Edexcel International A Level Physics · Unit 2 · Waves and Electricity
Series and parallel circuits and I–V characteristics
40 free, high-quality past-paper-style questions on Series and parallel circuits and I–V characteristics for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 134 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U2-7-A-01Unit 2 · ASEasy1 mark
Four wires meet at a junction P. A current of 2.4 A flows into P along the first wire and 0.9 A flows into P along the second wire. A current of 1.5 A flows out of P along the third wire. What is the current out of P along the fourth wire?
Figure: currents in the four wires at junction P
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Answer
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Notes
Total current in = total current out (conservation of charge). 2.4 + 0.9 = 1.5 + I, so I = 1.8 A.
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Answer B
Answer: B
Exam tip: Sort the currents into “in” and “out” before adding. Write ΣIin = ΣIout as an equation with the unknown on one side.
The total current into any junction in a circuit equals the total current out of it. This is a consequence of the conservation of which quantity?
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Notes
Current is the rate of flow of charge. Charge cannot be created, destroyed or stored at a junction, so the charge arriving each second equals the charge leaving each second: conservation of charge. Conservation of energy explains how p.d.s add around a loop.
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Answer A
Answer: A
Common error: Swapping the two conservation laws. Currents at junctions: charge. P.d.s around loops: energy.
The I–V characteristic of a component shows no current for any negative p.d. For positive p.d.s the current is negligible until the p.d. reaches about 0.6 V; the current then rises very steeply. Which component is this?
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Answer
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Conduction in one direction only, starting at a small forward p.d. of about 0.6 V, is the characteristic of a semiconductor diode. The other three conduct equally in both directions, so their graphs are symmetrical about the origin.
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Answer D
Answer: D
Exam tip: Check symmetry first: a graph that is not symmetrical about the origin can only be the diode among the four standard components.
In a model railway control board, a current of 1.25 A from the supply enters a junction. From the junction, 0.40 A flows to the signal lamps, 0.62 A flows to the train motor and the rest flows to the motors that move the points.
Figure: currents at the junction in the control board
(a)
State Kirchhoff's first law.
(1)
(b)
Name the quantity whose conservation leads to this law.
(1)
(c)
Calculate the current in the branch to the points motors.
(1)
(d)
Calculate the charge that flows through the train motor in 5.0 minutes.
(1)
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Answer
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Notes
Part (a)
the sum of the currents into a junction equals the sum of the currents out of it
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Part (b)
charge
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Part (c)
1.25 − 0.40 − 0.62 = 0.23 A
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Part (d)
Q = It = 0.62 × 300 = 190 C (186 C)
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Exam tip: Convert minutes to seconds before using Q = It: 5.0 minutes = 300 s.
The diagram shows part of a circuit. Two junctions, P and Q, are joined by a wire containing a resistor. The currents in three of the wires are shown. What is the current I?
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Answer
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At P: 1.6 A in = 0.50 A out + current in PQ, so the current from P to Q is 1.1 A. At Q: 1.1 A + 0.30 A flow in, so I = 1.4 A.
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Answer C
Answer: C
Exam tip: With two junctions, apply ‘current in = current out’ to one junction at a time, and write the current in the connecting wire on the diagram before moving on.
The diagram shows a network of four resistors connected between terminals X and Y. What is the resistance between X and Y?
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Answer
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Notes
Lower branch: 12 + 27 = 39 Ω. In parallel with 33 Ω: 33 × 39/(33 + 39) = 17.9 Ω. In series with 22 Ω: 22 + 17.9 = 40 Ω (39.9 Ω).
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Answer B
Answer: B
Exam tip: Work from the inside out: combine series resistors within a branch first, then the parallel branches, then add the series resistor outside them.
A student wants to measure the current in a filament lamp for p.d.s from zero up to the full battery p.d. The diagrams show four circuits, P, Q, R and S, that she could build. Which circuit should she use?
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Circuit Q uses the variable resistor as a potential divider, so moving the slider to the left-hand end of the track makes the lamp's p.d. zero. The ammeter is in series with the lamp and the voltmeter is across the lamp only.
