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Admissions tests / ESAT / Physics / Electricity and magnetism

Stretch. 15 questions, 15 marks, about 30 minutes.

ESAT Physics: Electricity and magnetism, set 4

Electrostatics, current, potential difference and resistance, series and parallel circuits, power and energy, magnets and magnetic fields, the motor effect and electromagnetic induction.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  • This is a stretch set: several questions combine two specification points, and the shortest route to the answer is rarely the one that grinds through every number.
  1. 11 mark

    A 2 ohm resistor and a 4 ohm resistor are connected in series across a 12 V battery. What is the ratio of the power dissipated in the 2 ohm resistor to the power dissipated in the 4 ohm resistor?

    1. A 2 : 1
    2. B 1 : 2
    3. C 4 : 1
    4. D 1 : 4
  2. 21 mark

    A person walks across a synthetic carpet wearing shoes with insulating rubber soles, and becomes charged by friction with the carpet. Reaching for a metal door handle, the person feels a small spark. A second person, wearing shoes with electrically conductive soles, walks across the same carpet but does not experience a spark when touching the handle. Which statement correctly explains the difference?

    1. A Neither type of sole makes any real difference to what is felt; the spark happens only because of the type of carpet each person walks across, not because of the shoes worn.
    2. B The insulating soles let the first person's charge flow away to earth easily, which is why a spark is still felt; truly conductive soles would trap the charge instead and prevent any spark at all.
    3. C The conductive soles generate a charge of their own that is opposite to the carpet's, cancelling out the charge produced by friction before it can build up.
    4. D The insulating soles prevent the charge that builds up by friction with the carpet from flowing away to earth through the floor, so it accumulates on the person until it discharges suddenly as a spark to the handle; the conductive soles let the second person's charge flow away to earth continuously instead, so no significant charge builds up.
  3. 31 mark

    In a circuit, a 5 ohm resistor and a 20 ohm resistor are connected in parallel with each other, and this parallel combination is connected to a battery. An ammeter (a circle marked A) is connected in series with the battery, before the circuit splits into the two branches, and a voltmeter (a circle marked V) is connected across the 20 ohm resistor only. The ammeter reads 5 A. What does the voltmeter read?

    1. A 20 V
    2. B 100 V
    3. C 125 V
    4. D 4 V
  4. 41 mark

    A phone charger moves a charge of 600 C onto a battery over 2 minutes, transferring 1800 J of energy to the battery in the process. What are (i) the average current during charging, and (ii) the potential difference at which the charge is transferred?

    1. A 300 A and 3 V
    2. B 300 A and 1/3 V
    3. C 5 A and 3 V
    4. D 5 A and 1/3 V
  5. 51 mark

    A filament lamp's V-I graph shows that when the current is 0.5 A, the potential difference across it is 2 V; when the current is 2 A, the potential difference across it is 12 V. What is the lamp's resistance (i) at 0.5 A and (ii) at 2 A, and is this consistent with how a filament lamp's resistance is known to behave?

    1. A 24 ohm at 0.5 A and 1 ohm at 2 A; resistance decreases with current
    2. B 4 ohm at 0.5 A and 6 ohm at 2 A; resistance increases with current, consistent with a filament lamp heating up
    3. C 1/4 ohm at 0.5 A and 1/6 ohm at 2 A; resistance decreases with current
    4. D 6 ohm at 0.5 A and 4 ohm at 2 A; resistance decreases with current
  6. 61 mark

    A light-dependent resistor (LDR) and a 200 ohm fixed resistor are connected in series across a 10 V battery. In bright light the LDR's resistance is 200 ohm; in darkness its resistance rises to 1800 ohm. What is the potential difference across the fixed resistor (i) in bright light, and (ii) in darkness?

    1. A 5 V in bright light and 1 V in darkness
    2. B 5 V in bright light and 9 V in darkness
    3. C 1 V in bright light and 5 V in darkness
    4. D 0.5 V in bright light and 0.1 V in darkness
  7. 71 mark

    An electromagnet is used to lift and then release scrap metal in a scrapyard, so it must lose its magnetism almost as soon as the current is switched off. Which statement about the choice of core material for this electromagnet is correct?

