Sinusoidal steady state and phasors
Introduzido| English | Significado |
|---|---|
| phasor/ˈfeɪzɔː/ | fasor |
| impedance/ɪmˈpiːdəns/ | impedância |
Zero a hundred times a second
- Mains voltage crosses zero a hundred times a second, yet a heater on it gives out steady warmth. Once the switch-on transient dies away, every voltage and current repeats the same sinusoid.
- At one frequency, a sinusoid is fixed by two numbers: its size and its timing. This lesson turns AC circuits into arithmetic on those two numbers.
The phasor: amplitude and phase
- $v(t) = V_{m}\cos(\omega t + \varphi)$ becomes the phasor 相量 $\mathbf{V} = V_{m}\angle\varphi$. The frequency $\omega$ is the same everywhere in the circuit, so it is left out.
- A phasor is a complex number: $V_{m}\angle\varphi = V_{m}\cos\varphi + jV_{m}\sin\varphi$. Use polar form to multiply and divide, rectangular form to add.
Write the phasor 10∠30° V in rectangular form. What is its real part, in volts?
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$10\cos 30^{\circ} = 8.66$ V; the imaginary part is $10\sin 30^{\circ} = 5$ V, so the phasor is $8.66 + j5$ V.
Calculus becomes multiplication
- Differentiating $\cos(\omega t + \varphi)$ gives a sinusoid $\omega$ times larger and 90° ahead. With phasors, $\dfrac{d}{dt}$ becomes multiplication by $j\omega$.
- So the inductor's $v = L\dfrac{di}{dt}$ becomes $\mathbf{V} = j\omega L\,\mathbf{I}$, and the capacitor's $i = C\dfrac{dv}{dt}$ becomes $\mathbf{I} = j\omega C\,\mathbf{V}$. The differential equations are gone.
Impedance: Ohm's law with complex numbers
- Impedance 阻抗 is $Z = \dfrac{\mathbf{V}}{\mathbf{I}}$: $Z_{R} = R$, $Z_{L} = j\omega L$, $Z_{C} = \dfrac{1}{j\omega C}$.
- The resistor is flat at 0°. The inductor sits at +90° and grows with frequency; the capacitor sits at −90° and shrinks. Dividers, nodal analysis and Thévenin all carry over unchanged.
Match each element to the angle of its impedance.
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$Z_{L} = j\omega L$ points straight up and lengthens as $\omega$ rises; $Z_{C} = \dfrac{1}{j\omega C} = -\dfrac{j}{\omega C}$ points straight down and shortens.
As frequency rises, the impedance of a capacitor
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$Z_{C} = \dfrac{1}{j\omega C}$: its size $\dfrac{1}{\omega C}$ falls with frequency at a fixed −90°. The inductor is the mirror image (+90°, grows).
What is the size of the impedance of a 100 µF capacitor at ω = 1000 rad/s, in ohms?
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$\dfrac{1}{\omega C} = \dfrac{1}{1000 \times 100 \times 10^{-6}} = 10\,\Omega$, at −90°.
Worked example: R and L in series
- $R = 30\,\Omega$ in series with $\omega L = 40\,\Omega$, driven by $100\angle 0^{\circ}\,\text{V}$. $Z = 30 + j40 = 50\angle 53.13^{\circ}\,\Omega$.
- $\mathbf{I} = \dfrac{100\angle 0^{\circ}}{50\angle 53.13^{\circ}} = 2\angle -53.13^{\circ}\,\text{A}$: divide the sizes, subtract the angles. So $i(t) = 2\cos(\omega t - 53.13^{\circ})$ A; the current lags.
Check in rectangular form: $\mathbf{I} = \dfrac{100(30 - j40)}{2500} = 1.2 - j1.6\,\text{A}$. Its size is $\sqrt{1.44 + 2.56} = 2$ and its angle is $-53.13^{\circ}$. The two forms agree.
A 30 Ω resistor in series with ωL = 40 Ω across 100∠0° V draws —
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$Z = 50\angle 53.13^{\circ}\,\Omega$; $\mathbf{I} = \dfrac{\mathbf{V}}{Z}$ divides the sizes and subtracts the angles, so the inductor makes the current lag. Rectangular check: $1.2 - j1.6$ A.
Worked example: R and L in parallel
- The same parts in parallel: $Z = \dfrac{30 \times j40}{30 + j40} = 19.2 + j14.4 = 24\angle 36.87^{\circ}\,\Omega$.
- The result is smaller than either branch (24 < 30 < 40), and its angle lies between the resistor's 0° and the inductor's 90°. Parallel always pulls both size and angle this way.
R = 30 Ω in parallel with j40 Ω. The equivalent impedance is
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Product over sum: $\dfrac{30 \times j40}{30 + j40} = 19.2 + j14.4 = 24\angle 36.87^{\circ}$. It is smaller than either branch, with its angle pulled towards the resistor.
Three common slips
- Adding sizes. $30\,\Omega$ and $j40\,\Omega$ in series make $50\,\Omega$, not $70\,\Omega$; only complex addition is correct.
- Calculator in the wrong mode. Degrees and radians mixed in one problem give nonsense angles.
- Mixing peak and RMS. Pick one for the whole problem, for both V and I.
A 30 Ω resistor in series with a j40 Ω inductor has an impedance of size 70 Ω.
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The two are at right angles, so the sizes add like the sides of a triangle: $\sqrt{30^{2} + 40^{2}} = 50\,\Omega$.
Recap
- At one frequency, a sinusoid is a phasor $V_{m}\angle\varphi$, and $\dfrac{d}{dt}$ becomes $j\omega$.
- $Z_{R} = R$, $Z_{L} = j\omega L$ (+90°), $Z_{C} = \dfrac{1}{j\omega C}$ (−90°); every DC method carries over.
- Check each answer in the other form, polar or rectangular.