Circuits with Resistors and Inductors (LR Circuits)
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| inductor/ɪnˈdʌktə/ | Cuộn cảm |
Flip the switch and the current takes its time to rise
- Close a switch on a plain resistor and the current jumps up instantly.
- Add an inductor 电感器 and the current climbs gradually instead.
- The inductor's back-EMF fights the sudden change, slowing the rise.
- A resistor and inductor together make an LR circuit — the magnetic cousin of RC.
The current rises exponentially
- Close the switch and the current climbs: $I(t) = \dfrac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right)$.
- It starts at zero, rises fast, then eases toward its final value $\varepsilon/R$.
- The inductor's opposition is strongest at the start and fades as $I$ steadies.
- The shape is the same exponential curve as an RC charging capacitor.
Long after the switch closes, the LR current settles at $\varepsilon/$ ____.
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The final steady current is $\varepsilon/R$ (inductor acts like a wire).
The time constant τ = L/R
- The natural timescale here is $\tau = \dfrac{L}{R}$.
- In one $\tau$, the current reaches about $63\%$ of its final value.
- After about $5\tau$ it is essentially at $\varepsilon/R$.
- A bigger $L$ (more inertia) or smaller $R$ makes the rise slower.
The time constant of an LR circuit is:
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For an LR circuit $\tau = L/R$.
An LR circuit has $L = 10\ \text{H}$ and $R = 5\ \Omega$. Find $\tau$ (in s).
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$\tau = L/R = 10/5 = 2\ \text{s}$.
The inductor's two extremes
- At the first instant ($t = 0$): the inductor blocks the change, acting like an open gap (zero current).
- After a long time ($t \to \infty$): the current is steady, so the inductor acts like a plain wire.
- Notice this is the opposite of a capacitor's behaviour.
- Those two limits let you check any LR answer quickly.
An LR circuit switching on
An inductor resists sudden change. Sort each fact by the moment it describes.
At the first instant a switch closes, an inductor acts like:
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It blocks the sudden change, so at $t=0$ the current is zero (open gap).
An inductor and a capacitor behave the same way at the start and end of charging.
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False. They are reversed: an inductor acts like an open circuit at first and a plain wire later, while a capacitor acts like a wire at first and an open circuit later.
Switching off can spark
- Open the switch and the current tries to drop to zero at once.
- The inductor fights that huge $dI/dt$ with a large back-EMF.
- That surge can jump the switch gap as a spark.
- It is why circuits with big coils need protection when switched off.
Select all true statements about LR circuits.
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Exponential rise, τ = L/R, blocked at t = 0. The rise is gradual, not instant.
An LR circuit has $L = 6\ \text{H}$ and $R = 2\ \Omega$. Find its time constant.
- $\tau = \dfrac{L}{R} = \dfrac{6}{2} = 3\ \text{s}$.
- After $3\ \text{s}$ the current is about $63\%$ of $\varepsilon/R$.
An inductor's limits are the reverse of a capacitor's. At $t = 0$ an inductor acts like an open gap (blocks current), and at $t \to \infty$ like a plain wire. Swapping these — or confusing them with the capacitor's — is the classic LR mistake.
In an LR circuit, the current rises as $I = \tfrac{\varepsilon}{R}(1 - e^{-t/\tau})$ with time constant $\tau = L/R$ (about $63\%$ per $\tau$). At $t=0$ the inductor acts like an open gap, and at $t\to\infty$ like a wire — the reverse of a capacitor.
Write from a time constant of 5 seconds
- An LR circuit has L = 10 henry, R = 2 ohms and a battery EMF of 8 volts.
- Find the time constant and the final current.
An LR circuit has inductance 10 henry, resistance 2 ohms and EMF 8 volts. What are the time constant and the final current, and what is the current at the first instant?
10 divided by 2 is 5 seconds. 8 divided by 2 is 4 amperes. The inductor opposes the initial rise.
What are the time constant and the final current?
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L over R is 5 seconds and EMF over R is 4 amperes. The start is an open-gap current of 0.