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Electric Circuits · ⁨Rangkaian Listrik⁩

AP Physics C: Electricity and Magnetism · ⁨AP Fisika C: Listrik dan Magnetisme⁩ · Topic 11 · ⁨Topik 11⁩

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8:44

Rangkaian Listrik

Anda menghidupkan sakelar, dan lampu menyala seketika. Jadi elektron harus berlari turun di kawat — benar? Tidak sama sekali. Di dalam kawat tembaga itu, pembawa muatan merayap…

English narration · English + 中文 subtitles burned in · ⁨Narasi bahasa Inggris · Subtitle bahasa Inggris + 中文 disematkan langsung⁩

11.1

Electric Current · ⁨Arus Listrik⁩

Syllabus · ⁨Silabus⁩
Learning ObjectiveEssential Knowledge

11.1.A
Describe the movement of electric charges through a medium.

  • 11.1.A.1 Current is the rate at which charge passes through a cross-sectional area of a wire.
    • Equation: $I = \dfrac{dq}{dt}$
    • 11.1.A.1.i Current within a conductor consists of charge carriers traveling through the conductor with an average drift velocity.
      • Equation: $I = nqv_d A$
    • 11.1.A.1.ii Electric charge moves in a circuit in response to an electric potential difference, sometimes referred to as electromotive force, or $\mathrm{emf}$ ($\mathcal{E}$).
    • 11.1.A.1.iii If the current is zero in a section of wire, the net motion of charge carriers in the wire is also zero, although individual charge carriers will not have zero speed.
  • 11.1.A.2 Current density is the flow of charge per unit area.
    • Equation: $I = \int \vec{J} \cdot d\vec{A}$
    • 11.1.A.2.i Current density is related to the motion of the charge carriers within a conductor.
      • Equation: $\vec{J} = nq\vec{v}_d$
    • 11.1.A.2.ii Current density is a vector quantity.
    • 11.1.A.2.iii A potential difference across a conductor creates an electric field within the conductor that is proportional to the resistivity of the conductor and the current density.
      • Equation: $\vec{E} = \rho\vec{J}$
  • 11.1.A.3 If a function of current density is given, the total current can be determined by integrating the current density over the area.
    • Equation: $I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}$
  • 11.1.A.4 Although current is a scalar quantity, it does have a direction. Because its direction is relative to the current carrier and not space, current does not obey the laws of vector addition and has no vector components.
    • 11.1.A.4.i The direction of conventional current is chosen to be the direction in which positive charge would move.
    • 11.1.A.4.ii In common circuits, the current is actually due to the movement of electrons (negative charge carriers).

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Electric current 电流 is the rate at which charge passes a cross-section of wire, $I=\dfrac{dq}{dt}$, measured in amperes 安培. Conventional current points the way positive charge would move. Microscopically, a current is a slow drift of many carriers:

$$I=nqv_dA,$$

with $n$ the number of carriers per volume, $q$ the charge each carries, $v_d$ the drift velocity 漂移速度, and $A$ the cross-sectional area 横截面积.

Worked example. A copper wire with $A=1.0\times10^{-6}\ \text{m}^2$ and $n=8.5\times10^{28}\ \text{m}^{-3}$ carries $1.7\ \text{A}$. Then $v_d=\dfrac{I}{nqA}=\dfrac{1.7}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})}\approx1.3\times10^{-4}\ \text{m/s}$ – the carriers drift slower than a snail, even though the signal travels near light speed.

Current density 电流密度 is charge flow per unit area, $\vec{J}=nq\vec{v}_d$, linked to the field driving it by $\vec{E}=\rho\vec{J}$. In general $I=\int\vec{J}\cdot d\vec{A}$; if $J(r)$ varies across the wire, integrate it over the cross-section to get the total current. One care point: current has a direction along its wire, but it is a scalar 标量 – currents do not add as vectors, and there are no "components of current".

Bahasa Indonesia

Arus listrik adalah laju muatan yang melewati penampang kawat, $I=\dfrac{dq}{dt}$, diukur dalam amperes. Arus konvensional menunjuk arah positif muatan akan bergerak. Secara mikroskopis, arus adalah aliran lambat banyak pembawa:

$$I=nqv_dA,$$

dengan $n$ jumlah pembawa per volume, $q$ muatan yang dibawanya, $v_d$ kecepatan hanyut, dan $A$ luas penampang.

Pembawa muatan mengalir lambat melalui konduktor untuk menghasilkan arus
Pembawa muatan melambat merayap melalui konduktor untuk menghasilkan arus

Contoh terpecahkan. Kawat tembaga dengan $A=1.0\times10^{-6}\ \text{m}^2$ dan $n=8.5\times10^{28}\ \text{m}^{-3}$ membawa $1.7\ \text{A}$. Maka $v_d=\dfrac{I}{nqA}=\dfrac{1.7}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})}\approx1.3\times10^{-4}\ \text{m/s}$ – pembawa bergerak lebih lambat dari siput, meskipun sinyal berjalan mendekati kecepatan cahaya.

