00:00
Read the passage, then answer the questions.
Electric charge is carried by charged particles called charge carriers. In a metal the charge carriers are free (delocalised) electrons; in an electrolyte they are positive and negative ions.
Charge is measured in coulombs (C). One electron carries a charge of −e, where e = 1.60×10−19 C, so a charge of 1 C is carried by 6.25×1018 electrons.
Electric current is the rate of flow of charge:
One ampere is a flow of one coulomb per second: 1 A = 1 C s−1. Rearranging gives the charge that passes in a time Δt: ΔQ = IΔt.
Conventional current is drawn from the positive terminal of the supply, round the circuit, to the negative terminal. Electrons are negative, so they drift the opposite way (Fig 1.1).
Charge is conserved: it cannot pile up or disappear anywhere in a circuit. In a series circuit the current is therefore the same at every point.
The number of charge carriers passing a point is the charge divided by the charge on each carrier:
A current of 0.25 A flows in a heater element for 3.0 minutes. Calculate the charge that passes through it.
A current of 0.40 A flows through a lamp for 10 s. Calculate the number of electrons that pass through the lamp.
Read the passage, then answer the questions.
As charge flows through a component, work is done and energy is transferred from the charge to the component, for example heating a resistor or lighting a lamp.
The potential difference (p.d.) V between two points is the work done (energy transferred) per unit charge moving between them:
The volt is one joule per coulomb: 1 V = 1 J C−1. A 12 V supply transfers 12 J to the components for every coulomb that flows round the circuit.
A cell, battery or power supply does work on the charge. The energy it transfers to electrical energy per coulomb is its electromotive force (emf), also measured in volts. Real supplies have some internal resistance; a supply of negligible internal resistance has a terminal p.d. equal to its emf. Lesson 8 develops both ideas.
A voltmeter measures p.d., so it is connected in parallel with (across) the component, as in Fig 1.1.
For a steady current the charge passing in a time t is Q = It, so combining this with W = QV gives the energy transferred: W = IVt. Lesson 6 develops this.
A 9.0 V battery moves 120 C of charge round a circuit. Calculate the energy it transfers.
Read the passage, then answer the questions.
The resistance R of a component is the p.d. across it divided by the current through it:
Resistance is measured in ohms (Ω): 1 Ω = 1 V A−1. A large resistance means a large p.d. is needed to drive each ampere of current.
In a metal, resistance arises because the free electrons collide with the vibrating positive ions of the lattice, transferring energy to them. Lesson 3 looks at what this depends on.
R = VI defines the resistance at any operating point. It does not mean that R is constant: for many components the resistance changes as the current changes (Lesson 2).
To find R from a graph of I against V, read the p.d. and the current at the required point and calculate VI.
A heating element draws a current of 4.0 A from a 230 V supply. Calculate its resistance.
Read the passage, then answer the questions.
A current–voltage (I–V) characteristic is a graph showing how the current through a component varies with the p.d. across it.
The ammeter is connected in series with the component and the voltmeter in parallel with it. A variable resistor in series lets the current and p.d. be changed in steps (Fig 2.1).
Unless a question says otherwise, meters are treated as ideal. An ideal ammeter has zero resistance, so it has no p.d. across it and does not change the current. An ideal voltmeter has infinite resistance, so no current passes through it.
Reversing the supply connections (or the component) gives readings for negative values of p.d. and current.
Read the passage, then answer the questions.
Characteristics may be plotted with I on the vertical axis or with V on the vertical axis, so always check the axes before using a graph.
To vary the p.d. right down to zero, connect the variable resistor as a potentiometer (a potential divider, Lesson 7) across the supply instead of in series. Switch off between readings so that the component does not heat up. A data logger records many readings quickly and plots them as they are taken, but it is not more accurate in itself.
The resistance at any point on the graph is VI for that point. The gradient of an I–V graph equals 1R only when the line is straight and passes through the origin.
| p.d. (V) | 0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 | 6.0 | 7.0 | 8.0 |
|---|---|---|---|---|---|---|---|---|---|
| current (A) | 0 | 0.003 | 0.007 | 0.010 | 0.013 | 0.017 | 0.020 | 0.023 | 0.026 |
Read the passage, then answer the questions.
Ohm's law: the current through a conductor is directly proportional to the p.d. across it, provided the physical conditions, such as temperature, remain constant.
Ohm's law is a special case. It describes ohmic conductors (for example a metal wire at constant temperature), whose I–V characteristic is a straight line through the origin. The definition R = VI applies to every component, ohmic or not.
Both lines in Fig 2.3 are ohmic. Line b has the greater resistance, because a larger p.d. is needed to give the same current.
