Senior 4 practice book · Unit 5 of 10

Direct Current Circuits

50 questions that complete the Senior 4 quiz for this unit: 24 core and 26 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

A cell gives out a fixed current whatever is connected to it.A cell provides an e.m.f.; the current it drives depends on the total resistance of the circuit, including its own internal resistance.
Electrons travel round the circuit at enormous speed, which is why the lamp lights at once.Electrons drift only a fraction of a millimetre per second; the lamp lights at once because the electric field pushes all the electrons in the wire at the same moment.
Adding more resistors to a circuit always increases the total resistance.That is true only in series; every extra resistor added in parallel gives the current another path, so the total resistance falls.
A fuse is there to protect the user from electric shock.A fuse melts when the current is too large, protecting the wiring and appliance from overheating; it is the earth wire together with the fuse that protects people.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Electric current and charge: I = Q/t, number of electrons, conventional current
Ohm's law and I–V characteristics (resistor, filament lamp, diode)
Resistivity: R = ρL/A and the factors affecting resistance
Resistors in series and in parallel
E.m.f., internal resistance and terminal p.d.
Cells in series and in parallel
Kirchhoff's current and voltage laws
Electrical energy, power and the cost of electricity
Ammeters and voltmeters: connection, resistance and loading; shunts
Potential dividers, rheostats and sensors (LDR, thermistor)
The Wheatstone bridge and metre bridge
Faults and safety: short circuits, fuses and earthing

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 26 harder ones are marked ★ Challenge.

1True or false
When a Wheatstone bridge is balanced, no current flows through its galvanometer.
Show answer
Answer: True
At balance the two ends of the galvanometer are at the same potential, so there is no p.d. across it and the current is zero.
Common mistakeSome learners think balance means equal currents in all four arms; it means the galvanometer current is zero.
2True or false · ★ Challenge
The resistance of a metal wire increases when its temperature rises.
Show answer
Answer: True
In a hotter metal the ions vibrate more strongly, so the drifting electrons collide with them more often and the resistance rises.
Common mistakeSome learners think heat "loosens" electrons and lowers resistance; that happens in semiconductors such as thermistors, not in metals.
3True or false · ★ Challenge
Connecting an extra resistor in parallel with a circuit always lowers the total resistance, even if the extra resistor is very large.
Show answer
Answer: True
The new resistor adds another path, so 1/R_total gets larger (by 1/R_new) and R_total gets smaller, though only slightly if R_new is large.
Common mistakeThe trap is to think a large resistor must raise the total; in parallel any extra path, however poor, lets a little more current through.
4True or false
Kirchhoff’s voltage (loop) law can only be used in circuits that contain a single cell.
Show answer
Answer: False
The loop law applies to any closed loop: the sum of e.m.f.s (with signs) equals the sum of the IR terms, however many cells there are.
Common mistakeThe error comes from only ever meeting single-cell circuits; Kirchhoff’s laws are most useful precisely when there are several cells.
5True or false · ★ Challenge
If one of three 1.5 V cells in a torch is put in the wrong way round, the total e.m.f. becomes 1.5 V.
Show answer
Answer: True
The reversed cell opposes the others: 1.5 + 1.5 − 1.5 = 1.5 V, so the bulb glows only dimly.
Common mistakeAnswering 3.0 V assumes the reversed cell simply does nothing; it actively works against the others.
6True or false
An ideal voltmeter has infinite resistance, so it takes no current from the circuit it is connected to.
Show answer
Answer: True
With infinite resistance no current is diverted through the voltmeter, so connecting it does not change the p.d. it measures.
Common mistakeSome learners think a voltmeter must let current pass to work; a real one takes a tiny current, and the ideal one takes none.
7True or false · ★ Challenge
When a battery delivers the maximum possible power to its load (load resistance = internal resistance), only half of the energy it supplies reaches the load.
Show answer
Answer: True
With R = r the same current flows through both, so equal power is dissipated in the load and inside the battery: efficiency = R/(R + r) = 50%.
Common mistakeMany learners think maximum power also means maximum efficiency; efficiency keeps rising as R becomes much larger than r, while the power falls.
8True or false
A semiconductor diode lets current flow easily in one direction but almost not at all in the reverse direction.