Kepadatan arus adalah aliran muatan per satuan area, $\vec{J}=nq\vec{v}_d$, terkait dengan medan yang menggerakkannya melalui $\vec{E}=\rho\vec{J}$. Secara umum $I=\int\vec{J}\cdot d\vec{A}$; jika $J(r)$ bervariasi di seluruh kawat, integralkan di atas penampang untuk mendapatkan arus total. Satu catatan: arus memiliki arah sepanjang kawatnya, tetapi ia adalah skalar – arus tidak dijumlahkan sebagai vektor, dan tidak ada "komponen arus".

Osiloskop: tegangan terhadap waktu menunjukkan bagaimana arus dan muatan berevolusi dalam sirkuit
Osiloskop: tegangan terhadap waktu mengungkap bagaimana arus dan muatan berevolusi dalam sebuah sirkuit
Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩
English Bahasa Indonesia
Electric current/ɪˈlektrɪk ˈkʌrənt/ Arus listrik
amperes/ˈæmpeəz/ ampere
drift velocity/drɪft vəˈlɒsɪti/ kecepatan rambat
cross-sectional area/krɒs ˈsekʃənl ˈeərɪə/ luas penampang
Current density/ˈkʌrənt ˈdensɪti/ Kepadatan arus
scalar/ˈskeɪlə/ skalar
circuit diagram/ˈsɜːkɪt ˈdaɪəɡræm/ Diagram rangkaian
series/ˈsɪəriːz/ seri
11.2

Electric Circuits · ⁨Sirkuit Listrik⁩

Syllabus · ⁨Silabus⁩
Learning ObjectiveEssential Knowledge

11.2.A
Describe the behavior of a circuit.

  • 11.2.A.1 A circuit is composed of electrical loops, which can include wires, batteries, resistors, lightbulbs, capacitors, inductors, switches, ammeters, and voltmeters.
  • 11.2.A.2 A closed electrical loop is a closed path through which charges may flow.
    • 11.2.A.2.i A closed circuit is one in which charges would be able to flow.
    • 11.2.A.2.ii An open circuit is one in which charges would not be able to flow.
    • 11.2.A.2.iii A short circuit is one in which charges would be able to flow with no change in potential difference.
  • 11.2.A.3 A single circuit element may be part of multiple electrical loops.
  • 11.2.A.4 Circuit schematics are representations used to describe and analyze electric circuits.
    • 11.2.A.4.i The properties of an electric circuit are dependent on the physical arrangement of its constituent elements.
    • 11.2.A.4.ii Circuit elements have common symbols that are used to create schematic diagrams. Variable elements are indicated by a diagonal strikethrough arrow across the standard symbol for that element. (Symbols: Battery, Bulb, Switch, Capacitor, Resistor, Ammeter, Voltmeter, Inductor.)

Boundary statement: Unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

A circuit is a set of closed loops built from wires, batteries, resistors, lightbulbs, capacitors, inductors, switches, and meters; charge can flow only around a closed path. One element can belong to several loops at once – that is what makes multi-loop problems interesting. Every analysis starts by reading the circuit diagram 电路图: trace each loop and identify which elements share the same current (series 串联) and which share the same potential difference (parallel 并联).

Bahasa Indonesia

Sebuah sirkuit adalah sekumpulan loop tertutup yang dibangun dari kawat, baterai, resistor, bohlam, kapasitor, induktor, sakelar, dan meter; muatan hanya dapat mengalir mengelilingi jalur tertutup. Satu elemen dapat menjadi bagian dari beberapa loop sekaligus – itulah yang membuat masalah multi-loop menarik. Setiap analisis dimulai dengan membaca diagram sirkuit: telusuri setiap loop dan identifikasi elemen mana yang memiliki arus yang sama (seri) dan mana yang memiliki beda potensial yang sama (paralel).

Explore · ⁨Jelajahi⁩

Build series and parallel circuits · ⁨Bangun rangkaian seri dan paralel⁩

In series the same current flows through every bulb and voltage divides; in parallel each branch gets the full voltage. Switch mode to see the bulbs' brightness change. · ⁨Dalam seri, arus yang sama mengalir melalui setiap bola lampu dan tegangan terbagi; dalam paralel, setiap cabang mendapat tegangan penuh. Ubah mode untuk melihat perubahan kecerahan bola lampu.⁩

11.3

Resistance, Resistivity, and Ohm's Law · ⁨Resistansi, Resistivitas, dan Hukum Ohm⁩

Syllabus · ⁨Silabus⁩
Learning ObjectiveEssential Knowledge

11.3.A
Describe the resistance of an object using physical properties of that object.

  • 11.3.A.1 Resistance is a measure of the degree to which an object opposes the movement of electric charge.
  • 11.3.A.2 The resistance of a resistor with uniform geometry is proportional to its resistivity and length and is inversely proportional to its cross-sectional area.
    • Equation: $R = \dfrac{\rho\ell}{A}$
    • 11.3.A.2.i Resistivity is a fundamental property of a material that depends on its atomic and molecular structure and quantifies how strongly the material opposes the motion of electric charge.
    • 11.3.A.2.ii The resistivity of a conductor typically increases with temperature.
    • 11.3.A.2.iii The total resistance of a resistor with uniform geometry, but that is made of a material whose resistivity varies along the length of the resistor, is given by $R = \int \dfrac{\rho(\ell)\,d\ell}{A}$.

