1. Circuit Symbols Switch (open): Switch (closed): Battery: Cell:
Filament bulb:
Voltmeter:
Ammeter:
Resistor:
4.2.1

2. Circuit Symbols Variable resistor: Thermistor:
Light-dependent resistor (LDR):
Fuse:
Diode:
Light-emitting diode (LED):
4.2.1

3. Conventional current: +  ‒
Charge & Current
Current is the flow of charge (electrons).
Faster flow of charge = higher current.
Current only flows if there’s a source of potential difference (e.g. cell).
The elementary charge, or charge of a single electron, is 1.6x10-19C.
Kirchoff’s 1st law: current flowing into a point is the same as current flowing out of that point.
Question: What is the charge if 12 mA flows in 15 minutes? How many electrons does this correspond to?
Q = I x t = 12x10-3 x 15 x 60 = 10.8C
n = 10.8/1.6x10-19 = 67.5x1018 electrons
Conventional current: +  ‒
Electron flow: ‒  +
Q = I × t
Charge in
Coulombs, C
Time in
seconds, s
Current in
Ampères (“Amps”), A
4.1.1

4. I = vAnq Mean Drift Velocity
Mean drift velocity is defined as the average displacement travelled by electrons along a wire per second.
Question: Calculate the electron number density in a copper wire of radius 1mm, carrying a current of 3A and where the mean drift velocity of the electrons is ~50mm/s.
I = vAnq
n = I/vAq
= 3/(50x10-3 x π(1x10-3)2 x 1.6x10-19)
= 1.2x1026m-3
I = vAnq
Charge of an electron = 1.6x10-19C
Current in Amps, A
Mean drift velocity in metres per second, m/s
Cross-sectional area of wire in metres squared, m2
Electron number density in per metres cubed, m-3
4.1.2

5. Work done (energy) in Joules, J
Potential Difference
Potential difference (voltage) is a measure of how much energy each electron has (technically, it’s the work done per unit charge).
The charges are given potential difference by the cell or battery.
Ohm’s law states that the potential difference across a metallic conductor is proportional to the current through it, provided the physical conditions do not change.”
Question: What is the potential difference when the current is 16.5A and resistance 98mΩ? Give your answer in mV to the nearest 100mV.
V = I x R = 16.5 x 0.098
= 1.617V = 1600mV
V = I × R
V = W ÷ Q
P.d. in
Volts, V
Resistance
in Ohms, Ω
P.d. in
Volts, V
Current in
Ampères (“Amps”), A
Work done (energy) in Joules, J
Charge in Coulombs, C
4.2.2

6. Resistance
All electrical components have a resistance.
Resistance alters the flow of charge: higher resistance = lower current.
Some components have a constant resistance, but others’ can change.
A component that has a constant resistance (at a constant temperature) is an Ohmic conductor.
Components whose resistance changes are non-Ohmic conductors.
, Question: Are these Ohmic conductors or not and why? Lamp, thermistor, LDR, diode.
Ohmic = constant resistance. So none are Ohmic because: lamp – resistance increases with temperature, thermistor – decreases with temp., LDR decreases with light intensity, diode – high resistance in reverse direction.
~0.6V
4.2.3

7. Series & Parallel Circuits
Voltmeters do not count as a branch…so a series circuit with a voltmeter is still series!
Series circuits have one continuous loop.
Parallel circuits have two or more loops. .
, Question: Give the potential difference, total resistance and current:
Potential difference = 12V
Total resistance = = 24Ω
Current = p.d./resistance = 12/24 = 0.5A
Series circuit
Parallel circuit
Current
Is the same through each component
Is shared between components
Potential difference
Is the same in total across each branch
Resistance
Rtot = R1 + R2 + …
Rtot-1 = R1-1 + R2-1 + …
4.3.1

8. Power
The amount of electrical energy an appliance uses over time is called its power. Power is measured in Watts. 1 Watt = 1 Joule of energy used every second.
, Question: A charge of 30C flows through a TV which is connected to the mains. What is the total energy transferred (in kJ to 1 significant figure)?
E = Q x V
= 30 x 230
= 6900J = 6.9kJ = 7kJ to 1 s.f.
P = E ÷ t
P = I × V
E = Q × V
Time in
seconds, s
Energy in
Joules, J
Power in
Watts, W
4.2.5

