Current, potential difference and resistance
| English | 中文 | Pinyin |
|---|---|---|
| current/ˈkʌrənt/ | 电流 | diàn liú |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | 电势差 | diàn shì chā |
What would explain this observation?
- A lamp becomes dimmer when another is added in series. Current 电流 and energy transfer depend on the whole circuit, not only one lamp.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Current is charge passing per time. Potential difference 电势差 is energy transferred per charge. Resistance is potential difference divided by current for a stated operating point.
- current: Rate of flow of charge; potential difference: Energy transferred per unit charge.
How is a voltmeter connected to measure voltage across a resistor?
Current is the same through components in series. Potential differences add around the series path. In parallel, branches share the same potential difference, while branch currents sum at a junction.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Current is the same through components in series. Potential differences add around the series path. In parallel, branches share the same potential difference, while branch currents sum at a junction.
- Place an ammeter in series and a voltmeter in parallel. For an I-V investigation, change voltage in steps, reverse polarity when appropriate and limit current to reduce heating.
Which two habits make the investigation or model in this case more defensible?
Place an ammeter in series and a voltmeter in parallel. For an I-V investigation, change voltage in steps, reverse polarity when appropriate and limit current to reduce heating.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: a resistor has 6.0 V across it and carries 0.30 A. Use R = V/I. R = 6.0/0.30 = 20 ohms. Power = VI. Power = 6.0×0.30 = 1.8 W. These quantities describe the same operating point.
A component has 12 V across it and carries 0.40 A. Find resistance. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A component has 12 V across it and carries 0.40 A. Find resistance.
The result is 30 Ω. Known: a resistor has 6.0 V across it and carries 0.30 A. Use R = V/I. R = 6.0/0.30 = 20 ohms. Power = VI. Power = 6.0×0.30 = 1.8 W. These quantities describe the same operating point.
Check the conclusion and its limits
- Current is not used up by a lamp. A filament heats up, so its resistance need not remain constant as voltage changes.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
A lamp uses up electric current as charge passes through it. This claim is false: Current is not used up by a lamp. A filament heats up, so its resistance need not remain constant as voltage changes.
Current, potential difference and resistance: Current is the same through components in series. Potential differences add around the series path. In parallel, branches share the same potential difference, while branch currents sum at a junction.
A lamp uses up electric current as charge passes through it.
Current is not used up by a lamp. A filament heats up, so its resistance need not remain constant as voltage changes.
Rate of flow of charge: write the technical term.
current means Rate of flow of charge.