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Answer B
Answer: B
Exam tip: A variable resistor in series can only reduce the current, never to zero p.d. Connected across the supply as a potential divider, it gives every p.d. from zero to the full supply p.d.
The diagram shows a circuit containing six ammeters, A1 to A6, which have negligible resistance. The readings on three of them are shown.
(a)
Determine the reading on A5.
(1)
(b)
Determine the reading on A3.
(1)
(c)
State the reading on A6 and explain your answer in terms of charge.
(2)
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Part (a)
0.90 − 0.40 = 0.50 A
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Part (b)
0.50 − 0.32 = 0.18 A (allow error carried forward from (a))
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Part (c)
0.90 A
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charge is conserved: no charge is stored or lost in the components, so the charge returning to the battery each second equals the charge leaving it each second
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Exam tip: Label the current in every wire on the diagram as you go. Each junction gives one equation, and the current returning to the supply always equals the current leaving it.
The diagram shows a 6.0 V battery of negligible internal resistance connected to a 12 Ω resistor R1, a lamp L and two resistors R2 and R3. The ideal voltmeters V1 and V2 show the readings given.
(a)
Calculate the p.d. across the lamp.
(1)
(b)
Calculate the p.d. across R3.
(1)
(c)
Calculate the current in R1.
(1)
(d)
Calculate the current in the lamp.
(2)
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Answer
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Notes
Part (a)
6.0 − 2.4 = 3.6 V
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Part (b)
3.6 − 1.5 = 2.1 V (allow error carried forward)
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Part (c)
2.4 ÷ 12 = 0.20 A
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Part (d)
current in R2 (and R3) = 1.5 ÷ 30 = 0.050 A
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lamp current = 0.20 − 0.050 = 0.15 A
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Exam tip: The branch containing R2 and R3 has the same p.d. as the lamp because both branches join the same two junctions. Find that p.d. first, then share it within the branch.
The I–V graph of a filament lamp passes through the points (3.0 V, 0.20 A) and (6.0 V, 0.30 A). What is the resistance of the lamp when the p.d. across it is 6.0 V?
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Resistance at a point is V/I using that point's coordinates: 6.0 ÷ 0.30 = 20 Ω. (At 3.0 V it is 15 Ω — the resistance rises as the filament gets hotter.)
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Answer C
Answer: C
Common error: Using ΔV/ΔI between two points. For a curved graph that is neither point's resistance: R = V/I at the point asked about.
A 12 V supply of negligible internal resistance is connected in series with a 1.5 kΩ resistor and a 4.5 kΩ resistor. An ideal voltmeter is connected across the 4.5 kΩ resistor. What is the reading on the voltmeter?
Figure: the circuit, with the voltmeter across the 4.5 kΩ resistor
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Answer
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Notes
Current = 12 ÷ 6000 = 2.0 mA, the same in both resistors. P.d. across 4.5 kΩ = 2.0 × 10−3 × 4500 = 9.0 V. Check: 3.0 V across the 1.5 kΩ resistor, and 3.0 + 9.0 = 12 V.
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Answer C
Answer: C
Exam tip: In series the p.d. divides in proportion to resistance: 4.5/(1.5 + 4.5) of 12 V.
Three identical 10 Ω resistors are available. Two are connected in parallel, and this pair is connected in series with the third. What is the total resistance?
Figure: two resistors in parallel, connected in series with the third
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Answer
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Notes
Two 10 Ω in parallel = 10/2 = 5.0 Ω. In series with 10 Ω: 5.0 + 10 = 15 Ω.
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Answer C
Answer: C
Exam tip: Reduce the parallel group to one resistor first, then add the series part.
An NTC thermistor is connected to a variable supply. The p.d. V across it is increased slowly from zero, and the current I through it warms the thermistor. Which describes the graph of I (vertical axis) against V (horizontal axis)?
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Answer
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Notes
As the thermistor warms, its resistance falls, so V/I decreases and each extra volt drives more extra current than the last: the gradient increases.