A filament lamp is non-ohmic. As the current increases the filament gets hotter, the positive ions in the metal lattice vibrate with larger amplitude, and the electrons collide with them more often (Fig 2.5). The resistance increases, so the graph curves towards the V axis (Fig 2.4).
The curve has the same shape for negative p.d.s, because the filament heats up whichever way the current flows.
Read the passage, then answer the questions.
A semiconductor diode allows current in one direction only. In forward bias almost no current flows until the p.d. reaches the threshold (about 0.6 V for a silicon diode); beyond this the resistance falls sharply and the current rises steeply.
In reverse bias the current is negligible because the resistance is very high (Fig 2.6).
Unless told otherwise, treat a conducting silicon diode as having about 0.6 V across it, and a reverse-biased diode as passing no current. When measuring a diode's characteristic, a protective resistor in series limits the forward current.
A negative temperature coefficient (NTC) thermistor is a semiconductor whose resistance decreases as its temperature rises, because more charge carriers are released, which outweighs the effect of the extra lattice vibration (Fig 2.7). This makes it useful as a temperature sensor.
A light-dependent resistor (LDR) is a semiconductor whose resistance decreases as the light intensity increases: absorbed light releases more charge carriers (Fig 2.8).
At a fixed temperature (or a fixed light level), a thermistor or LDR behaves like an ordinary ohmic resistor.
Read the passage, then answer the questions.
The resistance of a wire depends on its length L, its cross-sectional area A, the material it is made of and its temperature.
Doubling the length doubles the number of collisions each electron makes on its way through, so R is proportional to L. Doubling the cross-sectional area doubles the number of paths, so R is inversely proportional to A.
The resistivity ρ of a material combines these:
Resistivity is a property of the material at a given temperature, not of a particular sample, in the same way that density belongs to a material while mass belongs to an object. Two copper wires of different sizes have different resistances but the same resistivity.
Most conductors are wires, so the cross-sectional area is found from the diameter d: A = πd24.
| material | resistivity at 20 °C (Ω m) |
|---|---|
| copper | 1.7×10−8 |
| aluminium | 2.8×10−8 |
| constantan | 4.9×10−7 |
| nichrome | 1.1×10−6 |
| graphite | about 8×10−6 |
| silicon | about 2×103 |
Calculate the resistance of a 2.0 m length of constantan wire (ρ = 4.9×10−7 Ω m) of diameter 0.50 mm.
Read the passage, then answer the questions.
For two samples of the same material, ρ is the same, so R is proportional to LA. Comparing them is then a ratio calculation: the resistivity cancels and is not needed.
For wires, A is proportional to d2, so R is proportional to Ld2. Doubling the diameter quarters the resistance.
Changing the shape of a conductor changes L and A together. A wire stretched to twice its length (same volume) has half the area, so its resistance rises by a factor of 4. A wire folded in half has half the length and twice the area, so its resistance falls by a factor of 4.
A block or rod can conduct between any pair of opposite faces: L is the distance between those faces and A is the area of one of them. The smallest resistance is between the largest faces.
A wire of resistance 2.0 Ω is drawn out to three times its original length. Its volume does not change. Calculate its new resistance.
Read the passage, then answer the questions.
For a metal, resistivity increases with temperature: the lattice ions vibrate more, so electrons collide with them more often.
An NTC thermistor behaves in the opposite way (Lesson 2): the extra charge carriers released as it warms outweigh the effect of extra vibration.
Some materials become superconductors: their resistivity falls to zero at and below a critical temperature Tc (Fig 3.1). The critical temperature depends on the material.
Mercury becomes superconducting below 4.2 K. Some copper-oxide ceramics have critical temperatures above 77 K, the boiling point of liquid nitrogen, so they can be cooled cheaply. Even the highest critical temperatures at normal pressure (about 135 K, roughly −138 °C) are far below room temperature.
With zero resistance, a current flows without any energy being dissipated as heat.
Read the passage, then answer the questions.
The resistivity of a wire is found from ρ = RAL: the resistance of a measured length is found from V and I, and the cross-sectional area from the diameter.
The diameter is the measurement that needs most care, because A = πd24 depends on d2: the percentage uncertainty in A is twice the percentage uncertainty in d.
A micrometer has a resolution of 0.01 mm (Fig 4.1). The reading is the whole and half millimetres on the sleeve plus the hundredths on the thimble. First close the jaws and check the zero reading: if it is not zero the micrometer has a zero error, which must be subtracted from every reading.
