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A-Level Physics · guided topic map

Electric circuits for Cambridge International AS & A Level Physics

Electric circuits for A-Level Physics, organized into 3 syllabus topics and 13 mapped concept guides.

Syllabus topics
3
Mapped concept guides
13
Educational level
Cambridge International AS & A Level

Choose the exact concept

Work in order or jump to the concept named in your specification, course outline, or assignment.

9

Electricity

AS Level foundations

7 guides
  1. 01Electric current and charge carriersMapped lesson
  2. 02Potential difference and energy per chargeMapped lesson
  3. 03Electrical powerMapped lesson
  4. 04Resistance, Ohm's law, and I-V graphsMapped lesson
  5. 05ResistivityMapped lesson
  6. 06Non-ohmic componentsMapped lesson
  7. 07Thermistors and light-dependent resistorsMapped lesson
10

D.C. circuits

AS Level foundations

5 guides
  1. 01Practical circuit symbols and diagramsMapped lesson
  2. 02E.m.f. and internal resistanceMapped lesson
  3. 03Kirchhoff's lawsMapped lesson
  4. 04Potential dividers and sensorsMapped lesson
  5. 05Potentiometer circuits and null methodsMapped lesson
21

Alternating currents

A Level extension

1 guide
  1. 01Alternating currentsMapped lesson

Diagrams

Electric circuits as A-Level Physics draws it

The figures from the A-Level Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 10.1Electricity · D.C. circuitsA Level
A battery with internal resistance in series with a variable resistor and an ammeterRAe.m.f. 12.0 VrbatteryI

Figure comment

Fig. 10.1A single series loop. Along the bottom, a cell with its long plate on the left and a small resistor labelled r sit together inside a dashed rectangle labelled battery, the cell marked e.m.f. 12.0 V. From the dashed box the wire runs to the left, up the left-hand side and along the top through a resistor labelled R that has an arrow drawn across it to show that it is variable, then down the right-hand side through a circle marked A, and back along the bottom into the box. An arrow on the top wire labelled I shows the direction of the conventional current.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

Guided questions 5 parts

Reading cue. r sits inside the dashed box with the cell, so read the box as the whole source, and read 12.0 V as labelling that source, not the p.d. across its terminals.

  1. aState The ammeter reads 2.0 A. State the current in the resistor r inside the dashed box, and give the reason from Fig. 10.1.

    recall2 marks

    Check answer 2 marks
    1. 2.0 A
    2. the components form a single series loop, so the current is the same at every point in it
  2. bCalculate With R set to 4.0 Ω the ammeter reads 2.0 A. Calculate the charge that passes through the ammeter in 5.0 minutes, and the energy delivered to R in that time.

    routine3 marks

    Check answer 3 marks
    1. Q = It = 2.0 × 300 = 6.0 × 10² C
    2. p.d. across R = 2.0 × 4.0 = 8.0 V
    3. energy = VQ = 8.0 × 600 = 4.8 × 10³ J
  3. cDetermine The internal resistance of the battery is 2.0 Ω. Determine the setting of R that makes the ammeter read 1.5 A.

    demanding3 marks

    Check answer 3 marks
    1. e.m.f. = I(R + r), so 12.0 = 1.5 × (R + 2.0)
    2. R + 2.0 = 8.0 Ω
    3. R = 6.0 Ω
  4. dDeduce A second identical battery is connected in series with the first in the loop of Fig. 10.1, with R left at 4.0 Ω. Deduce whether the ammeter reading doubles.

    top of the paper4 marks

    Check answer 4 marks
    1. total e.m.f. = 24.0 V
    2. total resistance = 4.0 + 2.0 + 2.0 = 8.0 Ω
    3. I = 24.0 / 8.0 = 3.0 A
    4. the reading rises from 2.0 A to 3.0 A, not to 4.0 A, because the internal resistance in the loop has doubled as well

Transfer challenge

A 12 V car battery of internal resistance 0.020 Ω delivers 150 A to a starter motor. Calculate the terminal potential difference of the battery while the motor is turning, and explain why the headlamps dim at that moment.

Check answer 3 marks
  1. lost volts = Ir = 150 × 0.020 = 3.0 V
  2. terminal p.d. = 12 − 3.0 = 9.0 V
  3. the lamps are connected across the terminals, so the p.d. across them falls and they are less bright