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Answer D
Answer: D
Exam tip: On an I–V graph, rising gradient means falling resistance (thermistor heating); falling gradient means rising resistance (filament lamp heating).
A silicon diode is connected in series with a 220 Ω resistor and a 5.0 V supply of negligible internal resistance. The diode is forward biased and the p.d. across it is 0.70 V. What is the current in the circuit?
Figure: diode and resistor in series with the supply (diode forward biased)
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Answer
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Notes
P.d. across the resistor = 5.0 − 0.70 = 4.3 V (p.d.s in series add up to the supply p.d.). I = 4.3 ÷ 220 = 20 mA (19.5 mA); the diode carries the same current.
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Answer B
Answer: B
Common error: Dividing the whole supply p.d. by the resistor. The diode takes its share of the p.d., so subtract it first.
This question is about resistors connected in parallel.
Figure: two resistors in parallel across a supply of p.d. V
(a)
Two resistors of resistance R1 and R2 are connected in parallel. Show that their combined resistance R is given by 1/R = 1/R1 + 1/R2. State the physical principle used at each step.
(3)
(b)
A technician needs a resistance of 82 Ω. She has a 100 Ω resistor. Calculate the resistance she should connect in parallel with it.
(2)
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Answer
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Notes
Part (a)
the p.d. V is the same across each resistor, because both join the same two points (conservation of energy)
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the supply current I = I1 + I2, by conservation of charge at the junction
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V/R = V/R1 + V/R2, and dividing by V gives the result
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Part (b)
1/X = 1/82 − 1/100
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X = 460 Ω (456 Ω)
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Exam tip: In a derivation, write each physical statement in words before turning it into an equation; an answer that only manipulates symbols misses the physics.
A student measures the current I in a 12 V filament lamp for several values of the p.d. V across it. Her results are: 1.0 V — 0.190 A; 2.0 V — 0.308 A; 4.0 V — 0.444 A; 6.0 V — 0.522 A; 8.0 V — 0.571 A; 10.0 V — 0.606 A.
Figure: the student's results for the filament lamp
(a)
Calculate the resistance of the lamp at 2.0 V and at 10.0 V.
(2)
(b)
Another student calculates (10.0 − 8.0) ÷ (0.606 − 0.571) and calls this the resistance at 10.0 V. Calculate his value and explain why it is not the resistance at 10.0 V.
(2)
(c)
Explain why the resistance of the lamp increases as the p.d. increases.
(1)
(d)
State the current in the lamp when the p.d. across it is −6.0 V.
(1)
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Answer
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Notes
Part (a)
2.0 ÷ 0.308 = 6.5 Ω (6.49 Ω)
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10.0 ÷ 0.606 = 16.5 Ω
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Part (b)
2.0 ÷ 0.035 = 57 Ω
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resistance is V/I at the point, not the ratio of changes / the inverse gradient; they are equal only for a straight-line graph through the origin
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Part (c)
a larger current transfers more power in the filament, so its temperature rises, and the resistance of the metal filament increases with temperature
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Part (d)
−0.522 A (same size as at +6.0 V, opposite direction)
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Exam tip: A lamp's characteristic is symmetrical about the origin: reversing the p.d. reverses the current but does not change its size.
A student is given a semiconductor diode, a 6 V supply, a variable resistor, a protective resistor, a voltmeter, a milliammeter and connecting leads. She is asked to determine the I–V characteristic of the diode.
(a)
Describe how she could obtain the I–V characteristic of the diode for both forward and reverse p.d.s. Your answer should include a description of the circuit.
(6)
(b)
With a forward p.d. of 0.72 V the current is 16 mA. Calculate the resistance of the diode at this p.d.