11.3.B
Describe the electrical characteristics of elements of a circuit.

  • 11.3.B.1 Ohm's law relates current, resistance, and potential difference across a conductive element of a circuit.
    • Equation: $I = \dfrac{\Delta V}{R}$
    • 11.3.B.1.i Materials that obey Ohm's law have constant resistance for all currents and are called ohmic materials.
    • 11.3.B.1.ii The resistivity of an ohmic material is constant regardless of temperature.
    • 11.3.B.1.iii Resistors can also convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment.
    • 11.3.B.1.iv The resistance of an ohmic circuit element can be determined from the slope of a graph of the current in the element as a function of the potential difference across the element.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Resistance 电阻 measures how strongly an object opposes charge flow. It grows with the material's resistivity 电阻率 and the conductor's length, and shrinks with its area:

$$R=\frac{\rho\,\ell}{A}.$$

Ohm's law 欧姆定律 relates the current through an element to the potential difference across it:

$$I=\frac{\Delta V}{R}.$$

Worked example. Stretch a wire to double its length: the volume is fixed, so the area halves, and $R=\rho\ell/A$ becomes $\rho(2\ell)/(A/2)=4R$ – four times the resistance. An element is ohmic 欧姆性 if $R$ stays constant (a straight line through the origin on an $I$–$\Delta V$ graph); a lightbulb filament, which heats up, is not.

Bahasa Indonesia

Resistansi mengukur seberapa kuat suatu objek melawan aliran muatan. Resistansi meningkat seiring dengan resistivitas material dan panjang konduktor, serta menurun seiring dengan luas permukaannya:

$$R=\frac{\rho\,\ell}{A}.$$

Hukum Ohm menghubungkan arus melalui suatu elemen dengan beda potensial di atasnya:

$$I=\frac{\Delta V}{R}.$$
Konduktor yang lebih panjang memiliki resistansi lebih besar; yang lebih lebar memiliki lebih sedikit
Konduktor yang lebih panjang memiliki resistansi lebih besar; yang lebih lebar memiliki lebih sedikit

Contoh terpecahkan. Regangkan kawat hingga gandakan panjangnya: volumenya tetap, sehingga luasnya berkurang separuh, dan $R=\rho\ell/A$ menjadi $\rho(2\ell)/(A/2)=4R$ – empat kali resistansi. Suatu elemen disebut ohmik jika $R$ tetap konstan (garis lurus melalui titik asal pada grafik $I$–$\Delta V$); filamen bohlam, yang memanas, tidak.

Explore · ⁨Jelajahi⁩

Apply Ohm's law · ⁨Terapkan Hukum Ohm⁩

Ohm's law $V=IR$: for a fixed resistance, current is proportional to voltage. Raise the resistance and the same voltage pushes less current. · ⁨Hukum Ohm $V=IR$: untuk hambatan tetap, arus berbanding lurus dengan tegangan. Naikkan hambatan dan tegangan yang sama mendorong arus lebih sedikit.⁩

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩
English Bahasa Indonesia
resistivity/ˌriːzɪˈstɪvəti/ resistivitas
Ohm's law/əʊmz lɔː/ Hukum Ohm
ohmic/ˈəʊmɪk/ Ohmik
equivalent resistance/ɪˈkwɪvələnt rɪˈzɪstəns/ resistansi ekuivalen
ideal battery/aɪˈdɪəl ˈbætəri/ Baterai ideal
internal resistance/ɪnˈtɜːnl rɪˈzɪstəns/ resistansi dalam
terminal voltage/ˈtɜːmɪnl ˈvəʊltɪdʒ/ Tegangan terminal
11.4

Electric Power · ⁨Daya Listrik⁩

Syllabus · ⁨Silabus⁩
English

Learning Objective 11.4.A: Describe the transfer of energy into, out of, or within an electric circuit, in terms of power.

  • 11.4.A.1 The rate at which energy is transferred, converted, or dissipated by a circuit element depends on the current in the element and the electric potential difference across it.
    • Equation: $P = I\Delta V$
    • Equation: $P = I^2 R = \dfrac{\Delta V^2}{R}$
  • 11.4.A.2 The brightness of a lightbulb increases with power, so power can be used to qualitatively predict the brightness of lightbulbs in a circuit.

Boundary statement: AP Physics C: Electricity & Magnetism only expects students to analyze the transfer of mechanical and electrical energy, although students should be aware that electrical energy can also be dissipated in the form of thermal energy.

Bahasa Indonesia

Tujuan Pembelajaran 11.4.A: Jelaskan transfer energi masuk, keluar, atau di dalam sebuah sirkuit listrik, berdasarkan daya.

  • 11.4.A.1 Laju transfer, konversi, atau pelepasan energi oleh elemen sirkuit bergantung pada arus dalam elemen tersebut dan beda potensial listrik melintasinya.
    • Persamaan: $P = I\Delta V$
    • Persamaan: $P = I^2 R = \dfrac{\Delta V^2}{R}$
  • 11.4.A.2 Kehebatan mentol meningkat dengan kuasa, maka kuasa boleh digunakan untuk meramal kehebatan mentol dalam litar secara kualitatif.