Measure the diameter at several points along the wire and in different orientations, in case the wire is not uniform or not circular, and take the mean. The uncertainty is half the range of the repeated readings, or half the resolution (±0.005 mm) if the readings all agree.
A micrometer reads 0.01 mm when closed. Readings of a wire's diameter are 0.32 mm, 0.33 mm, 0.32 mm and 0.33 mm. Calculate the cross-sectional area of the wire.
| student | zero error (mm) | observed reading (mm) |
|---|---|---|
| A | +0.01 | 8.69 |
| B | −0.02 | 13.45 |
| C | −0.04 | 3.50 |
| D | +0.03 | 3.74 |
Read the passage, then answer the questions.
Connect the circuit in Fig 4.2. Clip a length L of the wire into the circuit, measure L between the clips with the metre ruler, and record V and I to find R = VI.
The voltmeter is connected across the measured length only, so that it does not include the p.d. across the leads, clips or ammeter. Keep the current small and disconnect the supply between readings: if the wire warms up, its resistivity increases. Never let the clips touch (zero length), as this short-circuits the supply.
Repeat for a range of lengths, for example 0.100 m to 1.000 m in 0.100 m steps, taking repeat readings at each length. The independent variable is L, the dependent variable is R, and the temperature, material and diameter are kept constant.
Read the passage, then answer the questions.
Since R = ρA × L, a graph of R against L is a straight line through the origin with gradient ρA, so ρ = gradient × A (Fig 4.3).
The graph uses many readings, so random errors average out, and any constant extra resistance (for example at the clips) moves the line up without changing its gradient. Take the gradient from a large triangle on the best-fit line, never from two data points.
The percentage uncertainty in ρ is the sum of the percentage uncertainties in the gradient and in A, and the percentage uncertainty in A is twice that in d.
The gradient of a graph of R against L is 6.4 Ω m−1 with an uncertainty of 3%. The wire's diameter is 0.28 mm ± 0.01 mm. Calculate the resistivity and its percentage uncertainty.
Read the passage, then answer the questions.
In a series circuit there is only one path. Charge is conserved, so the current is the same through every component.
Energy is conserved, so the p.d.s across components in series add up to the supply p.d.: V = V1 + V2 + … The same current passes through each resistor, so the supply p.d. is shared in proportion to the resistances: the largest resistor has the largest p.d.
In a parallel circuit each branch is connected across the same two points, so every branch has the same p.d.
At a junction the current divides. The branch currents add up to the total current: I = I1 + I2 + … (conservation of charge again). The branch with the lower resistance takes the larger share.
A useful picture is a water slide: in series everyone goes down every slide in turn and the drops add up to the total height; in parallel each person chooses one slide, and every slide drops the full height.
These two conservation rules are developed into Kirchhoff's laws in Lesson 7.
Read the passage, then answer the questions.
Series: the current I is the same in each resistor, and V = V1 + V2 + V3. Writing each p.d. as IR and dividing by I:
Parallel: the p.d. V is the same across each resistor, and I = I1 + I2 + I3. Writing each current as VR and dividing by V:
Adding a resistor in parallel always lowers the total resistance, because it adds another path; the total is always less than the smallest single resistance. A set of n identical resistors R in parallel have a total resistance Rn.
For a network, work from the inside out: add resistors in series within each branch, replace each parallel group by its equivalent resistance, and repeat until a single resistance remains.
Calculate the total resistance of 10 Ω, 20 Ω and 15 Ω resistors connected in parallel.
Read the passage, then answer the questions.
Cells in series: the emfs add when the cells are connected the same way round (positive terminal to negative). A cell connected the opposite way round subtracts its emf.
Identical cells in parallel: the emf of the combination equals the emf of one cell. Each cell supplies only a fraction of the current, so the combination lasts longer (Fig 5.5).
In this lesson cells are treated as having negligible internal resistance; Lesson 8 removes this simplification.
To analyse a circuit: find the total resistance, use the supply p.d. to find the supply current, then use V = IR for each part. In a parallel section the current divides so that the p.d. across each branch is the same.
A 9.0 V supply of negligible internal resistance is connected to a 2.0 Ω resistor in series with a parallel pair of 6.0 Ω and 12 Ω. The parallel pair is equivalent to 4.0 Ω, so the total resistance is 6.0 Ω. Calculate the supply current.
Read the passage, then answer the questions.
Power is the rate of energy transfer, measured in watts (1 W = 1 J s−1).
From V = WQ, the energy transferred is W = QV, and the charge flowing in time t is It, so the energy transferred, E, is:
Dividing by time gives the power, and substituting V = IR or I = VR gives two further forms:
Use P = I2R when you know the current through a resistor, and P = V2R when you know the p.d. across it. The energy transferred in time t is E = Pt.