(1)
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Answer
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Notes
Part (a)
Indicative content:
—
the variable resistor is connected across the supply as a potential divider, with the diode circuit connected between one end and the slider, so the p.d. can be varied from zero
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the voltmeter is connected in parallel with the diode only
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the milliammeter is connected in series with the diode
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the protective resistor is connected in series with the diode to limit the current
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the current is recorded for a range of p.d.s, with small steps between about 0.4 V and 0.8 V where the current rises rapidly
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the diode connections are reversed to obtain readings for reverse p.d.s, and I is plotted against V
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Part (b)
0.72 ÷ 0.016 = 45 Ω
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Model sketch
Figure: one suitable circuit: variable resistor as a potential divider, with milliammeter and protective resistor in series with the diode and voltmeter across the diode
Exam tip: In a method description, say where each meter goes and why the potential divider is used: it lets the p.d. start from zero, so the whole characteristic can be plotted.
The graphs W, X, Y and Z show how the potential difference V across four different components varies with the current I in them. Note that V is on the vertical axis. Which graph is for a filament lamp?
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Notes
With V on the vertical axis, V/I is the resistance, and the gradient grows as the resistance grows. A filament lamp's resistance rises as it heats up, for either direction of current, so its graph steepens away from the origin in both quadrants: Z. W is an ohmic conductor, X a thermistor and Y a diode.
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Answer D
Answer: D
Common error: Matching the shape from memory without reading the axes. A lamp's I–V graph flattens off, but with the axes swapped the same lamp gives a graph that gets steeper.
A battery of e.m.f. 9.0 V and negligible internal resistance is connected in series with resistor P, a filament lamp L and resistor Q. Two ideal voltmeters are connected as shown in the diagram. What is the p.d. across the lamp?
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Answer
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V1 = VP + VL = 6.2 V and V2 = VL + VQ = 5.0 V. Around the loop VP + VL + VQ = 9.0 V. Adding the two readings counts the lamp twice: 6.2 + 5.0 − 9.0 = 2.2 V. (Then VP = 4.0 V and VQ = 2.8 V.)
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Answer A
Answer: A
Exam tip: Check which components each voltmeter spans before using its reading. A voltmeter across two series components reads the sum of their p.d.s.
The graph shows the I–V characteristic of an NTC thermistor. The thermistor is connected in parallel with a 150 Ω resistor across a 6.0 V supply of negligible internal resistance. What is the current from the supply?
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Answer
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Notes
Both components have 6.0 V across them. From the graph (point T) the thermistor current is 60 mA. The resistor current is 6.0 ÷ 150 = 0.040 A = 40 mA. Supply current = 60 + 40 = 100 mA.
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Answer D
Answer: D
Exam tip: In parallel, every branch has the supply p.d. Read each branch current at that p.d. (from a graph or from I = V/R) and add them.
The diagram shows three identical 12 Ω resistors connected between terminals X and Y. A connecting wire of negligible resistance joins junction M directly to junction N. What is the resistance between X and Y?
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Answer
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The wire from M to N has zero resistance, so there is no p.d. between M and N and no current in R2 or R3: they are short-circuited. The only resistance left is R1, so R = 12 Ω.
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Answer C
Answer: C
Common error: Ignoring a plain wire drawn across part of a network. Any components connected between the two ends of a zero-resistance wire carry no current.
A 9.0 V battery of negligible internal resistance is connected in series with three resistors. The graph shows how the potential varies with position around the circuit, from the positive terminal A to the negative terminal D. Resistor R2 has a resistance of 30 Ω. What is the resistance of R1?
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Answer
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Notes
P.d. across R2 = 7.5 − 3.0 = 4.5 V, so the current is 4.5 ÷ 30 = 0.15 A, the same in every resistor. P.d. across R1 = 9.0 − 7.5 = 1.5 V, so R1 = 1.5 ÷ 0.15 = 10 Ω.
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Answer A
Answer: A
Exam tip: On a potential–position graph, the p.d. across a component is the drop in potential across it, not the potential at one end of it.
The diagram shows three resistors R1, R2 and R3 connected in series with a 9.0 V supply of negligible internal resistance. The current is I and the p.d.s across the resistors are V1, V2 and V3.
(a)
Show that the total resistance R of the three resistors is given by R = R1 + R2 + R3. State the conservation law used at each step.
(3)
(b)
Calculate V2.
