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Physics · 02 / 12

Electricity: voltage, current, resistance

Middle school · Worked example and 4 exercises with solutions, at your pace.

Need a foundation?Units: numbers have dimensionsEquations: keep the balance level
UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

Current is a rate of charge flow; voltage is a potential difference. An ideal ohmic conductor follows U = RI. Series components share current; parallel branches share voltage. A resistor receives power UI. These are classroom model relationships.

U=R*I
THE PICTURE TO REMEMBERElectricity: voltage, current, resistance
  1. 01Voltage U
  2. 02Resistance R
  3. 03Current I
  4. 04Power UI

WORKED EXAMPLE

From principle to reasoning

12 V across 120 Ω gives I = 0.1 A and P = 1.2 W.

Solve a problem in this subject

  1. Draw the system

    Define the studied object, reference frame and positive axis. A force diagram shows external interactions acting on the system.

  2. Sort quantities

    Separate data and unknowns; convert units before substitution. 36 km/h = 10 m/s, so never insert 36 into a formula expecting SI speed.

  3. Write the model

    Choose a law matching assumptions: constant acceleration, thin lens, ohmic resistor. Isolate the unknown symbolically before rounding.

  4. Interpret

    Check dimensions and signs. Negative velocity gives direction on the chosen axis. Discuss scale and model limitations.

YOUR TURN

01 · Try before checking

A 50 Ω resistor has 5 V across it. Find I.

One hint

Rearrange U = RI.

Reveal and understand the solution

I = 5/50 = 0.1 A = 100 mA. Multiplying resistance and current recovers 5 V.

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

Two resistors of 10 and 20 Ω are in series. Find equivalent resistance.

A hint to get started

The same current flows through both.

Step-by-step solution
  1. Series voltage drops add.
  2. Req = R1 + R2 = 30 Ω.
  3. This assumes ideal resistors.

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

A 30 Ω resistor has 6 V across it. Find the current.

A hint to get started

I = U/R, with voltage in volts.

Step-by-step solution
  1. I = 6/30 = 0.2 A.
  2. This equals 200 mA.
  3. Current and voltage are different quantities.

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

Two 10 Ω resistors are in parallel. Find Req and compare.

A hint to get started

Add their reciprocals.

Step-by-step solution
  1. 1/Req = 1/10 + 1/10 = 1/5.
  2. Req = 5 Ω.
  3. An extra current path lowers resistance, unlike a series connection.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

Do not confuse A with mA or assume every component is ohmic.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

Choosing a state saves it in this browser. This is your assessment, not a grade.

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