Circuit power, induction and charged-particle motion
| English | 中文 | Pinyin |
|---|---|---|
| flux linkage/flʌks ˈlɪŋkɪdʒ/ | 磁通链 | cí tōng liàn |
| root mean square | 均方根 | jūn fāng gēn |
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:
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:
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).
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.
Which answer fits this case?
Solve resistor networks and distinguish rms from peak AC quantities
An ideal single series diode can convert an AC sine wave into positive half-wave pulses, but not full-wave pulses.
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.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.