(2)
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Part (a)
the current I is the same in each resistor, because there are no junctions and charge is conserved
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the supply p.d. V = V1 + V2 + V3, because energy is conserved: the energy given to each coulomb equals the energy it transfers in the resistors
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IR = IR1 + IR2 + IR3, and dividing by I gives the result
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Part (b)
I = 9.0 ÷ 7.9 kΩ = 1.14 mA
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V2 = 1.14 × 10−3 × 2200 = 2.5 V (2.51 V)
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Exam tip: In a ‘show that’ derivation, write the physics in words before each equation. A line of algebra with no physical reason given is not a derivation.
A student measures the current in an NTC thermistor for p.d.s from 0 to 9.0 V. The graph shows her results, with points A and B marked on the curve.
(a)
Calculate the resistance of the thermistor at A and at B.
(2)
(b)
Explain why the resistance of the thermistor decreases as the p.d. across it increases.
(2)
(c)
The student reverses the connections to the thermistor and repeats the experiment. Describe the graph she obtains for negative p.d.s.
(1)
(d)
Suggest why the student waits for a short time after changing the p.d. before she reads the ammeter.
(1)
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Answer
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Notes
Part (a)
A: 3.0 ÷ 0.020 = 150 Ω
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B: 9.0 ÷ 0.180 = 50 Ω
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Part (b)
a larger current transfers more energy per second in the thermistor, so its temperature rises
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in the semiconductor, more charge carriers are released as the temperature rises, so the number of charge carriers per unit volume increases
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Part (c)
the same shape, rotated through 180° about the origin: at −9.0 V the current is −180 mA (the characteristic is symmetrical)
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Part (d)
the thermistor needs time to reach a steady temperature, so that the current stops changing
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Common error: Calculating resistance from the gradient of the curve. At any point, the resistance is V/I using that point's coordinates, with the current in amperes.
Three resistors are connected in a triangle, as shown in the diagram. Connections can be made to the corners A, B and C.
(a)
Calculate the resistance between A and B.
(2)
(b)
Calculate the resistance between A and C.
(1)
(c)
A 4.5 V supply of negligible internal resistance is connected between A and C. Calculate the current in the 6 Ω resistor and the p.d. across the 12 Ω resistor.
(2)
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Answer
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Part (a)
the 12 Ω and 18 Ω resistors are in series (30 Ω), and this path is in parallel with the 6 Ω resistor
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6 × 30 ÷ (6 + 30) = 5.0 Ω
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Part (b)
18 Ω in parallel with (6 + 12) = 18 Ω, so 18 ÷ 2 = 9.0 Ω
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Part (c)
the path A–B–C has the full 4.5 V across it: current = 4.5 ÷ 18 = 0.25 A
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p.d. across 12 Ω = 0.25 × 12 = 3.0 V
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Common error: Treating the triangle as three resistors in series or in parallel. Redraw it between the two terminals you are using: one resistor forms one path, the other two in series form the second path.
A student has a set of identical resistors, each of resistance R. She connects n of them in parallel and measures the combined resistance with an ohmmeter, for n = 1 to 5. The graph shows her results plotted against 1/n, with her line of best fit.
(a)
Write down an equation for the combined resistance RT of n identical resistors of resistance R in parallel.
(1)
(b)
Explain why the graph is a straight line through the origin.
(1)
(c)
Use the graph to determine R.
(2)
(d)
Calculate the combined resistance of eight of these resistors in parallel.
(1)
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Part (a)
RT = R/n (or 1/RT = n/R)
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Part (b)
RT = R × (1/n), so RT is directly proportional to 1/n, with gradient R
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Part (c)
gradient of the line of best fit, using a large triangle, e.g. 40 Ω ÷ 1.0
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R = 40 Ω (allow 39–41 Ω)
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Part (d)
40 ÷ 8 = 5.0 Ω (allow error carried forward)
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Exam tip: Take the gradient from the line of best fit, not from one data point, and use a triangle that covers at least half of the line.
A 9.0 V supply of negligible internal resistance is connected to an 8.0 Ω resistor in series with a parallel pair of resistors of 12 Ω and 24 Ω. What is the current in the 24 Ω resistor?
