电路

Electric current 电流 is the rate at which charge flows past a point, measured in amperes 安培 (A):

By convention, current points the way positive charge would move (opposite to the electrons in a wire). A steady current needs a complete loop and an energy source (a battery's electromotive force 电动势, or emf).

A circuit 电路 is a closed loop of conductors 导体, a source (battery), and components. In a series 串联 path the same current flows through each element; in a parallel 并联 path the same voltage is across each branch 支路. A circuit diagram 电路图 uses standard symbols; reading it correctly is the first step of any circuit problem.

Resistance 电阻 $R$ opposes current, measured in ohms. For an ohmic 欧姆性 component under fixed physical conditions, Ohm's law 欧姆定律 states that current is proportional to voltage. Its constant resistance relates the quantities:

A component's resistance depends on the material's resistivity 电阻率 $\rho$, its length, and its cross-sectional area 横截面积: $R=\dfrac{\rho L}{A}$ – longer and thinner means more resistance.

Worked example. A $2.0\ \text{A}$ current flows through a $6.0\ \Omega$ resistor 电阻器. The voltage across it is $V=IR=2.0\times6.0=12\ \text{V}$.

Electric power 电功率 is the rate a component converts electrical energy (to heat, light, motion):

Pick the form that uses the quantities you know. A resistor's power all becomes heat.

Worked example. The $6.0\ \Omega$ resistor above, carrying $2.0\ \text{A}$, dissipates 耗散 $P=I^2R=2.0^2\times6.0=24\ \text{W}$ – equivalently $P=IV=2.0\times12=24\ \text{W}$.

In a simple filament-bulb model, a light bulb 灯泡 glows as it dissipates electrical power. For comparable bulbs with similar light-conversion behaviour, their brightness 亮度 generally rises with power. Compare power using resistances at the actual operating conditions. In a series string every bulb carries the same current, so by $P=I^2R$ the bulb with the largest resistance is brightest; wired in parallel every bulb gets the full battery voltage, so by $P=V^2/R$ the bulb with the smallest resistance is brightest.

Combine resistors to find an equivalent resistance 等效电阻:

• Series: $R_{\text{eq}}=R_1+R_2+\cdots$ (resistances add).
• Parallel: $\dfrac{1}{R_{\text{eq}}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\cdots$ (the total is less than the smallest).

Reduce the network step by step to find the total current from the battery, then work back to each element.

Worked example. A $12\ \text{V}$ battery drives a $4.0\ \Omega$ and a $12\ \Omega$ resistor in parallel. First combine them: $\dfrac{1}{R_{\text{eq}}}=\dfrac14+\dfrac{1}{12}=\dfrac{4}{12}\Rightarrow R_{\text{eq}}=3.0\ \Omega$. The total current from the battery is $I=\dfrac{V}{R_{\text{eq}}}=\dfrac{12}{3.0}=4.0\ \text{A}$, which splits so that the smaller resistor carries the larger share ($3.0\ \text{A}$ through the $4\ \Omega$, $1.0\ \text{A}$ through the $12\ \Omega$).

Worked example (series and parallel together). A $12\ \text{V}$ battery with negligible internal resistance 内阻 drives $R_1 = 2.0\ \Omega$ in series with a parallel pair, $R_2 = 6.0\ \Omega$ and $R_3 = 3.0\ \Omega$. Find the current in each resistor and the power each one dissipates.

1. Parallel pair: $\dfrac{1}{R_{23}} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{3}{6.0}$, so $R_{23} = 2.0\ \Omega$.
2. Whole circuit: $R_{\text{eq}} = 2.0 + 2.0 = 4.0\ \Omega$, so $I = \dfrac{12}{4.0} = 3.0\ \text{A}$ through $R_1$ (and the battery).
3. Voltage across $R_1$: $3.0 \times 2.0 = 6.0\ \text{V}$. The loop rule leaves $12 - 6.0 = 6.0\ \text{V}$ across the pair.
4. Branch currents: $I_2 = \dfrac{6.0}{6.0} = 1.0\ \text{A}$ and $I_3 = \dfrac{6.0}{3.0} = 2.0\ \text{A}$. Junction 节点 check: $1.0 + 2.0 = 3.0\ \text{A}$.
5. Powers: $P_1 = 3.0^2(2.0) = 18\ \text{W}$, $P_2 = 1.0^2(6.0) = 6.0\ \text{W}$, $P_3 = 2.0^2(3.0) = 12\ \text{W}$. Total $36\ \text{W} = (12\ \text{V})(3.0\ \text{A})$, the battery's output.