Pernyataan sempadan: AP Physics C: Elektrik & Magnetisme hanya mengharapkan pelajar menganalisis pemindahan tenaga mekanikal dan elektrik, walaupun pelajar harus menyedari bahawa tenaga elektrik juga boleh terlarap dalam bentuk tenaga haba.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

A charge $q$ falling through a potential difference $\Delta V$ gives up energy $q\Delta V$, so the rate of energy transfer in an element is

$$P=I\,\Delta V=I^2R=\frac{(\Delta V)^2}{R}.$$

In a resistor all of it becomes heat. Use the form whose quantities you actually know – and use power to rank lightbulb brightness: brighter = more power, not necessarily more resistance. In series, the larger resistance is brighter ($P=I^2R$, same $I$); in parallel, the smaller one is ($P=\Delta V^2/R$, same $\Delta V$).

Bahasa Indonesia

Muatan $q$ yang jatuh melewati beda potensial $\Delta V$ melepaskan energi $q\Delta V$, sehingga laju transfer energi dalam suatu elemen adalah

$$P=I\,\Delta V=I^2R=\frac{(\Delta V)^2}{R}.$$

Dalam resistor semuanya berubah menjadi panas. Gunakan bentuk yang besaran-besarannya Anda ketahui sebenarnya – dan gunakan daya untuk peringkat kecerahan bohlam: lebih terang = lebih banyak daya, tidak selalu resistansi lebih besar. Dalam seri, resistansi yang lebih besar lebih terang ($P=I^2R$, arus $I$ sama); dalam paralel, yang lebih kecil yang lebih terang ($P=\Delta V^2/R$, beda potensial $\Delta V$ sama).

Explore · ⁨Jelajahi⁩

Read an I-V characteristic · ⁨Baca karakteristik I-V⁩

Power is $P=IV$. A resistor's I-V line is straight, but a lamp curves as it heats and its resistance rises. The area under I-V relates to the energy delivered. · ⁨Daya adalah $P=IV$. Garis I-V resistor lurus, tetapi lampu melengkung saat memanas dan hambatannya naik. Area di bawah I-V berkaitan dengan energi yang disalurkan.⁩

11.5

Compound Direct Current Circuits · ⁨Sirkuit Arus Langsung Majemuk⁩

Syllabus · ⁨Silabus⁩
Learning ObjectiveEssential Knowledge

11.5.A
Describe the equivalent resistance of multiple resistors connected in a circuit.

  • 11.5.A.1 Circuit elements may be connected in series and/or in parallel.
    • 11.5.A.1.i A series connection is one in which any charge passing through one circuit element must proceed through all elements in that connection and has no other path available. The current in each element in series must be the same.
    • 11.5.A.1.ii A parallel connection is one in which charges may pass through one of two or more paths. Across each path, the potential difference is the same.
  • 11.5.A.2 A collection of resistors in a circuit may be analyzed as though it were a single resistor with an equivalent resistance $R_{\text{eq}}$.
    • 11.5.A.2.i The equivalent resistance of a set of resistors in series is the sum of the individual resistances.
      • Equation: $R_{\text{eq},s} = \sum_{i} R_i$
    • 11.5.A.2.ii The inverse of the equivalent resistance of a set of resistors connected in parallel is equal to the sum of the inverses of the individual resistances.
      • Equation: $\dfrac{1}{R_{\text{eq},p}} = \sum_{i} \dfrac{1}{R_i}$
    • 11.5.A.2.iii When resistors are connected in parallel, the number of paths available to charges increases, and the equivalent resistance of the group of resistors decreases.

11.5.B
Describe a circuit with resistive wires and a battery with internal resistance.

  • 11.5.B.1 Ideal batteries have negligible internal resistance. Ideal wires have negligible resistance.
    • 11.5.B.1.i The resistance of wires that are good conductors may normally be neglected, because their resistance is much smaller than that of other elements of a circuit.
    • 11.5.B.1.ii The resistance of wires may only be neglected if the circuit contains other elements that do have resistance.
    • 11.5.B.1.iii The potential difference a battery would supply if it were ideal is the potential difference measured across the terminals when there is no current in the battery and is sometimes referred to as its $\mathrm{emf}$ ($\mathcal{E}$).
  • 11.5.B.2 The internal resistance of a nonideal battery may be treated as the resistance of a resistor in series with an ideal battery and the remainder of the circuit.
  • 11.5.B.3 When there is current in a nonideal battery with internal resistance $r$, the potential difference across the terminals of the battery is reduced relative to the potential difference when there is no current in the battery.
    • Equation: $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$

11.5.C
Describe the measurement of current and potential difference in a circuit.

  • 11.5.C.1 Ammeters are used to measure current at a specific point in a circuit.
    • 11.5.C.1.i Ammeters must be connected in series with the element in which current is being measured.
    • 11.5.C.1.ii Ideal ammeters have zero resistance so that they do not affect the current in the element that they are in series with.
  • 11.5.C.2 Voltmeters are used to measure electric potential difference between two points in a circuit.
    • 11.5.C.2.i Voltmeters must be connected in parallel with the element across which potential difference is being measured.
    • 11.5.C.2.ii Ideal voltmeters have infinite resistance so that no charge flows through them.
  • 11.5.C.3 Nonideal ammeters and voltmeters will change the properties of the circuit being measured.