Figure: the circuit
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Answer
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Notes
Parallel pair: 12 × 24/(12 + 24) = 8.0 Ω, so the total is 16 Ω and the supply current is 9.0 ÷ 16 = 0.5625 A. P.d. across the pair = 0.5625 × 8.0 = 4.5 V, so the current in 24 Ω is 4.5 ÷ 24 = 0.19 A (0.1875 A).
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Answer A
Answer: A
Exam tip: Find the p.d. across the parallel pair first; each branch current is then that p.d. divided by the branch resistance.
A 12 V supply of negligible internal resistance is connected in series with an ideal ammeter, a 40 Ω resistor P and a 60 Ω resistor Q. An ideal voltmeter is connected across Q. A switch S, when closed, connects a third resistor R of 60 Ω in parallel with Q. S is initially open.
Figure: the circuit, with switch S open
(a)
Calculate the readings on the ammeter and on the voltmeter with S open.
(2)
(b)
S is closed. Calculate the new readings on the ammeter and on the voltmeter.
(3)
(c)
Explain why the voltmeter reading falls when S is closed, even though the current from the supply increases.
(1)
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Answer
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Part (a)
I = 12 ÷ (40 + 60) = 0.12 A
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V = 0.12 × 60 = 7.2 V
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Part (b)
Q and R in parallel = 30 Ω, so total resistance = 70 Ω
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I = 12 ÷ 70 = 0.17 A (0.171 A)
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V = 0.171 × 30 = 5.1 V (5.14 V) (allow error carried forward)
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Part (c)
the larger current gives a larger p.d. across P, and the p.d.s across P and across Q must still add up to 12 V, so the p.d. across Q is smaller
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Common error: Assuming the p.d. across Q stays the same when R is added. Adding R changes the total resistance, so every current and p.d. in the circuit changes.
A 12.0 V supply of negligible internal resistance is connected in series with an ideal ammeter and an 18 Ω resistor. The rest of the circuit is a 30 Ω resistor in parallel with a resistor X of unknown resistance. The ammeter reads 0.50 A.
Figure: the circuit
(a)
Calculate the p.d. across the 18 Ω resistor.
(1)
(b)
Determine the p.d. across the 30 Ω resistor, and name the conservation law you use.
(2)
(c)
Calculate the resistance of X.
(3)
(d)
Resistor X breaks, leaving a gap in its branch. Calculate the new ammeter reading.
(1)
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Answer
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Notes
Part (a)
0.50 × 18 = 9.0 V
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Part (b)
12.0 − 9.0 = 3.0 V
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conservation of energy: the p.d.s around the loop add up to the supply p.d.
Exam tip: When one resistance is unknown, work from what you can find directly: the series resistor's p.d. from the ammeter reading, then the parallel p.d. by subtraction.
The I–V characteristic of a small filament lamp is given by these readings: 1.0 V — 0.143 A; 2.0 V — 0.222 A; 3.0 V — 0.273 A; 4.0 V — 0.308 A; 5.0 V — 0.333 A; 6.0 V — 0.353 A. The lamp is connected in series with a 13 Ω resistor and an 8.0 V supply of negligible internal resistance.
Figure: the lamp in series with the 13 Ω resistor
(a)
Calculate the resistance of the lamp when the p.d. across it is 4.0 V.
(1)
(b)
Show that the p.d. across the lamp in this circuit is 4.0 V.
(3)
(c)
Calculate the power transferred in the lamp.
(1)
(d)
The supply is changed to 4.0 V. A student predicts that the p.d. across the lamp will now be 2.0 V. Explain why the actual p.d. across the lamp is less than 2.0 V.
