Ohm's law, Kirchhoff's laws and reference directions
引入| English | 释义 | 拼音 |
|---|---|---|
| Ohm's law/əʊmz lɔː/ | 欧姆定律 | ōu mǔ dìng lǜ |
| resistor/rɪˈzɪstə/ | 电阻 | diàn zǔ |
| conductance | 电导 | diàn dǎo |
| Kirchhoff's current law/ˈkɜːkhɒfs ˈkʌrənt lɔː/ | 基尔霍夫电流定律 | jī ěr huò fū diàn liú dìng lǜ |
| node/nəʊd/ | 节点 | jié diǎn |
| Kirchhoff's voltage law/ˈkɜːkhɒfs ˈvəʊltɪdʒ lɔː/ | 基尔霍夫电压定律 | jī ěr huò fū diàn yā dìng lǜ |
| loop/luːp/ | 回路 | huí lù |
| series/ˈsɪəriːz/ | 串联 | chuàn lián |
| parallel/ˈpærəlel/ | 并联 | bìng lián |
| voltage divider/ˈvəʊltɪdʒ dɪˈvaɪdə/ | 分压器 | fēn yā qì |
| current divider/ˈkʌrənt dɪˈvaɪdə/ | 分流器 | fēn liú qì |
Which way does the current go?
- Two wires bring current into a junction and two carry it away. How much takes each exit?
- The answer needs Ohm's law and two laws named after Kirchhoff. Neither is new physics: one says charge does not pile up, the other says the energy books balance.
Ohm's law ties voltage to current
- A resistor 电阻 obeys Ohm's law 欧姆定律: $v = Ri$, with $R$ in ohms (Ω). Its conductance 电导 $G = \dfrac{1}{R}$, in siemens, is sometimes handier.
- A resistor always absorbs power, in three equivalent forms: $p = vi = i^{2}R = \dfrac{v^{2}}{R}$. Use whichever variable you already know.
6 V across 1.5 kΩ: $i = \dfrac{6}{1500} = 4\,\text{mA}$, and $p = 6 \times 0.004 = 24\,\text{mW}$. Check: $\dfrac{6^{2}}{1500} = 24\,\text{mW}$ too.
A 12 V supply is connected across a 2.4 kΩ resistor. What current flows, in mA?
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$i = \dfrac{v}{R} = \dfrac{12}{2400} = 0.005\,\text{A} = 5\,\text{mA}$.
Two laws do all the bookkeeping
- Kirchhoff's current law 基尔霍夫电流定律 (KCL): the signed currents leaving any node 节点 sum to zero. Charge moves; it is never used up at a junction.
- Kirchhoff's voltage law 基尔霍夫电压定律 (KVL): the signed voltages around any closed loop 回路 sum to zero. Climb a voltage in one direction and you descend it again on the way back.
At node A: 2 A and 0.5 A enter, 1 A leaves. Writing currents out as positive, $-2 - 0.5 + 1 + i_{4} = 0$, so $i_{4} = 1.5\,\text{A}$ leaves. Every ampere into A has somewhere to go.
At a node, 4 A and 1.5 A flow in, and 2 A flows out through a third branch. What does KCL give for the fourth branch?
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Signed sum out = 0 gives −4 − 1.5 + 2 + i₄ = 0, so i₄ = 3.5 A out. Nothing is consumed at the node.
Combine resistors, then divide the voltage
- In series 串联, resistances add and one current flows through all of them. In parallel 并联, $R_{1} \parallel R_{2} = \dfrac{R_{1}R_{2}}{R_{1}+R_{2}}$, which is smaller than either.
- A voltage divider 分压器 shares the source voltage in proportion: $v_{2} = v_{s} \cdot \dfrac{R_{2}}{R_{1}+R_{2}}$.
A 9 V source feeds 6 kΩ then 3 kΩ. Current $i = \dfrac{9\,\text{V}}{9\,\text{k}\Omega} = 1\,\text{mA}$, so $v_{2} = 1\,\text{mA} \times 3\,\text{k}\Omega = 3\,\text{V}$. Check by KVL: $-9 + 6 + 3 = 0$.
A 12 V source feeds 8 kΩ in series with 4 kΩ. What is the voltage across the 4 kΩ resistor?
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The divider gives 12 × 4/(8 + 4) = 4 V. The same current 12/12k = 1 mA flows through both, and 1 mA × 4 kΩ = 4 V.
Divide a current between parallel branches
- Two parallel resistors fed by a current share one voltage. The current divider 分流器 gives $i_{1} = i_{s} \cdot \dfrac{R_{2}}{R_{1}+R_{2}}$.
- Notice the swap: each branch takes the share set by the other resistor, so the smaller resistor carries the larger current.
2 A into 6 Ω in parallel with 3 Ω: the pair is $\dfrac{6 \times 3}{9} = 2\,\Omega$, so both see 4 V. Currents: $\dfrac{4}{6} \approx 0.67\,\text{A}$ and $\dfrac{4}{3} \approx 1.33\,\text{A}$, which add back to 2 A.
Match each arrangement to its rule.
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Series elements share one current, so voltages split by own resistance. Parallel elements share one voltage, so currents split by the other resistance.
A 3 A source feeds 4 Ω in parallel with 12 Ω. How much current flows through the 4 Ω resistor, in amperes?
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$i_{4} = 3 \times \dfrac{12}{4 + 12} = 2.25\,\text{A}$. The 12 Ω branch takes the other 0.75 A; the smaller resistor carries more.
Close the account with power
- The three power forms must agree for each resistor, and the power supplied must equal the power absorbed. That is a free check on your arithmetic.
- In the 9 V divider: the source gives $9 \times 1\,\text{mA} = 9\,\text{mW}$; the resistors take $6 + 3 = 9\,\text{mW}$.
A 9 V source with 1 Ω internal resistance feeds 6 Ω and 3 Ω in parallel. The loads combine to 2 Ω, so the source drives $\dfrac{9}{3} = 3\,\text{A}$ and the terminal voltage is 6 V. The powers close: $6 + 12 + 9\,(\text{internal}) = 27\,\text{W} = 9\,\text{V} \times 3\,\text{A}$.
A resistor carries 0.5 A and drops 8 V. Which set of power values does the three-way check give?
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R = 8/0.5 = 16 Ω, so vi = 4 W, i²R = 0.25 × 16 = 4 W and v²/R = 64/16 = 4 W. Agreement is the check.
Three common slips
- Adding current magnitudes at a node. Write "sum of currents leaving = 0" and let the signs do the work; a negative answer means the current flows against your arrow.
- Two blobs of wire treated as one node. They are one node only if a wire connects them.
- A parallel pair larger than its parts. The combination is always smaller than the smallest resistor.
100 Ω in parallel with 1 kΩ gives a combined resistance between 100 Ω and 1 kΩ.
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$\dfrac{100 \times 1000}{1100} \approx 90.9\,\Omega$, below the smaller resistor. Adding a parallel path can only lower the resistance.
Recap
- Ohm's law: $v = Ri$; power $p = vi = i^{2}R = \dfrac{v^{2}}{R}$.
- KCL: currents leaving a node sum to zero. KVL: voltages around a loop sum to zero.
- Dividers follow from these laws: voltage shares by own resistance, current by the other one.