Show answer
Answer: True
In forward bias (above about 0.7 V for silicon) its resistance is low; in reverse bias it is very high, so it is used to change a.c. into d.c.
Common mistakeSome learners think a diode is just a small resistor; its resistance depends strongly on the direction of the p.d.
9Multiple choice · ★ Challenge
A technician has three 6.0 Ω resistors and needs a total resistance of 4.0 Ω. How should she connect them?
1. AAll three in parallel with each other
2. BTwo in series, with the third in parallel across that pair
3. CAll three in series with each other
4. DTwo in parallel, with the third in series with that pair
Show answer
Answer: B. Two in series, with the third in parallel across that pair
Two in series give 12 Ω; 12 Ω in parallel with 6.0 Ω gives (12 × 6)/(12 + 6) = 4.0 Ω. (All parallel gives 2 Ω, all series 18 Ω, and the last arrangement 9 Ω.)
Common mistakeChoosing "two in parallel with one in series" gives 3 + 6 = 9 Ω; check each arrangement by calculating, not by guessing.
10True or false
In the external circuit, conventional current is taken to flow from the positive terminal to the negative terminal, the opposite way to the electrons.
Show answer
Answer: True
Conventional current was defined before electrons were discovered; electrons, being negative, actually move from − to + in the wires.
Common mistakeSome learners think conventional current and electron flow are the same; they point in opposite directions.
11Fill in the blank · ★ Challenge
A phone is charged with a current of 1.5 A for 40 minutes. The charge that flows into it is ______ C.
Show answer
Answer: 3600
Q = It = 1.5 × (40 × 60) = 1.5 × 2400 = 3600 C.
Common mistakeAnswering 60 C forgets to change minutes into seconds before using Q = It.
12Multiple choice
Which of these is a unit of energy?
1. AKilowatt
2. BAmpere-hour
3. CVolt per ampere
4. DKilowatt-hour
Show answer
Answer: D. Kilowatt-hour
A kilowatt-hour is the energy used by a 1 kW appliance in one hour, 3.6 × 10⁶ J. The kilowatt is a unit of power, the ampere-hour of charge and V/A of resistance.
Common mistakeChoosing the ampere-hour (printed on batteries) confuses charge with energy; you also need the voltage to find the energy.
13Multiple choice · ★ Challenge
In a street-light circuit, a light-dependent resistor (LDR) is in series with a fixed resistor across a steady supply. As darkness falls, what happens to the p.d. across the LDR?
1. AIt falls, as the LDR’s resistance rises and less current flows
2. BIt rises, as the LDR’s resistance falls and more current flows
3. CIt stays the same, since the supply voltage does not change
4. DIt rises, as the LDR’s resistance rises and it takes a larger share
Show answer
Answer: D. It rises, as the LDR’s resistance rises and it takes a larger share
In the dark an LDR’s resistance becomes large; in a potential divider the larger resistance takes the larger share of the supply p.d., so the p.d. across the LDR rises (this can switch the lamp on).
Common mistakeChoosing "it falls, less current" looks only at the current; although the current falls, the LDR’s share of the voltage grows.
14Fill in the blank
A rheostat connected as a potential divider (using all three terminals) can give an output voltage from zero up to the full ______ voltage.
Show answer
Answer: supply
With the slider at one end the output is 0 V; at the other end the output is the whole supply p.d.
Common mistakeAnswering "half the supply" confuses one slider position with the full range; the slider can be moved to any point.
15Short answer · ★ Challenge
A metal-cased kettle has an earth wire and a fuse. Explain how they protect the user if a loose live wire touches the inside of the case.
Show answer
Model answer: The earth wire connects the metal case to the ground through a very low resistance. When the live wire touches the case, a very large current flows from live to earth through this low-resistance path (a short circuit). This large current melts the fuse in the live wire, cutting off the supply. The case therefore never stays at a dangerous voltage, so a person touching it is not shocked.
Common mistakeSaying "the earth wire takes the shock instead of the person" is only half the story; the fuse must also melt to cut off the live supply.
16Fill in the blank
The resistance of a metal wire is directly proportional to its length and inversely proportional to its ______.
Show answer
Answer: cross-sectional area
R = ρL/A: a longer wire has more resistance, a thicker wire (larger A) has less.
