)}} \\
R = \frac{V}{I}
\end{gather}[/math]
Example:
What is the resistance of the resistor in a circuit which is electric current is 0.18 A while potential difference is 4 V.
Solution:
[math]\begin{gather}
\text{Resistance (}\Omega\text{)} = \frac{\text{Potential difference (V)}}{\text{Current (I)}} \\
R = \frac{V}{I} \\
R = \frac{4}{0.18} \\
R = 22 \, \Omega
\end{gather}[/math]
⇒ Ohm’s Law:
If the current is proportional to the voltage driving it through a component, this component is called an ohmic conductor, as this means it follows Ohm’s law.
[math]\begin{gather}
I \propto V \\
V = I \times R
\end{gather}[/math]
For an ohmic conductor, the answer to the calculation of resistance would be the same for all voltages and their corresponding current values (providing the temperature remains constant.
67)
(a) Understand how the distribution of current in a circuit is a consequence of charge conservation
(b) Understand how the distribution of potential differences in a circuit is a consequence of energy conservation
⇒ Current:
The total amount of charge within a circuit cannot increase or decrease when the circuit is functioning.
Figure 2 Ammeters must always be in series with the component they are measuring.
⇒ Current in Series Circuits:
Any group of components that follow in series in a circuit, with no junctions in the circuit, must have the same current through them all.
Figure 3 The current is constant throughout a series loop in a circuit.
⇒ Current in Parallel Circuits:
Whenever a current encounter a junction in a circuit, the charges can only go one way or the other, so the current must split.
The proportions that travel along each possible path will be in inverse proportion to the resistance along that path. If the path has high resistance, the current is less likely to go that way.
However, the total along the branches must add up to the original total current.
Figure 4 Current splits at circuit junctions, but the total current will remain the same, in order to conserve charge.
⇒ Voltage around Circuits:
Figure 5 Voltmeters must always be in parallel across the component they are measuring.
⇒ Voltages in Series Circuits:
Any group of emfs that follow in series in a circuit, with no junctions in the circuit, will have a total emf that is the sum of their individual values, that accounts for the direction of their positive and negative sides.
For example, two 1.5 V cells in a TV remote control will be in series, so that they supply an emf of 3.0 V. This is also true for any string of potential differences in series.
Figure 6 Emfs in series will add up. Take care to account for their direction, as cells in opposite directions oppose each other and cancel out the emf they would supply.
Figure 7 Potential differences in series will add up. What would be the total pd across the two resistors in this diagram?
⇒ Voltages in Parallel circuits:
If the total voltage across any branch of a circuit is known, then the voltage across any other branch in parallel with it will be identical.
Figure 8 Both voltmeters here will read 6 V: the voltage is the same across all parallel branches within a circuit.
68) Be able to derive the equations for combining resistances in series and parallel using the principles of charge and energy conservation, and be able to use these equations
⇒ Resistance:
Resistance can be calculated from Ohm’s law as [math]\frac{V}{I}[/math] This can be done for any individual component or a whole branch of a circuit, as long as the voltage and current are known. This also means that if we follow the rules for current and voltage variations around series and parallel circuits, we come up with simple rules for the total resistance in series and parallel combinations.
⇒ Resistance in Series:
Any group of resistances that follow in series in a circuit, with no junctions in the circuit, will have a total resistance that is the sum of their individual values. In figure 7, we could use the total pd across both resistors and the current through them to calculate their total resistance:
[math]\begin{gather}
R_{\text{total}} = \frac{V_{\text{total}}}{I} \\
R_{\text{total}} = \frac{6}{0.2} \\
R_{\text{total}} = 30 \, \Omega
\end{gather}[/math]
⇒ Derivation:
Resistors in series will have a total pd across them that is the sum of their individual pds. The current through them will be the same for each one. Take an example of three resistors in series:
[math]\begin{gather}
V_{\text{total}} = V_1 + V_2 + V_3 \\
IR_{\text{total}} = IR_1 + IR_2 + IR_3 \\
R_{\text{total}} = R_1 + R_2 + R_3
\end{gather}[/math]
⇒ Resistance in Parallel:
The total resistance of a group of resistors in parallel can be calculated from the equation:
[math]\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots[/math]
Figure 9 Resistors in parallel follow a reciprocal sum rule to find their total resistance.