If these were identical-type bulbs, $R_1$ would be brightest: it carries the whole current.

Worked example (internal resistance). A battery has emf $\varepsilon = 9.0\ \text{V}$ and internal resistance $r = 0.50\ \Omega$. It is connected to a $4.0\ \Omega$ resistor. The internal resistance is in series with the external one:

A voltmeter across the battery reads $8.0\ \text{V}$, not $9.0\ \text{V}$, while current flows. The "lost" $1.0\ \text{V}$ heats the battery.

Two meters read a circuit. An ammeter 电流表 measures the current at a point, so it must be wired in series – the current you want to measure has to flow through it. A voltmeter 电压表 measures the potential difference 电势差 between two points, so it is wired in parallel, bridging across the component whose voltage you want.

For a meter to read the true value it must barely disturb the circuit:

• an ideal ammeter has zero resistance, so putting it in series does not reduce the current it reads;
• an ideal voltmeter has infinite resistance, so almost no current is diverted through it.

A real, nonideal 非理想 meter is imperfect: a real ammeter has a small resistance (it slightly lowers the current), and a real voltmeter lets a little current leak through (it slightly lowers the voltage it reads). So connecting any meter changes, a little, the very quantity it is measuring.

Kirchhoff's loop rule 基尔霍夫电压定律 (energy conservation): around any closed loop, the voltage gains and drops sum to zero. Add the battery's emf and subtract each $IR$ drop as you go around. This gives one equation per independent loop.

Kirchhoff's junction rule 基尔霍夫电流定律 (charge conservation): the total current into any junction equals the total current out. Together with the loop rule, it lets you solve any multi-loop circuit for its unknown currents.

Worked example (two loops). In the circuit below, find the current in each resistor.

1. Junction (top middle): $I_2 = I_1 + I_3$.
2. Loop 1, clockwise from the bottom-left corner: up through the $12\ \text{V}$ battery ($+12$), along $R_1$ with $I_1$ ($-2.0I_1$), down $R_2$ with $I_2$ ($-4.0I_2$): $\;12 - 2.0I_1 - 4.0I_2 = 0$.
3. Loop 2, clockwise from the bottom of the middle branch: up $R_2$ against $I_2$ ($+4.0I_2$), along $R_3$ against $I_3$ ($+2.0I_3$), down through the $6.0\ \text{V}$ battery from $+$ to $-$ ($-6.0$): $\;6.0 - 2.0I_3 - 4.0I_2 = 0$.
4. Substitute $I_2$: $12 = 6.0I_1 + 4.0I_3$ and $6.0 = 4.0I_1 + 6.0I_3$. Multiply the first by 3 and the second by 2, then subtract: $36 - 12 = 18I_1 - 8.0I_1$, so $I_1 = 2.4\ \text{A}$. Then $I_3 = \dfrac{6.0 - 4.0(2.4)}{6.0} = -0.60\ \text{A}$ and $I_2 = 2.4 - 0.60 = 1.8\ \text{A}$.

The minus sign on $I_3$ means $0.60\ \text{A}$ flows to the right through $R_3$ and down into the $+$ terminal of the $6.0\ \text{V}$ battery: the bigger battery is charging the smaller one. Check loop 1: $12 - 2.0(2.4) - 4.0(1.8) = 12 - 4.8 - 7.2 = 0$.

An RC circuit RC电路 contains a resistor and a capacitor 电容器. When charging, the capacitor's voltage rises and the current falls, both exponentially, over a characteristic time $\tau=RC$. Key limits: at the first instant the uncharged capacitor acts like a plain wire (maximum current); after a long time it is fully charged and blocks current (acts like a break).

Worked example. For $R=10\ \text{k}\Omega$ and $C=100\ \mu\text{F}$, the time constant 时间常数 is $\tau=RC=(10\times10^{3})(100\times10^{-6})=1.0\ \text{s}$. After one time constant the capacitor reaches about $63\%$ of the supply voltage; after about $5\tau$ it is essentially fully charged.