9. Resistivity
ρ = RA ÷ L
Length in
metres, m
Resistivity in
Ohmmetres, Ωm
Resistance in Ohms, Ω
Question: Find the resistance of a wire of length 1.2m, radius 5mm and resistivity 7.3Ωm. Give your answer in milliohms and to 2 significant figures.
ρ = RA ÷ L  R = ρA ÷ L
= 7.3x(5x10-3)2xπ ÷ 1.2
= 4.77x10-4Ω
= 48mΩ
Area in metres squared, m2
A = πd2 ÷ 4 for a wire with diameter d in metres, m
4.2.4

10. Potential Dividers
The potential divider is made up of two resistors in series, connected to a cell or battery.
It is used to provide a variable potential difference.
The potential difference can be varied by changing the position of a sliding contact.
The ratio of the resistances of the resistors is equal to the ratio of the potential differences across them: R1/R2 = V1/V2.
Replacing one of the resistors with a thermistor can create a temperature sensor.
Doing the same thing with an LDR can create a light sensor.
Question: A potential divider is made using a 1kΩ and 5kΩ resistor. A battery provides 4.5V of potential difference. Calculate the potential difference across each resistor.
Ratio of Rs = ratio of Vs. So 4.5V is split in the ratio 1:5.
V = (4.5÷6)x1 = 0.75V across 1kΩ resistor. V2 = (4.5÷6)x5 = 3.75V across 5kΩ.
4.3.3

11. Internal Resistance
Every component in a circuit has a resistance. This includes the cell or battery. The resistance of a cell or battery is known as internal resistance.
The internal resistance of a cell or battery is equal to the loss in potential difference per unit charge when current passes through it.
In circuit diagrams, internal resistance may be shown as a resistor next to the cell or battery.
Question: A cell with EMF 24V and negligible internal resistance would provide how many μJ of electrical energy to each electron?
ε = E ÷ Q  E = εQ
= 24 x 1.6x10-19
= 3.8x10-18J
= 3.8x10-12μJ
4.3.2

12. ε = E ÷ Q ε = Ir + IR Electromotive Force
The electromotive force (ε), or EMF, of a cell or battery is the electrical energy per unit charge it produces:
The product Ir is known as the lost p.d. and the product IR is known as the terminal p.d..
Kirchoff’s 2nd law: in any loop (path) around a circuit, sum of the EMFs = sum of the p.d.s.
Question: Find the current flowing in the circuit shown in the diagram.
ε = Ir + IR = I(r + R)  I = ε ÷ (r + R)
= 3 ÷ (1 + 5)
= 0.5A
ε = E ÷ Q
ε = Ir + IR
Charge in
Coulombs, C
Internal resistance in Ohms, Ω
Total resistance of components in circuit
EMF in
Volts, V
EMF in
Volts, V
Electrical energy in Joules, J
4.2.2

13. Capacitance
A capacitor is an electrical component that stores charge.
It is usually made of two metal plates (conductors) separated by an air gap (insulator). When connected to a power source, one plate gains electrons (becomes negatively-charged) and the other loses electrons (becomes positively-charged).
Capacitance (C) is the charge stored per unit potential difference. Unit: farad (F).
Question: What is the p.d. across a capacitor of capacitance 20F holding a charge of 30kC?
C = Q ÷ V  V = Q ÷ C
= 30,000 ÷ 20
= 1,500V
C = Q ÷ V
measured in
Farads, F
measured in
Volts, V
measured in
Coulombs, C
6.1.1

14. W = ½QV Capacitance (cont’d.)
Total capacitance of capacitors in series is 1/Ctot = 1/C1 + 1/C2 + …
Total capacitance of capacitors in parallel is Ctot = C1 + C2 + …
On a p.d.-charge graph for a capacitor, the area under the graph gives the energy stored by the capacitor (W).
Question: What is the total capacitance of three capacitors in series, each with a capacitance of mF? Give your answer in mF.
1/Ctot = 1/C1 + 1/C2 + 1/C3 = 3 x (1/40x10-3) = 75
Ctot = 1/75 = 0.013F = 13mF.
W = ½QV
measured in
Joules, J
measured in
Volts, V
measured in
Coulombs, C
6.1.1 – 6.1.2

15. (Dis)charging Capacitors Using Resistors
Capacitors can be charged or discharged using resistors.
The final charge after charging/discharging (Q) is related to the initial charge (Q0) by the equation below.
The product RC (resistance x capacitance) is known as the time constant.
Question: What is the charge of a capacitor after 3 minutes when the initial charge is 0.5C and its time constant is given as 150s?
Q = Q0exp(-t/RC) = 0.5exp(-(3x60)/150)
= 0.15C
Q = Q0exp(-t/RC)
measured in
Coulombs, C
measured in
seconds, s
measured in
seconds, s
6.1.3