⇒ Derivation:
Resistors in parallel will have a total current through them that is the sum of their individual currents. The pd across them will be the same for each branch. Take an example of three resistors in parallel:
[math]\begin{gather}
I_{\text{total}} = I_1 + I_2 + I_3 \\
I = \frac{V}{R} \\
\frac{V}{R_{\text{total}}} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} \\
\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\end{gather}[/math]
69) Be able to use the equations P = VI, W = VIt and be able to derive and use related equations, e.g.
[math]P = I^2 R \quad \text{and} \quad P = \frac{V^2}{R}[/math]
⇒ Electrical work:
Work done has the symbol W, and as the equation defining potential difference includes a term for the amount of energy transferred, E, these two will be the same.
[math]W = E[/math]
The equation for pd, in rearranged form, gives us:
[math]W = V × Q[/math]
The definition of current, in rearranged form, is:
[math]Q = I × t[/math]
Combining these will give us the work done by a component in an electric circuit:
[math]W = V × I × t[/math]
⇒ Electrical Power:
Power, P, is the rate of transfer of energy, or the rate of doing work. In an electrical circuit, the energy is dissipated by component. The mathematical definition is:
[math]P = \frac{E}{t}[/math]
Or
[math]P = \frac{W}{t}[/math]
Incorporating the equation for work in a circuit from above:
[math]\begin{gather}
P = \frac{VIt}{t} \\
P = VI
\end{gather}[/math]
70) Understand how to sketch, recognize and interpret current-potential difference graphs for components, including ohmic conductors, filament bulbs, thermistors and diodes
⇒ Current-Voltage Characteristics:
For an electric circuit design, we need to know how components will react when the pd across them changes, in order to ensure that the circuit performs its intended function under all circumstances.
Part of the specification of any component is a graph of its I-V characteristics.
The graph for a simple resistor, and a metal at constant temperature would produce the same straight-line result.
The only difference would be that the gradient will be different in each case, as it corresponds to the specific resistance of the resistor or wire.
Figure 10 A filament lamp is an example of a non-ohmic conductor. This is because the current through it affects its own temperature – higher current means a higher temperature – and controlling temperature is part of the definition of Ohm’s law.
⇒ I-V Graph for a filament bulb:
For the filament lamp I-V graph in figure 10, For a small voltage, the current is proportional to it, as shown by the straight-line portion of the graph through the origin. At higher voltages, a larger current is driven through the lamp filament wire, and this heats it up.
At hotter temperatures, metals have higher resistance.
The gradient of the graph in figure 11 was given as the reciprocal of the resistance, and so the gradient here becomes less towards higher voltages: higher resistance means a lower gradient.
Figure 11 I-V graph for a resistor.
⇒ I-V graph for a diode:
For the thermistor in figure 12, its resistance reduces with the temperature. The gradient of the line increases with the heating effect of the increasing current.
The gradient represents the reciprocal of the resistance: larger gradient value means lower resistance. This is a result of its manufacture from semiconductor materials, whose atoms release more conduction electrons as the temperature rises.
Figure 12 A thermistor is designed to alter its resistance with temperature; in the reverse manner to a filament bulb.
71) Be able to use the equation
[math]R = \frac{ρl}{A}[/math]
⇒ Resistivity:
Resistance is the result of collisions between charge carriers and atoms in the current’s path. This effect will vary depending on the density of charge carriers and the density of fixed atoms, as well as the strength of the forces between them.
So, pieces of different materials with identical dimensions will have differing resistances.
The general property of a material to resist the flow of electric current is called resistivity, which has the symbol rho, ρ, and SI units ohm metres, Ωm. The resistance of an object is dependent on its dimensions and the material from which it is made.
[math]\begin{gather}
\text{Resistance (}\Omega\text{)} = \frac{\text{Resistivity (}\Omega\text{m)} \times \text{Sample length (m)}}{\text{Cross-sectional area (m}^2)} \\
R = \frac{\rho l}{A}
\end{gather}[/math]
Example:
What is the resistance of a piece of copper fuse wire if it is 0.40 mm in diameter and 2 cm long?
Solution:
Wire radius = [math]r = 0.2 mm = 2 × 10^{-4} m[/math]
Cross- sectional area A = [math]πr^2 = (3.14) (2 × 10^{-4})^2 = 1.3 × 10^{-7} m^2[/math]
[math]\begin{gather}
R = \frac{(1.7 \times 10^{-8})(0.02)}{1.3 \times 10^{-7}} \\
R = 2.6 \times 10^{-3} \, \Omega
\end{gather}[/math]