Boundary statement: Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Circuits with batteries of different potential differences connected in parallel will not be assessed.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Reduce resistor networks to an equivalent resistance 等效电阻: series resistances add ($R_{\text{eq}}=R_1+R_2+\cdots$), while parallel resistances add as reciprocals ($\tfrac{1}{R_{\text{eq}}}=\tfrac{1}{R_1}+\tfrac{1}{R_2}+\cdots$ – always less than the smallest branch). Collapse the network step by step to find the battery current, then expand back out to find each element's current and voltage.

Worked example. A $12\ \text{V}$ battery drives a $4.0\ \Omega$ and a $2.0\ \Omega$ resistor in series: $R_{\text{eq}}=6.0\ \Omega$, $I=2.0\ \text{A}$, the voltages split $8.0\ \text{V}$ and $4.0\ \text{V}$, and the $4.0\ \Omega$ resistor dissipates $P=I^2R=16\ \text{W}$.

Real batteries are not ideal. Model a battery as an ideal battery 理想电池 of emf $\varepsilon$ in series with its own internal resistance 内阻 $r$. When current flows, some emf is used up inside, so the terminal voltage 端电压 – what a voltmeter across the battery actually reads – drops:

$$\Delta V_{\text{terminal}}=\varepsilon-Ir.$$

Worked example. A battery with $\varepsilon=12\ \text{V}$ and $r=0.50\ \Omega$ supplies $2.0\ \text{A}$: the terminals sit at $\Delta V=12-2.0(0.50)=11\ \text{V}$. With no current, a voltmeter reads the full $12\ \text{V}$.

Meters: an ammeter 电流表 goes in series at the point whose current you want (ideal ammeter: zero resistance); a voltmeter 电压表 goes in parallel across the element (ideal voltmeter: infinite resistance). Non-ideal meters disturb the circuit they measure – a real ammeter adds series resistance, a real voltmeter steals current.

Bahasa Indonesia

Sederhanakan jaringan resistor menjadi resistansi ekuivalen: resistansi seri ditambahkan ($R_{\text{eq}}=R_1+R_2+\cdots$), sementara resistansi paralel ditambahkan sebagai resiprokal ($\tfrac{1}{R_{\text{eq}}}=\tfrac{1}{R_1}+\tfrac{1}{R_2}+\cdots$ – selalu kurang dari cabang terkecil). Runtuhkan jaringan langkah demi langkah untuk menemukan arus baterai, lalu kembangkan kembali untuk menemukan arus dan tegangan setiap elemen.

Contoh terpecahkan. Sebuah baterai $12\ \text{V}$ mendorong arus melalui resistor $4.0\ \Omega$ dan $2.0\ \Omega$ secara seri: $R_{\text{eq}}=6.0\ \Omega$, $I=2.0\ \text{A}$, tegangan terbagi menjadi $8.0\ \text{V}$ dan $4.0\ \text{V}$, dan resistor $4.0\ \Omega$ mendissipasi $P=I^2R=16\ \text{W}$.

Resistor dalam paralel bergabung menjadi resistansi ekuivalen yang lebih kecil
Resistor dalam paralel bergabung menjadi resistansi ekuivalen yang lebih kecil

Baterai nyata tidak ideal. Modelkan baterai sebagai baterai ideal dengan GGL $\varepsilon$ yang dirangkai seri dengan resistansi internal sendiri $r$. Ketika arus mengalir, sebagian GGL habis di dalam, sehingga tegangan terminal – apa yang dibaca voltmeter melintasi baterai sebenarnya – turun:

$$\Delta V_{\text{terminal}}=\varepsilon-Ir.$$

Contoh terpecahkan. Sebuah baterai dengan $\varepsilon=12\ \text{V}$ dan $r=0.50\ \Omega$ menyediakan $2.0\ \text{A}$: terminal-nya berada pada $\Delta V=12-2.0(0.50)=11\ \text{V}$. Tanpa arus, voltmeter menunjukkan $12\ \text{V}$ penuh.

Meteran: ammeter dipasang seri pada titik di mana arusnya ingin Anda ukur (ammeter ideal: nol resistansi); voltmeter dipasang paralel melintasi elemen (voltmeter ideal: resistansi tak terhingga). Meteran non-ideal mengganggu sirkuit yang diukurnya – ammeter nyata menambah resistansi seri, voltmeter nyata mengambil arus.

Vocabulary · ⁨Kosa kata⁩ Train · ⁨Latih⁩
English Bahasa Indonesia
parallel/ˈpærəlel/ paralel
Resistance/rɪˈzɪstəns/ Penolakan
ammeter/ˈæmiːtə/ amperemeter
voltmeter/ˈvəʊltmiːtə/ voltmeter
Kirchhoff's loop rule/ˈkɜːkhɒfs luːp ruːl/ Aturan loop Kirchhoff
Kirchhoff's junction rule/ˈkɜːkhɒfs ˈdʒʌŋkʃn ruːl/ Aturan titik simpul Kirchhoff
junction/ˈdʒʌŋkʃn/ sambungan
equivalent capacitance/ɪˈkwɪvələnt kəˈpæsɪtəns/ Kapasitansi ekuivalen
RC circuit/ˌɑː ˈsiː ˈsɜːkɪt/ sirkuit RC
time constant/taɪm ˈkɒnstənt/ konstanta waktu
steady state/ˈstedi steɪt/ keadaan tunak
11.6

Kirchhoff's Loop Rule · ⁨Aturan Loop Kirchhoff⁩

Syllabus · ⁨Silabus⁩
Learning ObjectiveEssential Knowledge

11.6.A
Describe a circuit or elements of a circuit by applying Kirchhoff's loop rule.