    1. A A steel core should be used, because steel is a hard magnetic material that becomes strongly magnetised and stays magnetised, giving the strongest possible lifting force at all times
    2. B Either material is equally suitable, because induced magnetism depends only on the current flowing through the coil, not on the properties of the core inside it
    3. C A soft iron core should be used, but only because soft iron is a stronger permanent magnet than steel once magnetised
    4. D A soft iron core should be used, because soft magnetic materials are easily magnetised by the coil's field and easily demagnetised again once the current stops
  8. 81 mark

    A straight wire of length 25 cm carries a current of 4 A at right angles to a uniform magnetic field of flux density 0.5 T. The current flows from left to right across the page, and the field points into the page. What are the magnitude and direction of the force on the wire?

    1. A 50 N, acting vertically upward on the page
    2. B 50 N, acting vertically downward on the page
    3. C 0.5 N, acting vertically upward on the page
    4. D 0.5 N, acting vertically downward on the page
  9. 91 mark

    A simple ac generator consists of a coil rotating at a constant speed in a uniform magnetic field. A graph is plotted of the generator's output voltage against time. Which statement correctly describes (i) the shape of this graph, and (ii) the effect of increasing the coil's rotation speed, with everything else unchanged?

    1. A The graph is a horizontal straight line, since the coil rotates at a steady speed, producing a constant output voltage
    2. B The graph is a smooth wave that alternates between positive and negative values, repeatedly crossing zero; increasing the rotation speed increases both the peak voltage and the frequency of the trace
    3. C The graph is a smooth wave that alternates between positive and negative values; increasing the rotation speed increases the frequency but leaves the peak voltage unchanged, since peak voltage depends only on the magnet's strength
    4. D The graph is a smooth wave that alternates between positive and negative values; increasing the rotation speed decreases the frequency, because each rotation takes less time, but increases the peak voltage
  10. 101 mark

    An ideal transformer has 100 turns on its primary coil and 400 turns on its secondary coil. The primary coil is connected to a 230 V ac supply and draws a current of 8 A. What are the secondary voltage and the secondary current?

    1. A 920 V and 2 A
    2. B 920 V and 32 A
    3. C 57.5 V and 8 A
    4. D 57.5 V and 32 A
  11. 111 mark

    Electrical power is transmitted along a cable of fixed resistance. The transmission voltage is increased by a factor of 10, and the current is correspondingly reduced to one tenth of its previous value, so that the power actually delivered to the far end stays the same. By what factor does the power lost as heat in the cable itself change?

    1. A It stays the same
    2. B It falls to 1/10 of its original value
    3. C It falls to 1/1000 of its original value
    4. D It falls to 1/100 of its original value
  12. 121 mark

    An ideal diode is connected in series with a resistor across an alternating current (ac) supply, which drives current first in one direction and then in the other through the circuit, once every cycle. Which statement correctly describes the current that flows in the resistor?

    1. A Current flows in both directions through the resistor, exactly as it would without the diode, because a diode's resistance is the same in either direction
    2. B No current flows in the resistor at all, because the diode blocks the current during every part of the cycle
    3. C Current flows in one direction only, during half of each cycle, and is zero during the other half; unlike the original ac supply, the current in the resistor never reverses direction
    4. D Current flows continuously in one direction throughout the whole cycle, at a constant, unchanging value, exactly like the current from a battery
  13. 131 mark

    A heating element of resistance 4 ohm is connected to an 8 V supply for 10 seconds. What are (i) the power delivered to the element, and (ii) the total energy transferred in that time?

    1. A 16 W and 160 J
    2. B 256 W and 2560 J
    3. C 10 W and 100 J
    4. D 16 W and 16 J
  14. 141 mark

    A straight length of wire, of length 0.5 m and negligible resistance, is placed at right angles to a magnetic field of flux density 0.2 T. The wire is connected in series with a 4 ohm resistor across a 12 V battery of negligible internal resistance. What is the magnitude of the force on the wire due to the magnetic field?

    1. A 1.2 N
    2. B 0.3 N
    3. C 1.5 N
    4. D 3 N
  15. 151 mark

    In a parallel circuit with two branches, a charge of 12 C flows through the first branch in 4 s, and a charge of 32 C flows through the second branch in the same 4 s. What is the total current supplied by the battery?