Material	Resistivity [math](ρ/Ωm) at 20℃[/math])	[math]\frac{\Delta \rho}{\Delta t} \quad \text{in } \% \, ^\circ\text{C}^{-1}[/math]
Silver	[math]1.6 × 10^{-8}[/math]	+0.38
Copper	[math]1.7 × 10^{-8}[/math]	+0.40
Aluminium	[math]2.8 × 10^{-8}[/math]	+0.38
Constantan	[math] 4.9 × 10^{-7}[/math]	+0.003
Germanium	[math]4.2 × 10^{-1}[/math]	-5.0
Silicon	[math]2.6 × 10^{3}[/math]	-7.0
Polyethene	[math]2 × 10^{11}[/math]
Glass	[math] ~ 10^{12}[/math]
Epoxy resin	[math]~ 10^{15}[/math]

72) CORE PRACTICAL 7: Determine the electrical resistivity of a material
?
73) Be able to use I = nqvA to explain the large range of resistivities of different materials
⇒ Conduction in metals:
The structure of metals has a regular lattice of metal atoms. These are bonded together through the sharing of electrons, which act as if they were associated with more than one atom. Many of the atoms also have an outer electron that is not needed for bonding between the atoms.

Figure 13 Conduction in metals happens as the free electrons add an overall movement along the direction of the voltage across the conductor (towards the positive anode) to their random collisions and vibrations.
o These free electrons have a random motion, which changes as they collide with atoms or other electrons, but on average the overall position of all the charge in the metal is stationary.
However, if a source of emf is connected across the metal, the electric field it sets up in the metal will push the negative electrons towards the positive end of the field.
The slow overall movement of the electrons is called their drift velocity.
⇒ The transport equation:
The value of the electric current in a metal can be calculated from the fundamental movement of the electrons.
[math]I = \frac{\Delta Q}{\Delta t}[/math]
If we consider the cylinder shaded on the diagram in figure 14 as the length of the wire that the charges move through in a time ∆t, then we need to calculate how much charge flows through it in that time.
There are n electrons per cubic meter of this metal, and the wire has a cross-sectional area, A. Their movement is at a drift velocity, v, and the distance this takes them along the wire in that time is ∆x. So:
[math]∆x = v∆t[/math]
The total charge will be the number of electrons multiplied by the charge on each, e. The number of electrons will be their density. n, multiplied by the volume of the cylinder, V, that they travel through in ∆t.
[math]\begin{gather}
\Delta Q = n \times V \times e \\
\Delta Q = n \times A \Delta x \times e \\
\Delta Q = n \times A v \Delta t \times e \\
I = \frac{\Delta Q}{\Delta t} \\
I = \frac{n A v \Delta t e}{\Delta t} \\
I = n A v e
\end{gather}[/math]
This is called the transport equation.
Figure 14 A section of metal wire with dimensions to show how to make calculations using the transport equation.
⇒ Example:
What is the drift velocity of the electrons in a copper wire which has a diameter of 0.22 mm and carries a current of 0.50 А?
Solution:
The number density of conduction electrons in copper,
[math]\begin{gather}
n = 8.5 \times 10^{28} \, \text{m}^{-3} \\
\text{Wire radius} = 0.11 \, \text{mm} = 1.1 \times 10^{-4} \, \text{m} \\
\text{Cross-sectional area, } A = \pi r^2 \\
A = (3.14)(1.1 \times 10^{-4})^2 \\
A = 3.8 \times 10^{-8} \, \text{m}^2 \\
I = n A v e \\
v = \frac{I}{n A e} \\
v = \frac{0.50}{(8.5 \times 10^{28})(3.8 \times 10^{-8})(1.6 \times 10^{-19})} \\
v = 9.67 \times 10^{-4} \, \text{m} \cdot \text{s}^{-1}
\end{gather}[/math]
74) Understand how the potential along a uniform current-carrying wire varies with the distance along it
75) Understand the principles of a potential divider circuit and understand how to calculate potential differences and resistances in such a circuit
⇒ Potential divider:
all the voltage supplied by emfs in a circuit loop must be used by components as potential differences at other points around the circuit loop.
The way that the voltage splits up is in proportion to the resistances of the components in the circuit loop, as shown in figure 15.
Figure 15 All of the 6 volts supplied by the battery must be used up by the resistors. The total pd of 6V is split into 2V and 4 V in proportion to their resistances. Ammeters have zero resistance, so transfer zero energy and have zero pd.
The voltages across the two resistors must add up to 6 V. The 20 resistors have twice the resistance, so takes twice the voltage of the 10 resistors. This is a potential divider circuit.
Ohm’s law explains why the voltage is split in proportion to the resistance.
Consider the resistances and voltages in figure 7 in more general terms, as R₁ (for the 10 resistor) and R₂, and their corresponding pds V₁ and V₂.
