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IGCSE Physics 0625: Electric Circuits, Current, Voltage and Resistance - Study Guide PDF with Answers

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Readnary original study guide

Electric Circuits and Resistance

Understand current, potential difference and series or parallel circuits with three worked examples and 12 original questions.

Original AI-assisted Readnary material. Not independently academically reviewed; answers may contain errors. Check important results against your course materials.

Selected content from Physics 0625 sections 4.2 and 4.3 for 2026-2028 exams. This guide uses idealised simple resistor circuits; detailed component characteristics and laboratory wiring are not covered.

QuantityMeaningUnit
Currentcharge passing per secondampere (A)
Potential differenceenergy transferred per unit chargevolt (V)
Resistancepotential difference / currentohm (ohm)

Understand the topic

Separate charge flow from energy transfer

Electric current is the rate of charge flow: I = Q/t, so 1 A means 1 C passes a point each second. Conventional current is drawn from the positive terminal around the external circuit to the negative terminal, although electrons in a metal move the other way. Potential difference is energy transferred per unit charge: V = E/Q. A 6 V supply transfers 6 J to each coulomb of charge across its terminals in an idealised account. Charge is not used up in a lamp.

Measure the right thing

An ammeter measures current through a branch and is placed in series with it. A voltmeter measures potential difference across a component and is placed in parallel with that component. Do not place a standard ammeter directly across a cell. For a fixed resistor at constant temperature, V = IR links potential difference, current and resistance. The ratio V/I need not stay constant for every component: a filament lamp changes temperature and resistance as it warms.

Series circuits have one path

In a series loop, current is the same at every point. The supply potential difference is shared among components, so component voltages add to the supply voltage. For resistors in series, total resistance is the sum of their resistances. Increasing series resistance with the same ideal supply generally reduces the loop current. A broken component opens the only path, so the whole series circuit stops carrying current.

Parallel circuits have branches

In a parallel circuit, each branch across the same two supply nodes has the same potential difference. The current entering a junction equals the sum of currents leaving it. Adding a branch can increase the total current drawn from an ideal fixed-voltage supply, while the current through an unchanged existing branch stays the same in the ideal model. A break in one branch need not stop current in other intact branches. State any ideal-supply assumption when applying these rules.

Worked examples

1. Current from charge

A charge of 24 C passes a point in 8 s. Find current.

  1. Current means charge per unit time: I = Q/t.
  2. I = 24/8 = 3 A.
  3. This says 3 C passes that point each second on average.

2. A fixed resistor

A fixed resistor has 12 V across it and carries 0.5 A. Find its resistance.

  1. Use R = V/I for these simultaneous readings.
  2. R = 12/0.5 = 24 ohm.
  3. The conclusion that this ratio stays 24 ohm assumes the resistor's conditions, especially temperature, remain suitable.

3. A junction and branches

A supply provides 2.0 A to two parallel branches. One branch carries 0.7 A. Find current in the other.

  1. At the junction, incoming current equals total outgoing current.
  2. I-other = 2.0 - 0.7 = 1.3 A.
  3. Both branches have the same potential difference if they connect across the same two nodes.

12 practice questions with explained answers

Try each question before opening its answer. Numerical answers should include your working.

1. 18 C passes a point in 6 s. Find current.

Show answer to question 1

I = Q/t = 18/6 = 3 A.

2. A current of 0.4 A flows for 20 s. Find charge.

Show answer to question 2

Q = I x t = 0.4 x 20 = 8 C.

3. A component transfers 30 J to 5 C passing through it. Find potential difference.

Show answer to question 3

V = E/Q = 30/5 = 6 V.

4. Where should an ammeter be connected to measure current through a lamp?

Show answer to question 4

In series with the lamp, so the same branch current passes through the meter.

5. Where should a voltmeter be connected to measure potential difference across a lamp?

Show answer to question 5

In parallel across the lamp's two terminals.

6. A resistor carries 2 A with 10 V across it. Find resistance.

Show answer to question 6

R = V/I = 10/2 = 5 ohm.

7. A 20 ohm fixed resistor has 6 V across it. Find current.

Show answer to question 7

I = V/R = 6/20 = 0.3 A.

8. Two series resistors are 3 ohm and 7 ohm. Find their total resistance.

Show answer to question 8

R-total = 3 + 7 = 10 ohm, assuming the resistors behave as stated.

9. A 9 V ideal supply feeds two series components. One has 4 V across it. Find the other voltage.

Show answer to question 9

Their potential differences add to 9 V, so the other is 9 - 4 = 5 V.

10. A 1.5 A current reaches a junction and divides into 0.6 A and a second branch. Find the second current.

Show answer to question 10

I-second = 1.5 - 0.6 = 0.9 A by conservation of charge.

11. Two components are connected in parallel directly across an ideal 6 V supply. What is the p.d. across each?

Show answer to question 11

6 V across each branch because each connects to the same two supply nodes.

12. Why might V/I for a filament lamp change as its p.d. increases?

Show answer to question 12

Its filament heats, changing its resistance; do not treat it automatically as a fixed resistor at constant temperature.

Revision checklist

  • Use I = Q/t, V = E/Q and R = V/I with appropriate units.
  • Place ammeters in series and voltmeters across components.
  • Add resistance and component p.d.s in a simple series circuit.
  • Apply equal branch p.d. and junction current sums in parallel circuits.

Common mistakes

Current is conserved at a junction; it is not consumed by a lamp. Equal voltage applies to parallel branches across the same nodes, not to arbitrary series components. The V/I ratio is not always constant for a warming filament.

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Source and review status

Cambridge IGCSE Physics 0625 syllabus, 2026-2028 (Version 2). Reference for topic scope only. No official exam questions or syllabus prose reproduced.

Edition: 2026-09-22. Original AI-assisted Readnary material. Not independently academically reviewed; answers may contain errors. Check important results against your course materials. Independent of Cambridge; not an official publication or a complete syllabus.

IGCSE Physics 0625: Electric Circuits, Current, Voltage and Resistance - Study Guide PDF with Answers

Physics · IGCSE · STUDY GUIDE

Understand current, potential difference and series or parallel circuits with three worked examples and 12 original questions.

Read the original Readnary guide in PDF view or use the web lesson above. Try the questions before revealing their explained answers.