  • 11.6.A.1 Energy changes in simple electrical circuits may be represented in terms of charges moving through electric potential differences within circuit elements.
    • Equation: $\Delta U_E = q\Delta V$
  • 11.6.A.2 Kirchhoff's loop rule is a consequence of the conservation of energy.
    • 11.6.A.2.i Kirchhoff's loop rule states that the sum of potential differences across all circuit elements in a single closed loop must equal zero.
      • Equation: $\sum \Delta V = 0$
    • 11.6.A.2.ii The values of electric potential at points in a circuit can be represented by a graph of electric potential as a function of position within a loop.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Charges moving through potential differences exchange energy ($\Delta U_E=q\Delta V$), and energy must balance around any closed path. That is Kirchhoff's loop rule 基尔霍夫回路定则:

$$\sum\Delta V=0\ \text{around any closed loop}.$$

Sign discipline wins these problems: crossing a battery from $-$ to $+$ is $+\varepsilon$; crossing a resistor with the assumed current is $-IR$ (against it, $+IR$). Write one equation per independent loop.

Bahasa Indonesia

Muatan yang bergerak melalui beda potensial bertukar energi ($\Delta U_E=q\Delta V$), dan energi harus seimbang di sekitar jalur tertutup apa pun. Itu adalah aturan loop Kirchhoff:

$$\sum\Delta V=0\ \text{around any closed loop}.$$

Disiplin tanda memenangkan masalah ini: menyeberangi baterai dari $-$ ke $+$ adalah $+\varepsilon$; menyeberangi resistor searah dengan arus yang diasumsikan adalah $-IR$ (melawannya, $+IR$). Tulis satu persamaan per loop independen.

11.7

Kirchhoff's Junction Rule · ⁨Aturan Simpul Kirchhoff⁩

Syllabus · ⁨Silabus⁩
English

Learning Objective 11.7.A: Describe a circuit or elements of a circuit by applying Kirchhoff's junction rule.

  • 11.7.A.1 Kirchhoff's junction rule is a consequence of the conservation of electric charge.
  • 11.7.A.2 Kirchhoff's junction rule states that the total amount of charge entering a junction per unit time must equal the total amount of charge exiting that junction per unit time.
    • Equation: $\sum I_{\text{in}} = \sum I_{\text{out}}$
Bahasa Indonesia

Tujuan Pembelajaran 11.7.A: Deskripsikan sebuah sirkuit atau elemen-elemen sirkuit dengan menerapkan aturan titik simpul Kirchhoff.

  • 11.7.A.1 Aturan titik simpul Kirchhoff merupakan konsekuensi dari kekekalan muatan listrik.
  • 11.7.A.2 Aturan titik simpul Kirchhoff menyatakan bahwa total jumlah muatan yang masuk ke sebuah titik simpul per satuan waktu harus sama dengan total jumlah muatan yang keluar dari titik simpul tersebut per satuan waktu.
    • Persamaan: $\sum I_{\text{in}} = \sum I_{\text{out}}$

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English

Kirchhoff's junction rule 基尔霍夫节点定则 is conservation of charge at a junction 节点:

$$\sum I_{\text{in}}=\sum I_{\text{out}}.$$

Together the two rules solve any multi-loop circuit: assign a current to each branch, write junction equations, then loop equations, and solve. A negative answer just means that current flows opposite to your assumed direction.

Worked example. In the circuit above, $\varepsilon_1=12\ \text{V}$ with $R_1=1.0\ \Omega$ on the left, $\varepsilon_2=9.0\ \text{V}$ with $R_2=1.0\ \Omega$ on the right, and a shared middle resistor $R_3=2.0\ \Omega$ carrying $I_3=I_1+I_2$ (junction rule). The two loop equations are

$$12=I_1+2(I_1+I_2)=3I_1+2I_2,\qquad 9=I_2+2(I_1+I_2)=2I_1+3I_2.$$

Solving: $I_1=3.6\ \text{A}$, $I_2=0.60\ \text{A}$, so $I_3=4.2\ \text{A}$ through the middle. Check with the second loop: $2(3.6)+3(0.60)=9.0$ ✓.

Bahasa Indonesia

Aturan simpul Kirchhoff adalah konservasi muatan di simpul:

$$\sum I_{\text{in}}=\sum I_{\text{out}}.$$
Arus terbagi di simpul: apa yang masuk sama dengan apa yang keluar
Arus terbagi di simpul: apa yang masuk sama dengan apa yang keluar

Bersama-sama kedua aturan tersebut menyelesaikan sirkuit multi-loop apa pun: tentukan arus untuk setiap cabang, tulis persamaan simpul, lalu persamaan loop, dan selesaikan. Jawaban negatif hanya berarti arus mengalir berlawanan dengan arah asumsi Anda.