Common mistakeAnswering "diameter" is a slip: R depends on the area, which goes with the SQUARE of the diameter.
17Fill in the blank · ★ Challenge
A 12 V cell of negligible internal resistance drives 1.5 A round a single loop containing a 2.0 Ω resistor, a 4.0 Ω resistor and an unknown resistor. The unknown resistor is ______ Ω.
Show answer
Answer: 2.0
Loop law: 12 = 1.5 × (2.0 + 4.0 + R), so 2.0 + 4.0 + R = 8.0 and R = 2.0 Ω.
Common mistakeAnswering 8.0 Ω gives the total resistance of the loop; the two known resistors must be subtracted.
18Multiple choice
Three identical lamps are connected so that lamps A and B are in parallel with each other, and this pair is in series with lamp C across a battery. Which lamp is brightest?
1. AA and B, because they are in parallel
2. BAll three are equally bright
3. CA, because it is nearest the battery
4. DC, because it carries the whole current
Show answer
Answer: D. C, because it carries the whole current
The full current passes through C and then splits equally between A and B, so C carries twice the current of each of the others and is brightest.
Common mistakeChoosing "A and B, because they are in parallel" applies the idea that parallel lamps get the full voltage, which is true only when they are connected directly across the supply.
19Multiple choice · ★ Challenge
In a metre bridge, an unknown resistor X is in the left gap and a 6.0 Ω standard resistor is in the right gap. The balance point is 40.0 cm from the left end of the wire. What is X?
1. A9.0 Ω
2. B2.4 Ω
3. C4.0 Ω
4. D15 Ω
Show answer
Answer: C. 4.0 Ω
X/6.0 = l₁/l₂ = 40.0/60.0, so X = 6.0 × 40.0 ÷ 60.0 = 4.0 Ω.
Common mistakeChoosing 9.0 Ω pairs X with the wrong length (60/40); each resistor goes with the length of wire on its own side.
20Fill in the blank
The power dissipated in a resistor can be written P = VI, P = I²R or P = ______.
Show answer
Answer: V²/R
Substituting I = V/R into P = VI gives P = V × V/R = V²/R.
Common mistakeWriting P = V/R² is a slip in the algebra; check the units: V²/R gives V × A = W.
21Multiple choice · ★ Challenge
An electric iron is rated 1.2 kW, 230 V. Fuses of 3 A, 5 A and 13 A are available. Which fuse should be fitted?
1. A13 A, since the normal current is about 5.2 A
2. B5 A, since the normal current is about 5 A
3. C3 A, since the smallest fuse is always safest
4. D13 A, since the normal current is 13 A
Show answer
Answer: A. 13 A, since the normal current is about 5.2 A
I = P/V = 1200 ÷ 230 = 5.2 A. The fuse must be the smallest rating ABOVE the normal current, so 5 A is too small and 13 A is chosen.
Common mistakeChoosing 5 A rounds 5.2 A down; a 5 A fuse would melt every time the iron is switched on.
22Multiple choice
A metal wire is pulled through a die so that its length doubles while its volume stays the same. Its resistance becomes:
1. A2 times as large
2. Bthe same as before
3. C4 times as large
4. Dhalf as large
Show answer
Answer: C. 4 times as large
Doubling L at constant volume halves A; R = ρL/A then increases by 2 × 2 = 4.
Common mistakeChoosing "2 times" accounts for the longer length but forgets that the wire also becomes thinner.
23Short answer · ★ Challenge
Explain why power lines carrying electricity from hydroelectric stations are thick aluminium cables, while the heating element of a kettle is a thin coil of nichrome.
Show answer
Model answer: Power lines must waste as little energy as possible, so they need a very low resistance: a thick cable (large A) of a metal with low resistivity; aluminium is chosen because it has low resistivity and is much lighter and cheaper than copper. A heating element must turn electrical energy into heat, so it needs a fairly large resistance in a short length: nichrome has a high resistivity and does not oxidise when red-hot, and a thin wire increases R further.
Common mistakeSaying "nichrome conducts better" reverses the idea: the element is chosen for its HIGH resistivity, so it gets hot.
24Multiple choice
In a household plug, why must the fuse be fitted in the live wire?
