回路電力、誘起および帯電粒子の運動
導入済み| English | 意味 |
|---|---|
| flux linkage/flʌks ˈlɪŋkɪdʒ/ | 磁束結合 |
| root mean square/ruːt miːn skweə/ | 二乗平均平方根 |
A decision before an answer
- A resistor in a parallel branch sees the branch voltage, which can differ from the battery voltage because another resistor is in series.
- Your goal: Solve resistor networks and distinguish rms from peak AC quantities.
Read the relationship
- For steady ideal resistor circuits, series resistances add and parallel conductances add. Reduce a network while preserving which nodes share the same voltage. With a 24 V source, a 2 Ω series resistor and parallel 3 Ω and 6 Ω branches, the parallel equivalent is 2 Ω. Total current is 6 A, branch voltage is 12 V and branch currents are 4 A and 2 A. The current through every component is not necessarily the total current; verify Kirchhoff current and voltage sums after reduction.
- Apply flux linkage, Lenz direction and the current-loop field.
In the worked 24 V resistor network, current through the 6 Ω branch is:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The parallel branches have 12 V, so the 6 Ω branch carries 12/6=2 A.
Use the defining rule
- For sinusoidal current I=Imax sin(ωt−φ), Irms=Imax/sqrt(2). A series RLC circuit has impedance magnitude sqrt(R²+(ωL−1/(ωC))²) and average dissipated power Irms²R=Vrms Irms cosφ. Ideal inductors/capacitors exchange stored energy with zero average dissipation. If rms current is already given, do not insert another factor 1/2. An ideal rectifying diode conducts in its forward direction and blocks in reverse: a suitably oriented series diode clips the negative half of a sinusoidal resistor voltage; it does not create a full-wave rectifier by itself.
- Use charge-to-mass ratios and balance crossed-field forces.
A series RLC circuit has Irms=3 A and R=4 Ω. Its average dissipated power is:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Irms²R=9·4=36 W; no extra half factor applies to rms current.
Check the conditions
- Magnetic flux is ∫B·dA, or BA cosθ for a uniform field with θ measured from the area normal. Faraday emf is −N dΦ/dt. For changing perpendicular field and fixed loop geometry, its average magnitude is NA|ΔB|/Δt. Determine the sign or current direction using opposition to the flux change, not opposition to the field itself. The magnetic field at the centre of one circular loop is μ0I/(2R), obtained from Biot–Savart: every element contributes in the same axial direction. An N-turn compact coil multiplies this result by N.
- Use charge-to-mass ratios and balance crossed-field forces.
In the 24 V network, 2 Ω in series with (3 Ω parallel 6 Ω) gives total 4 Ω, current 6 A and the 6 Ω branch current 2 A. A 100-turn coil of area 0.020 m² has perpendicular field rise by 0.030 T in 0.10 s: average emf magnitude is 0.60 V. A velocity selector with E=6000 V/m and B=0.20 T selects speed 30000 m/s.
At fixed B, an ion with proton charge and three times proton mass has cyclotron frequency ____ times the proton frequency (fraction).
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Angular frequency is |q|B/m.
Apply the task format
- For a nonrelativistic charge with velocity perpendicular to uniform B, magnetic force |q|vB provides centripetal force mv²/r. Thus r=mv/(|q|B) and cyclotron angular frequency is |q|B/m, independent of speed in this approximation. Charge sign sets rotation direction. A deuteron has approximately twice proton mass and the same charge, while an alpha particle has four times its mass and twice its charge; their frequencies are each half the proton value. In crossed E and B fields, undeflected motion requires electric and magnetic forces opposite and v=E/B. Check vector directions before using this magnitude.
- Use charge-to-mass ratios and balance crossed-field forces.
Do not apply battery voltage to every branch or halve an rms-current power again. Flux angle uses the area normal; a selector also needs opposing force directions.
このケースに当てはまる答えはどれか？
抵抗器回路を解き、ACの実効値と最大値を区別する
An ideal single series diode can convert an AC sine wave into positive half-wave pulses, but not full-wave pulses.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
It conducts in one polarity and blocks in the other.
Keep the distinctions
- root mean square 均方根 — Square root of the mean squared value, used for effective AC current or voltage.
- flux linkage 磁通链 — Sum of magnetic flux through a coil’s turns.
- Solve resistor networks and distinguish rms from peak AC quantities.
- Apply flux linkage, Lenz direction and the current-loop field.
- Use charge-to-mass ratios and balance crossed-field forces.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.