Dua persamaan loop dan satu persamaan simpul menyelesaikan sirkuit dua-baterai ini
Dua persamaan loop dan satu persamaan simpul menyelesaikan sirkuit dua-baterai ini

Contoh terpecahkan. Pada sirkuit di atas, $\varepsilon_1=12\ \text{V}$ dengan $R_1=1.0\ \Omega$ di sebelah kiri, $\varepsilon_2=9.0\ \text{V}$ dengan $R_2=1.0\ \Omega$ di sebelah kanan, dan resistor tengah bersama $R_3=2.0\ \Omega$ yang membawa $I_3=I_1+I_2$ (aturan simpul). Dua persamaan loop adalah

$$12=I_1+2(I_1+I_2)=3I_1+2I_2,\qquad 9=I_2+2(I_1+I_2)=2I_1+3I_2.$$

Menyelesaikan: $I_1=3.6\ \text{A}$, $I_2=0.60\ \text{A}$, sehingga $I_3=4.2\ \text{A}$ melalui bagian tengah. Periksa dengan loop kedua: $2(3.6)+3(0.60)=9.0$ ✓.

11.8

Resistor-Capacitor (RC) Circuits · ⁨Sirkuit Resistor-Kapasitor (RC)⁩

Syllabus · ⁨Silabus⁩
Learning ObjectiveEssential Knowledge

11.8.A
Describe the equivalent capacitance of multiple capacitors.

  • 11.8.A.1 A collection of capacitors in a circuit may be analyzed as though it was a single capacitor with an equivalent capacitance $C_{\text{eq}}$.
    • 11.8.A.1.i The inverse of the equivalent capacitance of a set of capacitors connected in series is equal to the sum of the inverses of the individual capacitances.
      • Equation: $\dfrac{1}{C_{\text{eq},s}} = \sum_{i} \dfrac{1}{C_i}$
    • 11.8.A.1.ii The equivalent capacitance of a set of capacitors in series is less than the capacitance of the smallest capacitor.
    • 11.8.A.1.iii The equivalent capacitance of a set of capacitors in parallel is the sum of the individual capacitances.
      • Equation: $C_{\text{eq},p} = \sum_{i} C_i$
  • 11.8.A.2 As a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate.

11.8.B
Describe the behavior of a circuit containing combinations of resistors and capacitors.

  • 11.8.B.1 The charge on a capacitor or the current in a resistor in an RC circuit can be described by a fundamental differential equation derived from Kirchhoff's loop rule.
    • Equation: $\mathcal{E} = \dfrac{dq}{dt}R + \dfrac{q}{C}$
  • 11.8.B.2 The time constant ($\tau$) is a significant feature of an RC circuit.
    • 11.8.B.2.i The time constant of an RC circuit is a measure of how quickly the capacitor will charge or discharge and is defined as $\tau = R_{\text{eq}}C_{\text{eq}}$.
    • 11.8.B.2.ii For a charging capacitor, the time constant represents the time required for the capacitor's charge to increase from zero to approximately 63 percent of its final asymptotic value.
    • 11.8.B.2.iii For a discharging capacitor, the time constant represents the time required for the capacitor's charge to decrease from fully charged to approximately 37 percent of its initial value.
  • 11.8.B.3 The potential difference across a capacitor and the current in the branch of the circuit containing the capacitor each change over time as the capacitor charges and discharges, but both will reach a steady state after a long time interval.
    • 11.8.B.3.i Immediately after being placed in a circuit, an uncharged capacitor acts like a wire, and charge can easily flow to or from the plates of the capacitor.
    • 11.8.B.3.ii As a capacitor charges, changes to the potential difference across the capacitor affect the charge on the plates of the capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor.
    • 11.8.B.3.iii The potential difference across a capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor all change with respect to time and asymptotically approach steady state conditions.
    • 11.8.B.3.iv After a long time, a charging capacitor approaches a state of being fully charged, reaching a maximum potential difference at which there is zero current in the circuit branch in which the capacitor is located.
    • 11.8.B.3.v Immediately after a charged capacitor begins discharging, the amount of charge on the capacitor and the energy stored in the capacitor begin to decrease.
    • 11.8.B.3.vi As a capacitor discharges, the amount of charge on the capacitor, the potential difference across the capacitor, and the current in the circuit branch in which the capacitor is located all decrease until a steady state is reached.
    • 11.8.B.3.vii After either charging or discharging for times much greater than the time constant, the capacitor and the relevant circuit branch may be modeled using steady-state conditions.

Source: College Board AP Course and Exam Description · ⁨Sumber: Deskripsi Kursus dan Ujian College Board AP⁩

English
Discharging a capacitor: τ = RC
Charging a capacitor (RC)

Capacitor networks reduce like resistors but with the rules swapped: parallel capacitances add ($C_{\text{eq}}=C_1+C_2$), series add as reciprocals – and capacitors in series must carry the same charge on each plate, by conservation of charge. Use the equivalent capacitance 等效电容 to analyse the network, then expand back.