1. AWhen it melts, the appliance is cut off from the high voltage
2. BThe neutral wire carries no current, so a fuse there could not melt
3. CThe live wire is thicker, so it has room to hold the fuse
4. DThe earth wire carries the current, so the fuse is not needed there
Show answer
Answer: A. When it melts, the appliance is cut off from the high voltage
If the fuse were in the neutral wire, a blown fuse would stop the current but leave the appliance connected to the live voltage, which could still give a shock.
Common mistakeChoosing "the neutral carries no current" is wrong; in normal use the same current flows in live and neutral, so the reason is safety, not current.
25Short answer · ★ Challenge
Explain why a metre bridge can give a more accurate value of resistance than an ammeter–voltmeter method, and why the balance point should be near the middle of the wire.
Show answer
Model answer: The bridge is a null method: you look for zero galvanometer current, so the resistance of the meters and their calibration errors do not matter, and the cell’s e.m.f. does not need to be steady. With the balance point near the middle, l₁ and l₂ are both large, so the same reading error of about 1 mm gives a smaller percentage error in each length (and end effects at the clips matter less).
Common mistakeSaying the bridge is better "because it has more resistors" misses the key idea of a null (zero-current) reading.
26Multiple choice
A torch bulb needs 4.5 V. How should 1.5 V dry cells be arranged to light it properly?
1. AThree cells in series, + to −
2. BThree cells in parallel
3. CTwo in series and one in parallel
4. DFour cells in parallel
Show answer
Answer: A. Three cells in series, + to −
E.m.f.s in series add: 1.5 + 1.5 + 1.5 = 4.5 V. Cells in parallel still give only 1.5 V.
Common mistakeChoosing "three in parallel" expects the e.m.f.s to add; parallel cells share the current but give the e.m.f. of just one cell.
27Short answer · ★ Challenge
When a driver starts a car with the headlights on, the headlights dim for a moment. Explain this using the idea of internal resistance.
Show answer
Model answer: The starter motor draws a very large current from the battery. The p.d. lost inside the battery, Ir, becomes large, so the terminal p.d. V = ε − Ir falls sharply. The headlights, connected across the same terminals, receive a smaller p.d. and dim until the starter motor stops.
Common mistakeSaying "the motor takes the electricity away from the lights" is vague; the key is that the large current increases the volts lost inside the battery.
28Fill in the blank
A resistor carries a current of 0.25 A when the p.d. across it is 6.0 V. Its resistance is ______ Ω.
Show answer
Answer: 24
R = V/I = 6.0 ÷ 0.25 = 24 Ω.
Common mistakeAnswering 1.5 Ω multiplies V by I; that product is the power in watts, not the resistance.
29Short answer · ★ Challenge
A solar home system battery bank is made of many identical cells: groups of cells joined in series, with several such groups joined in parallel. Explain the purpose of each kind of connection.
Show answer
Model answer: Cells in series add their e.m.f.s, so each group gives the voltage the lamps and charger need (for example 12 V). Joining identical groups in parallel keeps the same voltage but lowers the total internal resistance and shares the current between groups, so a larger current can be supplied and the bank lasts longer before it needs recharging.
Common mistakeSaying parallel groups "double the voltage" is a common slip; identical groups in parallel give the same e.m.f. as one group.
30True or false
A short circuit is a path of very low resistance, so it can draw a dangerously large current from the supply.
Show answer
Answer: True
With almost no resistance, I = V/R becomes very large; the wires heat up quickly, which is why fuses and circuit breakers are needed.
Common mistakeSome learners think a short circuit stops the current; in fact the current becomes much larger, but it bypasses the appliance.
31Multiple choice · ★ Challenge
Two cells, each of e.m.f. 1.5 V and internal resistance 0.40 Ω, are connected in series to a 2.2 Ω bulb. What current flows?
1. A1.4 A
2. B1.2 A
3. C0.50 A
4. D1.0 A
Show answer
Answer: D. 1.0 A
Total e.m.f. = 3.0 V; total resistance = 2.2 + 0.40 + 0.40 = 3.0 Ω; I = 3.0 ÷ 3.0 = 1.0 A.
Common mistakeChoosing 1.2 A counts only one internal resistance (3.0 ÷ 2.6); in series both cells’ internal resistances add.