    1. A 44 A
    2. B 5.5 A
    3. C 8 A
    4. D 11 A

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: B

    1. The two resistors are in series, so exactly the same current I flows through both of them.
    2. Using P = I^2 R with a common current, the ratio of the two powers is simply the ratio of the two resistances: P(2 ohm) : P(4 ohm) = R(2 ohm) : R(4 ohm) = 2 : 4 = 1 : 2.
    3. This shortcut needs no knowledge of the battery voltage or the actual current at all, since the shared current cancels out of the ratio; a full calculation (total R = 6 ohm, I = 2 A, giving 8 W and 16 W) confirms the same 1 : 2 ratio but takes far longer.
    4. Therefore the answer is B.
    • Why not A: This assumes the two resistors have the same potential difference across them, as they would in parallel, and uses P = V^2/R; that formula is not appropriate here because a series circuit gives the two resistors the same current, not the same voltage, and using it flips the true ratio the wrong way round.
    • Why not C: This both inverts the ratio and squares it, as though power depended on the square of the resistance and the smaller resistance dissipated more power; neither assumption is correct for two resistors sharing the same current.
    • Why not D: This wrongly assumes power is proportional to the square of the resistance rather than to the resistance itself, squaring the true 1 : 2 ratio to get 1 : 4.

  2. Question 2Answer: D

    1. Walking across the carpet transfers charge to the person by friction, exactly as in any charging-by-friction situation.
    2. An insulating sole has very few free electrons able to move, so it cannot let this charge escape to earth through the floor; the charge instead accumulates on the person until it is large enough to jump as a spark to the earthed metal handle.
    3. A conductive sole has many free electrons able to move, so it provides a continuous path for the charge to flow away to earth as it forms, keeping the charge, and the risk of a spark, small.
    4. Therefore the answer is D.
    • Why not A: This denies that the shoes make any difference, but both people walk on the same carpet; the only difference described between them is the sole material, and it is exactly this difference, insulating versus conductive, that explains why one person sparks and the other does not.
    • Why not B: This has conductors and insulators the wrong way round. An insulating sole has very few free electrons and cannot let a built-up charge escape to earth, which is why charge accumulates and later discharges as a spark; a conductive sole is what would let the charge flow away continuously and prevent one.
    • Why not C: This invents an extra mechanism, the conductive sole generating its own opposing charge, rather than the real reason it prevents a spark: a conductor simply lets any charge that forms flow away to earth as fast as it builds up, rather than cancelling it out with an opposite charge.

  3. Question 3Answer: A

    1. The two resistors are in parallel, so their combined resistance is found from R = (R1 x R2)/(R1 + R2) = (5 x 20)/(5 + 20) = 100/25 = 4 ohm.
    2. The ammeter sits before the split, so its 5 A reading is the total current entering the parallel combination, and the voltmeter reads the potential difference across that combination, which is the same across each branch since parallel branches share the same p.d.
    3. Using V = IR with the total current and the combined resistance, V = 5 x 4 = 20 V; this is quicker than separately finding each branch current from the current-divider rule.
    4. Therefore the answer is A.
    • Why not B: This treats the ammeter's reading as though it were the current flowing through the 20 ohm branch alone, when in fact the ammeter is positioned before the split and so measures the TOTAL current entering both branches together.
    • Why not C: This combines the two resistors as though they were in series, adding them (5 + 20 = 25 ohm) instead of using the parallel combination formula; the two resistors are connected in parallel here, not in series.
    • Why not D: This correctly combines the resistors in parallel to get 4 ohm, but then reports this resistance value itself as the voltmeter reading, forgetting to multiply by the current to obtain a potential difference (V = IR).

  4. Question 4Answer: C

    1. Current is charge per unit time, I = Q/t. The charging time must be in seconds: 2 minutes = 120 s, so I = 600/120 = 5 A.
    2. Potential difference is energy transferred per unit charge, V = E/Q, so V = 1800/600 = 3 V.
    3. Both relationships use the same charge (600 C), so once the time has been correctly converted to seconds, the two calculations are independent short divisions with no calculator needed.
    4. Therefore the answer is C.
    • Why not A: This correctly finds the potential difference but forgets to convert the charging time from minutes to seconds, dividing the charge by 2 instead of by 120, which makes the calculated current 60 times too large.
    • Why not B: This combines two errors: the charging time is left in minutes instead of being converted to seconds when finding the current, and the energy relationship is inverted, charge divided by energy instead of energy divided by charge, when finding the potential difference.
    • Why not D: This correctly converts the time and finds the current, but inverts the energy relationship, calculating charge divided by energy (600/1800) instead of energy divided by charge (1800/600).