Electricity bills use the kilowatt-hour (kWh): the energy transferred by a 1 kW appliance in 1 hour, 1 kWh = 1000 W × 3600 s = 3.6×106 J.
A heater of resistance 24 Ω is connected to the 230 V mains. Calculate its power.
Read the passage, then answer the questions.
Appliances carry a power rating at a stated p.d., for example '230 V, 750 W'. The normal operating current is I = PV and the resistance is R = V2P.
A fuse protects the flex: choose the lowest standard rating (3 A, 5 A or 13 A) that is above the normal operating current.
If the resistance stays constant, the power is proportional to V2, so a small fall in the supply p.d. gives a larger fractional fall in power.
Battery capacity is often given in milliamp-hours. A 2400 mAh cell can supply 2400 mA for one hour, which is a charge of 2.4 A × 3600 s = 8640 C.
A 1200 W iron runs from the 230 V mains. Calculate its operating current and choose a suitable fuse.
Read the passage, then answer the questions.
A current in a resistor transfers energy to the lattice as thermal energy at a rate I2R. This is useful in heaters and kettles but wasteful in cables.
Electricity is transmitted at a very high p.d. For a fixed power P = IV, raising V lowers I, and the power wasted in the cables, I2R, falls with the square of the current.
Superconducting cables would remove these resistive losses completely (Lesson 3).
A heating element links electrical and thermal energy: the energy supplied, E = IVt, raises the temperature of a mass m by Δθ, where ΔE = mcΔθ (assuming no energy is lost to the surroundings).
A power station sends 5.0 MW at 25 kV along cables of total resistance 4.0 Ω. Calculate the power wasted in the cables.
| V (V) | V2 (V2) | P (W) |
|---|---|---|
| 1.00 | 1.0 | 0.21 |
| 1.71 | 2.9 | 0.58 |
| 2.25 | 1.01 | |
| 2.67 | 1.43 | |
| 3.00 | 9.0 | 1.80 |
| 3.27 | 10.7 | 2.18 |
| 3.50 | 12.3 | 2.43 |
Read the passage, then answer the questions.
Kirchhoff's first law: the total current into a junction equals the total current out of it. This follows from conservation of charge (Fig 7.1).
Kirchhoff's second law: around any closed loop in a circuit, the sum of the emfs equals the sum of the p.d.s (the IR terms). This follows from conservation of energy: the energy given to each coulomb by the sources equals the energy it transfers in the components.
Choose a direction round the loop. An emf counts as positive if it drives current in that direction and negative if it opposes it; an IR term is positive when the current flows in the chosen direction.
If a calculated current comes out negative, the current flows in the opposite direction to the arrow you drew.
For a circuit with two loops, apply the first law at a junction and the second law to two loops, then solve the equations together.
In Fig 7.2 the 9.0 V and 3.0 V cells oppose each other and have negligible internal resistance. Calculate the current.
Read the passage, then answer the questions.
A potential divider is two (or more) resistors in series across a supply (Fig 7.4). The same current passes through each, so the supply p.d. is shared in the ratio of the resistances:
Fixed resistors give a constant output p.d. Making one of them a variable resistor gives a variable output p.d.
A potentiometer is a resistance track or wire with a sliding contact (Fig 7.5). It divides the supply in the ratio of the lengths either side of the slider, so the output can be varied smoothly from zero up to the full supply p.d. A rheostat in series with a load cannot bring the p.d. down to zero.
Connecting a load across the output puts it in parallel with R2. This lowers the resistance of the lower section, so the output p.d. falls. The effect is small only if the load's resistance is much larger than R2.
A 12 V supply is connected across R1 = 3.3 kΩ and R2 = 6.8 kΩ in series. Calculate the output p.d. across R2.
Read the passage, then answer the questions.
Replacing R2 with an NTC thermistor makes a temperature sensor. As the temperature rises, the thermistor's resistance falls, it takes a smaller share of the supply p.d., and Vout across it falls. Taking the output across the fixed resistor instead gives an output that rises with temperature.
Using an LDR as R2 gives a light sensor. In darkness the LDR's resistance is high, so Vout across it is high; this can switch on a street lamp.
Making R1 a variable resistor lets the user set the temperature or light level at which the output reaches the switching p.d.
In a design calculation, find the p.d. needed across each component at the switching point, then use V1V2 = R1R2.
Read the passage, then answer the questions.
The electromotive force (emf) ε of a source is the energy transferred to electrical energy per unit charge passing through the source:
Like p.d., emf is measured in volts (J C−1). Emf describes energy given to the charge by a source; p.d. describes energy transferred from the charge to a component.