32Multiple choice · ★ Challenge
A copper wire 20 m long has a cross-sectional area of 1.0 mm². The resistivity of copper is 1.7 × 10⁻⁸ Ω m. What is its resistance?
1. A0.34 Ω
2. B3.4 × 10⁻⁷ Ω
3. C340 Ω
4. D0.017 Ω
Show answer
Answer: A. 0.34 Ω
1.0 mm² = 1.0 × 10⁻⁶ m². R = ρL/A = 1.7 × 10⁻⁸ × 20 ÷ 1.0 × 10⁻⁶ = 0.34 Ω.
Common mistakeChoosing 3.4 × 10⁻⁷ Ω leaves the area as "1"; 1 mm² must be converted to 10⁻⁶ m² because ρ is in Ω m.
33Fill in the blank
To convert a sensitive galvanometer into an ammeter for large currents, a small resistance called a ______ is connected in parallel with it.
Show answer
Answer: shunt
The shunt carries most of the current, so only a small, known fraction passes through the galvanometer.
Common mistakeConnecting the extra resistance in series makes a voltmeter, not an ammeter; an ammeter needs a low-resistance parallel path.
34Short answer
Give three reasons why the lamps in a house are connected in parallel rather than in series.
Show answer
Model answer: (1) Each lamp gets the full mains voltage, so it works at its normal brightness. (2) Each lamp can be switched on and off on its own. (3) If one lamp breaks, the others still work, because each has its own complete path for the current.
Common mistakeSaying parallel "uses less current" is wrong; the total current from the supply is actually larger, since each lamp draws its own current.
35Short answer · ★ Challenge
A poultry farmer wants an alarm to sound when her incubator gets too hot. Describe how a thermistor (whose resistance falls as temperature rises) and a fixed resistor can be used in a potential divider to give a voltage that rises with temperature.
Show answer
Model answer: Connect the thermistor and a fixed resistor in series across the supply, and take the output across the FIXED resistor. When the temperature rises, the thermistor’s resistance falls, so it takes a smaller share of the supply voltage and the fixed resistor takes a larger share: the output voltage rises. When it passes a set value it can switch on the alarm (through a transistor or relay); the fixed resistor can be variable to set the temperature.
Common mistakeTaking the output across the thermistor gives a voltage that FALLS as it gets hotter; the output must be taken across the component whose share increases.
36Short answer · ★ Challenge
Electrons drift along a copper wire at less than 1 mm per second, yet a classroom lamp 10 m away lights almost the instant the switch is closed. Explain.
Show answer
Model answer: The wire is already full of free electrons. Closing the switch sets up an electric field along the whole circuit almost instantly (at nearly the speed of light), so the electrons everywhere, including those in the lamp filament, start drifting at the same time. The lamp does not have to wait for electrons from the switch to reach it.
Common mistakeSaying "the electrons move at the speed of light" is wrong; it is the signal (the field) that travels fast, while the electrons themselves drift slowly.
37Short answer
A learner connects an ammeter in parallel with a lamp instead of in series. Describe and explain what happens.
Show answer
Model answer: The ammeter has a very low resistance, so it short-circuits the lamp: almost all the current goes through the ammeter instead of the lamp. The lamp goes out, and because the total resistance of the circuit is now very small, a very large current flows, which can blow the ammeter’s fuse or damage it and flatten the cell.
Common mistakeSaying "the ammeter just reads the lamp current" ignores its low resistance; meters change the circuit if connected the wrong way.
38Multiple choice · ★ Challenge
A learner tests two components. X: V = 1.0, 2.0, 3.0 V gives I = 0.20, 0.40, 0.60 A. Y: the same voltages give I = 0.20, 0.32, 0.40 A. Which conclusion is correct?
1. AX is ohmic (5.0 Ω); Y’s resistance rises from 5.0 Ω to 7.5 Ω
2. BY is ohmic (5.0 Ω); X’s resistance rises from 5.0 Ω to 7.5 Ω
3. CBoth are ohmic, because both give 5.0 Ω at 1.0 V
4. DX is ohmic (0.20 Ω); Y’s resistance falls from 0.20 Ω to 0.13 Ω
Show answer
Answer: A. X is ohmic (5.0 Ω); Y’s resistance rises from 5.0 Ω to 7.5 Ω
X: V/I = 1.0/0.20 = 2.0/0.40 = 3.0/0.60 = 5.0 Ω, constant. Y: 1.0/0.20 = 5.0 Ω but 3.0/0.40 = 7.5 Ω, so its resistance increases (like a lamp).