  5. Question 5Answer: B

    1. Resistance is R = V/I at each point on the graph. At 0.5 A: R = 2/0.5 = 4 ohm (dividing by a half is the same as multiplying by 2, so 2 x 2 = 4).
    2. At 2 A: R = 12/2 = 6 ohm.
    3. Resistance rises from 4 ohm to 6 ohm as the current rises from 0.5 A to 2 A, which matches the known behaviour of a filament lamp: it heats up as more current flows, and a hotter filament has a higher resistance, curving the V-I graph away from a straight line.
    4. Therefore the answer is B.
    • Why not A: This pairs each current value with the OTHER point's potential difference (12 V with 0.5 A, and 2 V with 2 A) instead of reading the two values from the same point on the graph, giving resistances that do not correspond to either real point.
    • Why not C: This inverts the resistance formula, calculating current divided by potential difference instead of potential difference divided by current, and so also gets the direction of change backwards.
    • Why not D: This calculates the two resistance values correctly (4 ohm and 6 ohm) but assigns them to the wrong current, concluding wrongly that resistance decreases as current increases.

  6. Question 6Answer: A

    1. In a series circuit, the potential difference across a component is its own resistance divided by the TOTAL resistance, multiplied by the supply voltage: V(fixed) = V(supply) x R(fixed)/(R(fixed) + R(LDR)).
    2. In bright light, R(LDR) = 200 ohm equals R(fixed) = 200 ohm, so the total is 400 ohm and V(fixed) = 10 x 200/400 = 5 V.
    3. In darkness, R(LDR) rises to 1800 ohm, so the total is 2000 ohm and V(fixed) = 10 x 200/2000 = 1 V; as the LDR's resistance rises in the dark, its own share of the fixed supply voltage rises, so the fixed resistor's share must fall to compensate.
    4. Therefore the answer is A.
    • Why not B: In bright light the LDR and fixed resistor happen to have equal resistance (200 ohm each), so this option's bright-light figure is coincidentally correct; but for darkness it reports the potential difference across the LDR itself (10 x 1800/2000 = 9 V) rather than across the fixed resistor that the question actually asks about.
    • Why not C: This swaps the two lighting conditions round, reporting the darkness value for bright light and the bright-light value for darkness.
    • Why not D: This correctly finds the fraction of the total resistance taken up by the fixed resistor in each case (200/400 and 200/2000) but forgets to multiply this fraction by the 10 V supply voltage to get an actual potential difference.

  7. Question 7Answer: D

    1. Soft magnetic materials, such as iron, magnetise easily when a field is applied but also lose their induced magnetism easily once the field is removed; hard magnetic materials, such as steel, magnetise less easily but retain their magnetism well, which is why they are used for permanent magnets.
    2. A scrapyard electromagnet needs to release its load as soon as the current (and so the magnetic field driving it) is switched off, which means it needs to demagnetise almost instantly.
    3. A soft iron core provides exactly this: strong induced magnetism while the current flows, and almost none once it stops; a steel core would keep the scrap metal held even with the current off.
    4. Therefore the answer is D.
    • Why not A: This swaps soft and hard magnetic behaviour round for this job: steel's ability to retain its magnetism, which makes it useful for permanent magnets, is exactly the property that would stop this electromagnet releasing scrap metal when the current is switched off.
    • Why not B: This ignores that the core material is itself one of the factors affecting an electromagnet's field and behaviour; induced magnetism depends on how easily a material magnetises and demagnetises, which differs between soft and hard magnetic materials such as iron and steel.
    • Why not C: This gets the reasoning backwards: soft iron is suitable here because it loses its magnetism easily, not because it is a strong permanent magnet once magnetised. A material that really is a strong permanent magnet, such as steel, would be the wrong choice for a job that needs demagnetising.