(2)
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Answer
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Notes
Part (a)
4.0 ÷ 0.308 = 13 Ω (13.0 Ω)
✓ 1
Part (b)
the current is the same in the lamp and the resistor
✓ 1
the p.d.s must add up to 8.0 V: Vlamp + 13I = 8.0 V
✓ 1
at 4.0 V: 4.0 + 13 × 0.308 = 8.0 V, whereas 3.0 V gives 6.5 V and 5.0 V gives 9.3 V
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Part (c)
P = VI = 4.0 × 0.308 = 1.2 W
✓ 1
Part (d)
the current is smaller, so the filament is cooler and its resistance is less than 13 Ω (e.g. 9.0 Ω at 2.0 V)
✓ 1
the p.d. in a series circuit is shared in the ratio of the resistances, so the lamp now takes less than half of the 4.0 V
✓ 1
Exam tip: With a non-ohmic component in series, use the same current for both components and test values from the characteristic until the p.d.s add up to the supply p.d.
A light-emitting diode (LED) is to be lit from a 5.0 V supply of negligible internal resistance. When it is forward biased with a current of 20 mA, the p.d. across the LED is 2.0 V. A resistor is connected in series with the LED.
Figure: LED and series resistor connected to the supply (LED forward biased)
(a)
Calculate the p.d. across the series resistor when the LED current is 20 mA.
(1)
(b)
Calculate the resistance of the series resistor.
(1)
(c)
The required resistance is made by connecting a 220 Ω resistor in parallel with a second resistor. Calculate the resistance of the second resistor.
(2)
(d)
The LED is now connected the other way round. It does not break down. State the current in the circuit and the p.d. across the LED.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
5.0 − 2.0 = 3.0 V
✓ 1
Part (b)
3.0 ÷ 0.020 = 150 Ω
✓ 1
Part (c)
1/R = 1/150 − 1/220
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R = 470 Ω (471 Ω) (allow error carried forward)
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Part (d)
the current is (practically) zero, because the LED is reverse biased
✓ 1
so the p.d. across the resistor is zero and the p.d. across the LED is 5.0 V
✓ 1
Common error: Thinking that a reverse-biased diode has no p.d. across it because no current flows. With no current there is no p.d. across the resistor, so the whole supply p.d. is across the diode.
The graph shows the I–V characteristics of two filament lamps, P and Q. The lamps are connected in series with a 10.0 V supply of negligible internal resistance. What is the power transferred in lamp Q?
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Answer
Mark
Notes
The current is the same in both lamps and their p.d.s must add up to 10.0 V. At 0.25 A, P has 4.0 V and Q has 6.0 V: 4.0 + 6.0 = 10.0 V. Power in Q = 6.0 × 0.25 = 1.5 W.
✓ 1
Answer D
Answer: D
Exam tip: For non-ohmic components in series, slide along a horizontal line (one current) until the two p.d.s add up to the supply p.d. Do not assume the p.d. is shared equally.
In the circuit shown, the battery has negligible internal resistance and the voltmeter is ideal. What is the reading on the voltmeter?
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Answer
Mark
Notes
The right-hand branch is 20 + 40 = 60 Ω; in parallel with 30 Ω this gives 20 Ω, so the total is 30 Ω and the battery current is 6.0 ÷ 30 = 0.20 A. P.d. across the parallel section = 0.20 × 20 = 4.0 V. Current in the right-hand branch = 4.0 ÷ 60 = 0.067 A, so the p.d. across 40 Ω = 0.067 × 40 = 2.7 V (2.67 V).
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Answer B
Answer: B
Exam tip: Once you know the p.d. across a parallel section, each branch is a simple series problem: that p.d. is shared between the resistors in the branch in proportion to their resistances.
A student sets up a circuit to measure the I–V characteristic of a filament lamp. The 6.0 V battery has negligible internal resistance and is connected across the whole track LM of a variable resistor, which has a uniform track. The slider touches the track 0.40 of the way from L to M. The meters are ideal. The diagram shows how she has connected the lamp and the meters.
(a)
State the reading on the ammeter. Explain your answer.
(2)
(b)
Determine the reading on the voltmeter.
(2)
(c)
State how the student should change the connections of the meters.
(1)
(d)
After she corrects the circuit, with the slider in the same position, the voltmeter reads less than 2.4 V. Explain why.