Real sources have internal resistance r: some energy is dissipated inside the source as charge passes through it, so a battery warms up when it supplies a large current.
The p.d. across the source's terminals when it supplies a current is the terminal p.d. V. The difference ε − V = Ir is called the lost volts: the energy per coulomb dissipated inside the source.
When no current flows (open circuit) there are no lost volts, so a high-resistance voltmeter across the terminals reads the emf.
Multiplying by I gives the power balance: εI = I2R + I2r. The power supplied by the source equals the power delivered to the external resistor plus the power wasted inside the source.
A battery of emf 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. Calculate the current.
Read the passage, then answer the questions.
Treat the internal resistance as an extra resistor in series with an ideal source of emf ε.
Cells in series: the emfs add and the internal resistances add.
n identical cells in parallel: the emf of the combination is ε and its internal resistance is rn. Each cell carries In, so the terminal p.d. is V = ε − Irn.
As the external resistance increases, the current falls, the lost volts fall, and the terminal p.d. rises towards ε.
Two cells, each of emf 1.5 V and internal resistance 0.50 Ω, are connected in series to a 4.0 Ω resistor. Calculate the current.
Read the passage, then answer the questions.
Multiplying ε = IR + Ir by I gives the power balance: εI = I2R + I2r. The power supplied by the source equals the power delivered to the external circuit plus the power wasted as heat inside the source.
The efficiency of the transfer is the fraction of the power that reaches the external circuit:
A large external resistance gives a high efficiency but a small current and little power. The power delivered to R is greatest when R = r; the efficiency is then only 50%, because half the power is wasted inside the source. (The maximum power result is not required by the specification, but it helps with some exam questions.)
This is why a car battery is built with a very low internal resistance: it must deliver hundreds of amps to the starter motor without a large drop in terminal p.d. A school EHT supply is built the opposite way, with a very high internal resistance, so that the current it can deliver is too small to be dangerous.
A battery of emf 12 V and internal resistance 0.50 Ω is connected to a 3.5 Ω resistor. Calculate the power delivered to the resistor and the efficiency of the transfer.
Read the passage, then answer the questions.
Required Practical 6 finds the emf and internal resistance of a cell from the way its terminal p.d. changes with the current it supplies. The variable resistor sets the current (the independent variable) and the voltmeter records the terminal p.d. (the dependent variable).
Opening the switch between readings stops the cell heating up and running down, either of which would change ε and r during the experiment.
A protective resistor in series limits the maximum current, so the cell and ammeter are safe even when the variable resistor is set to zero. The voltmeter must still be connected directly across the cell's terminals.
The voltmeter has a very high resistance, so it draws negligible current; the ammeter has negligible resistance, so it does not add to the external resistance.
Read the passage, then answer the questions.
Rearranging ε = V + Ir gives V = −rI + ε, which has the form y = mx + c.
Plot the terminal p.d. V (vertical axis) against the current I (horizontal axis). The y-intercept is ε and the gradient is −r (Fig 9.2).
Take the gradient from a large triangle on the best-fit line. The gradient is negative, so r = −gradient. If the current axis does not start at zero the intercept cannot be read directly: use ε = V + Ir with a point on the line instead.
A best-fit line passes through the points (0.20 A, 1.40 V) and (0.80 A, 1.04 V). Calculate the internal resistance and the emf of the cell.
| current I (A) | 0.10 | 0.20 | 0.30 | 0.40 | 0.50 |
|---|---|---|---|---|---|
| terminal p.d. V (V) | 1.43 | 1.33 | 1.23 | 1.15 | 1.04 |
Read the passage, then answer the questions.
Random errors, such as fluctuating meter readings, scatter the points either side of the line. Repeating each reading and drawing a best-fit line through many points averages them out.
Systematic errors shift every reading the same way. A zero error on the voltmeter shifts the whole line up or down, so ε is wrong but the gradient, and therefore r, is unaffected. A zero error on the ammeter shifts the line sideways, so again r is unaffected.
To find the uncertainty in r and ε, draw the steepest and shallowest lines that still pass through the error bars (worst-fit lines). Half the difference between their gradients is the uncertainty in r; half the difference between their intercepts is the uncertainty in ε.
The method assumes ε and r stay constant, which is why the switch is opened between readings and the current is kept below about 1 A for a dry cell. A cell that warms up or runs down gives points that drift off the line.
An alternative method uses a set of known resistors R as the load. Since 1I = R + rε, a graph of 1I against R is a straight line with gradient 1ε and intercept rε.