  8. Question 8Answer: C

    1. The length must be in metres before it is used in F = BIL: 25 cm = 0.25 m.
    2. F = BIL = 0.5 x 4 x 0.25 = 0.5 N.
    3. Using Fleming's left-hand rule (First finger = Field, into the page; Second finger = Current, to the right; thuMb = force): with the field into the page and the current to the right, the thumb points vertically upward on the page, so that is the direction of the force.
    4. Therefore the answer is C.
    • Why not A: This gets the direction right but forgets to convert the wire's length from 25 cm to 0.25 m before substituting into F = BIL, making the calculated force 100 times too large.
    • Why not B: This combines two errors: the length is left in centimetres instead of being converted to metres, and the direction is found using the right-hand rule, appropriate for generators and induced current, instead of the left-hand rule that applies to the motor effect.
    • Why not D: This gets the magnitude right but reverses the direction, as if using the right-hand rule instead of Fleming's left-hand rule, which is the rule that applies to a current-carrying wire in a field.

  9. Question 9Answer: B

    1. A rotating coil generator produces an alternating output because the angle between the coil and the field constantly changes and reverses each half turn, so the rate of cutting field lines rises, falls and reverses sign throughout each rotation, giving a smooth wave that crosses zero regularly.
    2. The size of an induced voltage depends on how quickly the field lines are cut, which is affected by the coil's rotation speed as well as by the field strength; a faster rotation increases the peak voltage.
    3. A faster rotation also means more complete rotations happen every second, and frequency is the number of complete cycles per second, so the frequency increases too.
    4. Therefore the answer is B.
    • Why not A: This confuses 'constant speed of rotation' with 'constant output'. Even though the coil turns at a steady rate, the angle between the coil and the field keeps changing throughout each rotation, so the rate of cutting field lines, and so the induced voltage, rises and falls and reverses, producing an alternating output rather than a constant one.
    • Why not C: This wrongly assumes the peak induced voltage depends only on the magnet's field strength. The size of an induced voltage depends on how quickly the field lines are cut, which is set by the rotation speed as well as the field strength, so a faster rotation increases the peak voltage too, not just the frequency.
    • Why not D: This has frequency and period the wrong way round: a shorter time for each rotation means more rotations happen every second, which is a higher frequency, not a lower one, even though the peak voltage is correctly identified as increasing.

  10. Question 10Answer: A

    1. The transformer has more turns on the secondary coil than the primary (400 against 100), so it steps the voltage up: Vs = Vp x (ns/np) = 230 x 4 = 920 V.
    2. An ideal transformer conserves power, so Vp Ip = Vs Is. Rearranging, Is = Ip x (np/ns) = 8 x (100/400) = 8/4 = 2 A: the current steps down by the same factor that the voltage steps up.
    3. Both calculations use only whole-number ratios, multiplying or dividing by 4, so neither needs a calculator.
    4. Therefore the answer is A.
    • Why not B: This correctly finds the secondary voltage but then scales the current by the same factor (multiplying by 4) instead of using the inverse relationship required by conservation of power (Vp Ip = Vs Is); stepping the voltage up must step the current down by the same factor, not up.
    • Why not C: This divides by the turns ratio instead of multiplying when finding the voltage, as though this were a step-down transformer despite it having more turns on the secondary coil, and then leaves the current completely unchanged from its primary value instead of recalculating it for the secondary side.
    • Why not D: This combines two errors: the turns ratio is inverted when finding the voltage, treating a transformer with more secondary turns as a step-down type, and the current is then scaled in the same wrong direction rather than the inverse direction that conservation of power requires.

  11. Question 11Answer: D

    1. The power lost as heat in a cable of resistance R carrying current I is P(loss) = I^2 R, one of the three equivalent expressions for power, P = IV = I^2 R.
    2. The cable's resistance R has not changed, only the current, which has fallen to one tenth of its previous value.
    3. Since P(loss) depends on the square of the current, it falls to (1/10)^2 = 1/100 of its original value; this is exactly why power is transmitted at high voltage and low current, even though the power actually delivered (P = IV, kept constant by raising V as I falls) is unchanged.
    4. Therefore the answer is D.
    • Why not A: This confuses the power actually delivered to the far end, which is designed to stay the same, with the power lost as heat in the cable's own resistance, which is a separate, unwanted quantity governed by a different relationship (P = I^2 R) and does change when the current changes.
    • Why not B: This treats the power lost in the cable as directly proportional to the current, as though using P = IV for the cable itself; the power lost as heat in a resistor is P = I^2 R, which depends on the square of the current, not the current alone.
    • Why not C: This double-counts the square relationship, multiplying the correct 1/100 factor by a further, spurious 1/10, as though the square relationship needed applying twice instead of once.

