Current, potential difference and resistance
도입됨| English | 의미 |
|---|---|
| current/ˈkʌrənt/ | 전류 |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | 전위차 |
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.
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?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
A voltmeter measures the potential difference between two points, so it connects in parallel across the component. In series it would add its very large resistance to the circuit and the current would collapse.
Match each technical term to its precise meaning.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
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.
In the lab: 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?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- 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.
Your turn: A component has 12 V across it and carries 0.40 A. Find resistance.
One operating point: voltage, current and resistance
$I=\frac{V}{R};\quad P=VI$
At the worked operating point, 6 V across 20 ohms produces 0.30 A and 1.8 W.
A component has 12 V across it and carries 0.40 A. Find resistance.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
The result is 30 Ω. Known: V = 12 V; I = 0.40 A. Use R = V/I. R = 12/0.40 = 30 ohms. This describes the component at that operating point; a filament lamp need not keep this resistance at other voltages.
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.
Not true: A lamp uses up electric current as charge passes through it. Current is a flow of charge, and charge is conserved: the same current enters and leaves a lamp. The lamp transfers energy from the charges to heat and light; it does not consume the charges.
- Current is charge passing per time.
- Current is the same through components in series.
- Watch for: current is not used up by a lamp.
Write from a resistance of 36 ohms
- A component has 9 V across it and carries 0.25 A.
- Find the resistance and the power. A lamp does not use up current.
A lamp uses up electric current as charge passes through it.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Current is a flow of charge, and charge is conserved: the same current enters and leaves a lamp. The lamp transfers energy from the charges to heat and light; it does not consume the charges.
Rate of flow of charge: write the technical term.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
current means Rate of flow of charge.
A component has 9 V across it and carries 0.25 A. What are the resistance and the power, and why does a lamp not use up current?
Resistance is potential difference divided by current. Power is potential difference times current.
What is the resistance?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
9 divided by 0.25 is 36 ohms. Other options — 30 ohms (the lesson component at 12 V and 0.40 A); 2.25 W (the power).