This will make the calculation valid for all values of resistances, and calculating the current through each one gives:
[math]I_1 = \frac{V_1}{R_1} \quad \text{and} \quad I_2 = \frac{V_2}{R_2}[/math]
However, they both have the same current passing through them, so
[math]\begin{gather}
I_1 = I_2 \\
\frac{V_1}{R_1} = \frac{V_2}{R_2} \\
\frac{V_1}{V_2} = \frac{R_1}{R_2}
\end{gather}[/math]
So, the voltage is split in the same proportion as the ratio of the resistances for any values of resistance.
⇒ The Potential divider equation:
As the pd in a potential divider circuit is split in proportion to the resistances of the components, we can calculate the pd across them mathematically.
Figure 16 Generalized potential divider circuit.
In the circuit diagram of figure 16, we have the general terminology for the values used in calculations of a potential divider circuit: V_out, R₁, R₂ and V_in.
Using these, we can calculate the pd across the component of interest, which is usually drawn as the second resistance, R₂, using the potential divider equation:
[math]V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2}[/math]
⇒ Example:
Using the values in figure 8, we can find the pd across resistor R₂:
[math]\begin{gather}
V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2} \\
V_{\text{out}} = 12 \times \frac{500}{100 + 500} \\
V_{\text{out}} = 12 \times \frac{5}{6} \\
V_{\text{out}} = 10 \, \text{V}
\end{gather}[/math]
76) Be able to analyse potential divider circuits where one resistance is variable including thermistors and light dependent resistors (LDRs)
The voltage is split in the same proportion as the ratio of the resistances for any values of resistance.
[math]\frac{V_1}{V_2} = \frac{R_1}{R_2}[/math]
This means that for a known emf supply voltage, we can use carefully chosen resistances to share the voltage and provide a specific value of pd on one component.
If we then also remember that all parallel branches in a circuit must have the same voltage, then any branch we set up in parallel with that specific pd will also have the same voltage.
So, we can set up a circuit to provide an exactly chosen value of voltage to the parallel branch. Using variable resistors, this set up can then be used to provide whatever voltage we choose.
This can be particularly useful if our emf source is a fixed value, like a battery, which can only supply, say, 6 V.
Figure 17 Adjusting the variable resistor alters the proportion of the voltage it takes, and so the pd that is left for the resistance wire can be varied to any value we choose.
We saw graph of current against potential difference for various components.
In order to undertake the investigation to produce these data, a potential divider circuit could be used. Figure 17 shows a piece of resistance wire for which the current and voltage can be measured.
If the variable resistor is set to a high resistance, then it will have a high potential difference, and only a small voltage will be left for the test wire.
By decreasing the resistance of the variable resistor, the pd across the test wire will increase. We can then adjust the variable resistor to set the values of pd on the test wire to those we need for the I-V plot.
77) Know the definition of electromotive force (e.m.f.) and understand what is meant by internal resistance and know how to distinguish between e.m.f. and terminal potential difference
The best ammeters have such a small resistance that the effect on their current measurements can be ignored, but it is not actually zero. Sources of electromotive force (emf), such as batteries, will also have a small, but non-zero, resistance.
The resistance of an emf source is called its internal resistance. In most situations, the internal resistance of an emf source will not affect the performance of a circuit. It may not even be noticeable.
However, as the power lost – energy wasted as heat – through a component increases with the current through it.
The net effect of this energy transfer by the emf source itself will be for it to supply a smaller voltage to the rest of the circuit, which could affect the circuit’s performance.
A large current through the internal resistance could also damage the emf source as a result of ohmic (resistance) heating.
Figure 18 Any emf source could be considered to be a pure emf with a small resistive potential difference acting in series with the emf.
The effect of having an internal resistance is that an emf will never be able to fully supply its maximum voltage.
There will always be a small drop in voltage over the internal resistance and this drop will be bigger with a higher current. The pd over the internal resistance, r, can be found from Ohm’s law:
[math]V_{\text{internal}} = I r[/math]
This pd is sometimes referred to as ‘lost volts’, as the circuit appears to have fewer volts in use by the load than the emf should be supplying.
A measurement of voltage across the terminals of a cell powering a circuit would not measure the emf; it would measure the emf minus the lost volts. This ‘terminal pd’ would then be:
[math]V_{\text{internal}} = \varepsilon – I r[/math]
As all parallel branches have the same voltage across them, the terminal pd would also equal the pd across the load resistance.
[math]V_{\text{internal}} = V_{\text{load}} = \varepsilon – I r[/math]
Figure 19 Circuit for demonstrating the effect of internal resistance.
By considering energy conservation, we can draw up an equation for the circuit that includes the internal resistance.
[math]\begin{gather}
\Sigma \mathcal{E} = \Sigma V \\
\varepsilon = V_{\text{out}} + V_{\text{internal}} \\
\varepsilon = V_{\text{out}} + I r \\
\varepsilon = I R + I r
\end{gather}[/math]
The circuit shown in figure 19 demonstrates what the mathematics means practically. The voltmeter will be measuring the voltage actually supplied by the cell, so that will be its pure emf minus the pd across the internal resistance, or lost volts.
However, as the ammeter should have zero pd, the voltmeter will also be reading the value of the potential difference across the variable resistor, R.
[math]ε – V_r = V_R[/math]
Or, in terms of the ammeter reading, I, and the voltmeter reading, V:
[math]\begin{gather}
\varepsilon – I r = V \\
\varepsilon = V + I r
\end{gather}[/math]
Which is, again, a statement of energy conservation in this circuit.
78) CORE PRACTICAL 8: Determine the e.m.f. and internal resistance of an electrical cell
⇒ Investigating internal resistance:
We can find the internal resistance for a source of emf such as a cell in an experiment. Using a circuit like that shown in figure 12, we adjust the variable resistor to change the current through the circuit, and at each current value we record the voltmeter reading.
Figure 12 Investigating the internal resistance
The equation we have previously seen for the quantities in this circuit is:
[math]ε – Ir = V[/math]
Or
[math]V = -Ir + ε[/math]
Comparing this with the equation for a straight-line graph y = mx + c shows us that a graph plotted with V on the y-axis and I on the x-axis will have a gradient equal in magnitude to the internal resistance, r.
The y-intercept, c, here represents the voltmeter reading when no current flows. If there is no current in the circuit, the internal resistance can take no energy from the circuit, so there will be zero lost volts and the voltmeter will read the full value of the emf.
So, the y-intercept is the value of the emf, ε
Figure 20 Graph of internal resistance investigation results.
79) Understand how changes of resistance with temperature may be modelled in terms of lattice vibrations and number of conduction electrons and understand how to apply this model to metallic conductors and negative temperature coefficient thermistors
80) Understand how changes of resistance with illumination may be modelled in terms of the number of conduction electrons and understand how to apply this model to LDRs.
⇒ Potential Dividers in Sensor Circuits:
If one of the resistors in a potential divider is a sensor component (such as thermistor or LDR) then we can use its changing resistance to control an external circuit.
Figure 21 A sense and control circuit which makes the light brighter as the light level in the room drops.
In figure 13, the lamp is in parallel with the LDR, so will be powered by the same voltage as that across the LDR.
The resistance of a LDR increases if the ambient light level drops (darker surroundings make its resistance go up).
In bright conditions, the LDR has a very low resistance, so it takes a very small proportion of the 12 V supplied by the battery.
This will make the lamp very dim, probably so much that it will appear to be off; it is not needed if the surroundings are bright. As conditions become darker, the LDR resistance goes up, increasing the proportion of the 12 V supplied that drop across it.
This increases the voltage across the bulb, and makes it brighter. This circuit increases the brightness of the lamp as the surroundings become darker.
Many electronic components have a switch-on voltage of 5 V. These could be used with a sensor in a potential divider circuit, and would only switch on when the sensor’s resistance reached a level where its pd was 5 V.
Altering the resistor in this circuit would then control the sensor level at which the electronic component was switched on.