(1)
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Answer
Mark
Notes
Part (a)
zero
✓ 1
the voltmeter, which has (practically) infinite resistance, is in series with the lamp, so there is no current in that part of the circuit
✓ 1
Part (b)
no current is drawn from the track, so the p.d. between L and the slider is 0.40 of the battery p.d.
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0.40 × 6.0 = 2.4 V
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Part (c)
swap the meters: ammeter in series with the lamp and voltmeter in parallel with the lamp
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Part (d)
the lamp now draws current and is in parallel with the section of track between L and the slider, so that section has a smaller combined resistance and takes a smaller share of the 6.0 V
✓ 1
Common error: Thinking that an ideal voltmeter in series reads zero. It stops the current, so there is no p.d. across anything else in that loop, and it reads the whole p.d. across the loop.
The graph shows the I–V characteristics of a filament lamp and of a 20 Ω resistor. The supplies used in this question have negligible internal resistance.
(a)
The lamp and the resistor are connected in parallel across a 6.0 V supply. Use the graph to determine the current from the supply.
(2)
(b)
Calculate the combined resistance of the lamp and the resistor in parallel.
(1)
(c)
The two graphs cross at 4.0 V. State what this shows about the lamp and the resistor at 4.0 V.
(1)
(d)
The lamp and the resistor are now connected in series with an 8.0 V supply. Show that the current is 0.20 A.
(2)
(e)
Explain why the resistance of the lamp in the parallel circuit is greater than its resistance in the series circuit.
Three lamps X, Y and Z are connected to a 9.0 V battery of negligible internal resistance, as shown in the diagram. Assume that the resistance of each lamp stays constant at 12 Ω.
(a)
Calculate the current in lamp X.
(2)
(b)
Calculate the power transferred in lamp X and in lamp Y.
(2)
(c)
The filament of lamp Y breaks. Explain what happens to the brightness of lamp X and of lamp Z. Support your answer with calculations.
(3)
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Answer
Mark
Notes
Part (a)
total resistance = 12 + 12/2 = 18 Ω
✓ 1
I = 9.0 ÷ 18 = 0.50 A
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Part (b)
X: P = I2R = 0.502 × 12 = 3.0 W
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Y carries 0.25 A: 0.252 × 12 = 0.75 W
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Part (c)
total resistance becomes 24 Ω, so the current in X and Z is 9.0 ÷ 24 = 0.375 A
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the p.d. across X falls from 6.0 V to 4.5 V (power 3.0 W → 1.7 W), so X becomes dimmer
✓ 1
the p.d. across Z rises from 3.0 V to 4.5 V (power 0.75 W → 1.7 W), so Z becomes brighter
✓ 1
Exam tip: When one branch of a circuit breaks, recalculate the total resistance first. Every current and p.d. in the circuit changes, not just those in the broken branch.
The graph shows the I–V characteristic of a component.
(a)
Identify the component. Give a reason for your answer.
(2)
(b)
Calculate the resistance of the component at P.
(1)
(c)
The component is connected in parallel with a 40 Ω resistor. A current of 50 mA enters the parallel combination. Show that the p.d. across the combination is 0.80 V.
(2)
(d)
The current entering the combination is doubled to 100 mA. Explain why the p.d. across the combination increases by less than 0.1 V.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
a (semiconductor) diode
✓ 1
there is no current for negative p.d.s, so it conducts in one direction only, and the current is negligible until about 0.6 V
✓ 1
Part (b)
0.80 ÷ 0.030 = 27 Ω (26.7 Ω)
✓ 1
Part (c)
the p.d. is the same across both, and the two currents must add up to 50 mA
✓ 1
at 0.80 V: resistor current = 0.80 ÷ 40 = 20 mA, component current = 30 mA from the graph, and 20 + 30 = 50 mA
✓ 1
Part (d)
the current in the component rises very steeply with p.d. near 0.8 V, so a very small increase in p.d. (about 0.06 V) lets it take almost all of the extra current
✓ 1
Exam tip: With a non-ohmic component in parallel, use the same p.d. for both branches and check values from the graph until the branch currents add up to the total.
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