In an RC circuit RC电路, Kirchhoff's loop rule gives the differential equation

$$\varepsilon=R\frac{dq}{dt}+\frac{q}{C},$$

whose solutions are exponentials with time constant 时间常数 $\tau=RC$:

$$q(t)=Q\big(1-e^{-t/RC}\big)\ \text{(charging)},\qquad q(t)=Q\,e^{-t/RC}\ \text{(discharging)},\qquad i(t)=\frac{\varepsilon}{R}e^{-t/RC}.$$

The current is largest at the first instant and decays – it never "waits" for the capacitor. Learn the two limits: at $t=0$ an uncharged capacitor acts like a plain wire (maximum current); after a long time it is fully charged, no current flows in its branch, and it acts like a break. In any steady state 稳态, cover the capacitor branch with your finger, solve the resistor circuit, then read the capacitor's voltage from the element it sits across.

Worked example. With $R=5.0\ \text{k}\Omega$, $C=200\ \mu\text{F}$, and a $10\ \text{V}$ battery: $\tau=RC=1.0\ \text{s}$; initial current $\dfrac{\varepsilon}{R}=2.0\ \text{mA}$; after one time constant the charge is $q=CV(1-e^{-1})\approx1.3\times10^{-3}\ \text{C}$, about $63\%$ of full charge, and the current has fallen to $37\%$ of its initial value.

Bahasa Indonesia
Mengosongkan kapasitor: τ = RC
Mengisi kapasitor (RC)

Jaringan kapasitor disederhanakan seperti resistor tetapi dengan aturan yang ditukar: kapasitansi paralel ditambahkan ($C_{\text{eq}}=C_1+C_2$), seri ditambahkan sebagai resiprokal – dan kapasitor dalam seri harus membawa muatan yang sama di setiap pelat, oleh karena konservasi muatan. Gunakan kapasitansi ekuivalen untuk menganalisis jaringan, kemudian kembangkan kembali.

Dalam sirkuit RC RC, aturan loop Kirchhoff memberikan persamaan diferensial

$$\varepsilon=R\frac{dq}{dt}+\frac{q}{C},$$

yang solusinya adalah eksponensial dengan konstanta waktu $\tau=RC$:

$$q(t)=Q\big(1-e^{-t/RC}\big)\ \text{(charging)},\qquad q(t)=Q\,e^{-t/RC}\ \text{(discharging)},\qquad i(t)=\frac{\varepsilon}{R}e^{-t/RC}.$$

Arus terbesar pada saat pertama dan meluruh – ia tidak pernah "menunggu" kapasitor. Pelajari dua batas: pada $t=0$ kapasitor tanpa muatan bertindak seperti kawat biasa (arus maksimum); setelah waktu lama ia terisi penuh, tidak ada arus mengalir di cabangnya, dan ia bertindak seperti pemutusan. Dalam keadaan stabil apa pun, tutup cabang kapasitor dengan jari Anda, selesaikan sirkuit resistor, lalu baca tegangan kapasitor dari elemen yang dilintasi olehnya.

Muatan pada kapasitor meluruh secara eksponensial saat melepaskan muatan
Muatan pada kapasitor meluruh secara eksponensial saat melepaskan muatan

Contoh terpecahkan. Dengan $R=5.0\ \text{k}\Omega$, $C=200\ \mu\text{F}$, dan baterai $10\ \text{V}$: $\tau=RC=1.0\ \text{s}$; arus awal $\dfrac{\varepsilon}{R}=2.0\ \text{mA}$; setelah satu konstanta waktu muatannya adalah $q=CV(1-e^{-1})\approx1.3\times10^{-3}\ \text{C}$, sekitar $63\%$ dari muatan penuh, dan arus telah turun ke $37\%$ dari nilainya awal.

Sirkuit elektronik yang dibangun di atas papan prototipe
Sirkuit nyata di papan breadboard: resistor dan kapasitor bersama-sama menentukan seberapa cepat tegangan naik dan turun
11.8

Exam tips · ⁨Tips ujian⁩

English
  • Relate current to charge flow $I=\tfrac{dQ}{dt}$ and use $J=\sigma E$, $R=\tfrac{\rho L}{A}$ for resistance.
  • Apply Kirchhoff's laws (junction: charge; loop: energy) with consistent sign conventions.
  • Analyse RC circuits with calculus: charging/discharging give exponentials with time constant $\tau=RC$.
  • Combine resistors (series add, parallel reciprocal) and track power $P=IV=I^2R$.
  • At $t=0$ a capacitor acts like a wire; after a long time ($t\to\infty$) it acts like an open branch.
Bahasa Indonesia
  • Hubungkan arus dengan aliran muatan $I=\tfrac{dQ}{dt}$ dan gunakan $J=\sigma E$, $R=\tfrac{\rho L}{A}$ untuk hambatan.
  • Terapkan hukum Kirchhoff (simpul: muatan; loop: energi) dengan konvensi tanda yang konsisten.
  • Analisis sirkuit RC dengan kalkulus: pengisian/pengosongan menghasilkan eksponensial dengan konstanta waktu $\tau=RC$.
  • Gabungkan resistor (seri tambahkan, paralel resiprokal) dan lacak daya $P=IV=I^2R$.
  • Pada $t=0$ kapasitor bertindak seperti kawat; setelah waktu lama ($t\to\infty$) ia bertindak seperti cabang terbuka.

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