  12. Question 12Answer: C

    1. An ac supply periodically reverses the direction in which it drives current, once every cycle; a dc current never reverses direction.
    2. An ideal diode has almost no resistance to current flowing in one direction, but effectively infinite resistance to current in the other direction.
    3. During the half of each cycle when the supply tries to drive current in the diode's high-resistance direction, no current flows at all; during the other half, current flows normally. The result is a current that is sometimes zero, but whenever it does flow, always flows the same way, unlike the original ac supply.
    4. Therefore the answer is C.
    • Why not A: This wrongly treats the ideal diode as though it had the same very low resistance in both directions, like a plain wire, ignoring its defining property: an ideal diode has almost no resistance to current in one direction, and, in the ideal case, infinite resistance in the other.
    • Why not B: This has the diode blocking every part of the cycle, but an ideal diode only blocks current during the half of the cycle when the supply is trying to drive current in its high-resistance direction; during the other half, it conducts almost as freely as a plain wire.
    • Why not D: This correctly identifies that the current becomes one-directional, but wrongly claims it also becomes constant and continuous throughout the whole cycle; the diode still blocks the current completely for half of each cycle, so the current cannot be continuous, only one-directional whenever it does flow.

  13. Question 13Answer: A

    1. Rearranging R = V/I gives the current: I = V/R = 8/4 = 2 A.
    2. Power is P = IV = 2 x 8 = 16 W.
    3. Energy transferred over 10 seconds is E = Pt = 16 x 10 = 160 J; this is the same result as using E = VIt directly (8 x 2 x 10 = 160 J), without needing to find the power as a separate step first.
    4. Therefore the answer is A.
    • Why not B: This finds the current by multiplying the voltage and resistance (8 x 4 = 32) instead of dividing (I = V/R = 8/4 = 2 A), which is the correct rearrangement of R = V/I; carrying this wrong current through the rest of the calculation makes both the power and the energy far too large.
    • Why not C: This adds the current and voltage together instead of multiplying them, as though power were a sum rather than a product; P = IV is a multiplication of the two quantities, not an addition.
    • Why not D: This correctly finds the power (16 W) but then reports this figure directly as the energy transferred, forgetting that power is only the rate of energy transfer; the actual energy transferred also depends on how long the device runs, and only multiplying by the time, E = Pt, gives it.

  14. Question 14Answer: B

    1. Since the wire itself has negligible resistance, the current in this series circuit is set entirely by the 4 ohm resistor: I = V/R = 12/4 = 3 A.
    2. The force on a wire carrying current at right angles to a magnetic field is F = BIL.
    3. F = 0.2 x 3 x 0.5 = 0.3 N.
    4. Therefore the answer is B.
    • Why not A: This uses the battery's voltage (12 V) directly as the current in F = BIL, skipping the step of using R = V/I to find that the 4 ohm resistor actually limits the current in this series circuit to 3 A.
    • Why not C: This correctly finds the current as 3 A but leaves the magnetic flux density out of the calculation, computing F = IL (3 x 0.5 = 1.5) instead of F = BIL, as if the strength of the field made no difference to the force.
    • Why not D: This misplaces the decimal point in the flux density, using 2 T instead of the 0.2 T given, which makes the calculated force ten times too large.

  15. Question 15Answer: D

    1. Current in each branch is I = Q/t: branch 1 gives I1 = 12/4 = 3 A, and branch 2 gives I2 = 32/4 = 8 A.
    2. In a parallel circuit, the total current supplied by the battery is the sum of the currents in the separate branches, not their average and not just the larger of the two.
    3. I(total) = I1 + I2 = 3 + 8 = 11 A.
    4. Therefore the answer is D.
    • Why not A: This adds the two charges together (12 + 32 = 44) without dividing by the time, ignoring that current is charge per unit time, I = Q/t, not charge on its own.
    • Why not B: This wrongly averages the two branch currents, (3 + 8)/2 = 5.5, instead of adding them; the two branches are separate additional paths for charge, so the total current supplied by the battery is the sum of the branch currents, not their average.
    • Why not C: This correctly finds each branch current (3 A and 8 A) but then applies the series rule that current is the same everywhere in the circuit, reporting only the larger branch current as the total instead of adding the two branch currents together, as the parallel current rule requires.

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