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Pearson Edexcel Physics

Unit 2: Waves and Electricity

2.4 Electric Circuits

Pearson Edexcel Physics

Unit 2: Waves and Electricity

2.4 Electric Circuits

64) Understand that electric current is the rate of flow of charged particles and be able to use the equation

65) Understand how to use the equation

66) Understand that resistance is defined by [math]R = \frac{V}{I}[/math] and that Ohm’s law is a special case when I ∝ V for constant temperature

67)

(a) Understand how the distribution of current in a circuit is a consequence of charge conservation

(b) Understand how the distribution of potential differences in a circuit is a consequence of energy conservation

68) Be able to derive the equations for combining resistances in series and parallel using the principles of charge and energy conservation, and be able to use these equations

69) Be able to use the equations P = VI, W = VIt and be able to derive and use related equations, e.g.

[math]P = I^2 R \quad \text{and} \quad P = \frac{V^2}{R}[/math]

70) Understand how to sketch, recognize and interpret current-potential difference graphs for components, including ohmic conductors, filament bulbs, thermistors and diodes

71) Be able to use the equation

[math]R = \frac{ρl}{A}[/math]

72) CORE PRACTICAL 7: Determine the electrical resistivity of a material

?

73) Be able to use I = nqvA to explain the large range of resistivities of different materials

74) Understand how the potential along a uniform current-carrying wire varies with the distance along it

75) Understand the principles of a potential divider circuit and understand how to calculate potential differences and resistances in such a circuit

76) Be able to analyse potential divider circuits where one resistance is variable including thermistors and light dependent resistors (LDRs)

77) Know the definition of electromotive force (e.m.f.) and understand what is meant by internal resistance and know how to distinguish between e.m.f. and terminal potential difference

78) CORE PRACTICAL 8: Determine the e.m.f. and internal resistance of an electrical cell

79) Understand how changes of resistance with temperature may be modelled in terms of lattice vibrations and number of conduction electrons and understand how to apply this model to metallic conductors and negative temperature coefficient thermistors

80) Understand how changes of resistance with illumination may be modelled in terms of the number of conduction electrons and understand how to apply this model to LDRs.

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