Ohm's law, Kirchhoff's laws and reference directions
Introducido| English | Significado |
|---|---|
| Ohm's law/əʊmz lɔː/ | The rule that current is proportional to potential difference across an ohmic conductor when physical conditions such as temperature are constant. |
| resistor/rɪˈzɪstə/ | A component that opposes current, so the voltage across it is proportional to the current. |
| conductance | conductance |
| Kirchhoff's current law/ˈkɜːkhɒfs ˈkʌrənt lɔː/ | ley de corrientes de Kirchhoff |
| node/nəʊd/ | nodo |
| Kirchhoff's voltage law/ˈkɜːkhɒfs ˈvəʊltɪdʒ lɔː/ | ley de tensiones de Kirchhoff |
| loop/luːp/ | malla |
| series/ˈsɪəriːz/ | en serie |
| parallel/ˈpærəlel/ | en paralelo |
| voltage divider/ˈvəʊltɪdʒ dɪˈvaɪdə/ | divisor de tensión |
| current divider/ˈkʌrənt dɪˈvaɪdə/ | Resistors in parallel that share a current, with more current through the smaller resistance. |
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?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$i = \dfrac{v}{R} = \dfrac{12}{2400} = 0.005\,\text{A} = 5\,\text{mA}$.
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?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
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?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
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.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
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?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$i_{4} = 3 \times \dfrac{12}{4 + 12} = 2.25\,\text{A}$. 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?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
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.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
$\dfrac{100 \times 1000}{1100} \approx 90.9\,\Omega$, 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.