Common mistakeChoosing "both ohmic" checks only one reading; to test Ohm’s law, V/I must stay the same for EVERY reading.
39Fill in the blank
When no current is being drawn from a cell, a very high-resistance voltmeter across its terminals reads its ______.
Show answer
Answer: e.m.f. (electromotive force)
With I = 0 there are no lost volts (Ir = 0), so the terminal p.d. equals the e.m.f.
Common mistakeAnswering "internal resistance" mixes up quantities: a voltmeter measures a p.d. in volts, not a resistance.
40Short answer · ★ Challenge
Describe how you would obtain the I–V characteristic of a resistor in the school laboratory, and state what the graph shows if the resistor obeys Ohm’s law.
Show answer
Model answer: Connect a cell (or power supply), a rheostat, an ammeter and the resistor in series, with a voltmeter in parallel across the resistor. Change the rheostat to get several values of V and record I each time; reverse the supply to get negative values. Plot I against V. If Ohm’s law is obeyed the graph is a straight line through the origin, and the resistance is 1 ÷ gradient (or V/I). Keep the current small so the resistor does not warm up.
Common mistakePutting the voltmeter in series stops the current (it has a very high resistance); the voltmeter must go in parallel with the component.
41Multiple choice
Four wires meet at a junction. 3.0 A and 1.5 A flow into it along two wires, and 2.5 A flows out along the third. What is the current in the fourth wire?
1. A2.0 A into the junction
2. B2.0 A out of the junction
3. C7.0 A out of the junction
4. D1.0 A out of the junction
Show answer
Answer: B. 2.0 A out of the junction
Current in = current out: 3.0 + 1.5 = 2.5 + I, so I = 2.0 A flowing out.
Common mistakeChoosing 7.0 A adds all three currents, ignoring that 2.5 A is already leaving the junction.
42Multiple choice · ★ Challenge
A car battery has an e.m.f. of 12.6 V. While the starter motor draws 150 A, the p.d. across the battery terminals falls to 9.6 V. What is the internal resistance of the battery?
1. A0.064 Ω
2. B0.084 Ω
3. C0.020 Ω
4. D50 Ω
Show answer
Answer: C. 0.020 Ω
Lost volts = 12.6 − 9.6 = 3.0 V = Ir, so r = 3.0 ÷ 150 = 0.020 Ω.
Common mistakeChoosing 0.064 Ω divides the terminal p.d. by the current, which gives the resistance of the starter motor, not of the battery.
43Multiple choice
A Wheatstone bridge is balanced with P = 10 Ω, Q = 20 Ω and R = 15 Ω, where P/Q = R/S. What is the unknown resistance S?
1. A7.5 Ω
2. B30 Ω
3. C45 Ω
4. D300 Ω
Show answer
Answer: B. 30 Ω
P/Q = R/S, so S = R × Q/P = 15 × 20 ÷ 10 = 30 Ω.
Common mistakeChoosing 7.5 Ω turns the ratio upside down (S = R × P/Q); check that S/R must equal Q/P = 2.
44Short answer · ★ Challenge
A thin extension cable has a total resistance of 0.40 Ω and carries 9.0 A to a heater. Calculate the power wasted in the cable and explain why a thin cable can be dangerous.
Show answer
Model answer: P = I²R = 9.0² × 0.40 = 81 × 0.40 = 32.4 W is turned into heat in the cable. Because the power depends on I², a large current in a thin (higher-resistance) cable makes it hot; the insulation can melt and start a fire. A thicker cable has less resistance and wastes much less power.
Common mistakeUsing P = V²/R with the mains voltage (230 V) is the trap; the 230 V is across the heater, not across the cable.
45Multiple choice
A potential divider is made of a 2.0 kΩ resistor and a 4.0 kΩ resistor in series across a 9.0 V supply. What is the p.d. across the 4.0 kΩ resistor?
1. A3.0 V
2. B4.5 V
3. C6.0 V
4. D18 V
Show answer
Answer: C. 6.0 V
V_out = V_in × R₂/(R₁ + R₂) = 9.0 × 4.0 ÷ 6.0 = 6.0 V. The larger resistor takes the larger share.
Common mistakeChoosing 3.0 V gives the p.d. across the 2.0 kΩ resistor; the voltage divides in proportion to the resistances.
46Multiple choice · ★ Challenge
A family replaces a 60 W filament lamp with a 10 W LED lamp that gives the same light. The lamp is used 6 hours a day for 30 days, and electricity costs 250 FRW per kWh. How much money is saved?
1. A2700 FRW
2. B2250 FRW
3. C450 FRW
4. D9 FRW
Show answer
Answer: B. 2250 FRW
Power saved = 60 − 10 = 50 W = 0.050 kW; time = 6 × 30 = 180 h; energy saved = 0.050 × 180 = 9.0 kWh; saving = 9.0 × 250 = 2250 FRW.
Common mistakeChoosing 2700 FRW is the cost of running the old lamp alone; the saving is the DIFFERENCE between the two lamps.
47Multiple choice
Which description fits the I–V graph of a filament lamp?
1. AA straight line through the origin, as R is constant
2. BNo current until about 0.7 V, then a very steep rise
3. CA curve that gets less steep as V rises, because R rises
4. DA curve that gets steeper as V rises, because R falls
Show answer
Answer: C. A curve that gets less steep as V rises, because R rises
As the current grows the filament gets hotter, its resistance rises, and each extra volt gives less extra current, so the I–V curve bends towards the V-axis.
Common mistakeChoosing the straight line describes an ohmic resistor at constant temperature; a lamp filament changes temperature a lot.
48Multiple choice · ★ Challenge
A learner uses an ammeter of resistance 1.0 Ω to measure the current from a 2.0 V cell (no internal resistance) through a 4.0 Ω resistor. What does the ammeter read, compared with the current without the ammeter?
1. A0.40 A instead of 0.50 A
2. B0.50 A, the same as without it
3. C0.67 A instead of 0.50 A
4. D2.0 A instead of 0.50 A
Show answer
Answer: A. 0.40 A instead of 0.50 A
Without the ammeter I = 2.0 ÷ 4.0 = 0.50 A; with it the total resistance is 5.0 Ω, so I = 2.0 ÷ 5.0 = 0.40 A.
Common mistakeChoosing "the same" assumes an ideal ammeter; a real ammeter adds resistance in series, which is why ammeters must have very low resistance.
49Multiple choice
A current of 0.32 A flows in a torch bulb. How many electrons pass through the bulb each second? (e = 1.6 × 10⁻¹⁹ C)
1. A5.0 × 10⁻¹⁹
2. B2.0 × 10¹⁸
3. C2.0 × 10¹⁹
4. D5.0 × 10¹⁸
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Answer: B. 2.0 × 10¹⁸
Charge per second = 0.32 C; number of electrons = 0.32 ÷ 1.6 × 10⁻¹⁹ = 2.0 × 10¹⁸.
Common mistakeChoosing 5.0 × 10⁻¹⁹ divides the wrong way round; a huge number of tiny charges is needed to make 0.32 C.
50Short answer · ★ Challenge
A 9.0 V cell and a 3.0 V cell, each with internal resistance 0.50 Ω, are connected in series with a 5.0 Ω resistor so that their e.m.f.s oppose each other. Find the current, say which cell is being charged, and find the p.d. across the terminals of the 3.0 V cell.
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Model answer: Loop law: net e.m.f. = 9.0 − 3.0 = 6.0 V = I(5.0 + 0.50 + 0.50), so I = 6.0 ÷ 6.0 = 1.0 A. The current is driven by the 9.0 V cell and flows into the + terminal of the 3.0 V cell, so the 3.0 V cell is being charged. Its terminal p.d. = ε + Ir = 3.0 + 1.0 × 0.50 = 3.5 V.
Common mistakeUsing V = ε − Ir for the cell being charged gives 2.5 V; when current is forced INTO a cell’s positive terminal, its terminal p.d. is larger than its e.m.f.