Connected to a $9.0\ \text{V}$ battery, this circuit starts with the capacitor uncharged, acting as a wire: $I_0 = \dfrac{9.0}{10 \times 10^3} = 0.90\ \text{mA}$. A long time later no current flows, the capacitor has the full $9.0\ \text{V}$ across it, and it stores $Q = CV = (100 \times 10^{-6})(9.0) = 9.0 \times 10^{-4}\ \text{C}$.

Answer each before reading the answers below.

1. A lamp is rated $60\ \text{W}$ at $120\ \text{V}$. Find the current through it and its resistance when it works normally.
2. Resistors of $2.0\ \Omega$, $3.0\ \Omega$ and $6.0\ \Omega$ are connected in parallel. Find the equivalent resistance. What is it if they are in series instead?
3. A $9.0\ \text{V}$ battery (no internal resistance) drives a $3.0\ \Omega$ resistor in series with a parallel pair of $4.0\ \Omega$ and $12\ \Omega$. Find the current from the battery and the current in each branch of the pair.
4. A battery of emf $6.0\ \text{V}$ has a terminal voltage of $5.0\ \text{V}$ while it delivers $2.0\ \text{A}$. Find its internal resistance and the external resistance.
5. A student says: "Adding another bulb in parallel makes every bulb dimmer, because the current now has to be shared." Correct the student.

1. $I=\dfrac{P}{V}=\dfrac{60}{120}=0.50\ \text{A}$, and $R=\dfrac{V}{I}=\dfrac{120}{0.50}=240\ \Omega$.
2. Parallel: $\dfrac1{R}=\dfrac12+\dfrac13+\dfrac16=1$, so $R=1.0\ \Omega$, less than the smallest resistor. Series: $2.0+3.0+6.0=11\ \Omega$.
3. Pair: $\dfrac1{R}=\dfrac14+\dfrac1{12}=\dfrac4{12}$, so $3.0\ \Omega$. Total $6.0\ \Omega$, so $I=9.0/6.0=1.5\ \text{A}$. The pair has $1.5\times3.0=4.5\ \text{V}$ across it: $4.5/4.0=1.1\ \text{A}$ in the $4.0\ \Omega$ branch and $4.5/12=0.38\ \text{A}$ in the $12\ \Omega$ branch. Check: $1.125+0.375=1.5\ \text{A}$.
4. The lost voltage is $6.0-5.0=1.0\ \text{V}=Ir$, so $r=1.0/2.0=0.50\ \Omega$. External: $R=5.0/2.0=2.5\ \Omega$.
5. With an ideal battery, every parallel branch has the full battery voltage across it, so each bulb carries the same current as before and is just as bright. The battery simply supplies more total current. (A real battery's internal resistance makes the bulbs a little dimmer, because a larger total current loses more voltage inside it.)

• In series the current is the same throughout; in parallel the voltage is the same across each branch — never mix these up.
• Combine resistors: series add; parallel $1/R_{\text{eq}}=\sum 1/R_i$ (the total is less than the smallest).
• Apply Kirchhoff's rules: junction (current in = current out, charge conserved) and loop (voltages sum to zero, energy conserved).
• Pick the power form that fits your knowns: $P=IV=I^2R=V^2/R$.
• An ammeter goes in series (ideal: zero resistance); a voltmeter goes in parallel (ideal: infinite resistance). A real meter slightly disturbs the circuit it measures.
• A bulb is brighter when it dissipates more power – in series the biggest resistance ($I^2R$) glows brightest; in parallel the smallest ($V^2/R$) does.
• In an RC circuit the capacitor acts like a plain wire the instant it starts charging and like a break once fully charged.

Common mistakes

• Adding parallel resistances directly, or forgetting to invert at the end. $\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = 0.50\ \Omega^{-1}$ gives $R_{\text{eq}} = 2.0\ \Omega$, not $0.50\ \Omega$.
• Using the battery's emf as the voltage across one resistor in a combination circuit. Find the voltage across that part from $IR$ or the loop rule.
• Saying the current is "used up" along a series circuit. The current is the same everywhere in a series path; energy is transferred.
• Changing a sign halfway round a Kirchhoff loop. Fix the direction of travel first; crossing a resistor with the assumed current is a drop ($-IR$), against it is a rise ($+IR$).
• Treating a negative answer from Kirchhoff's rules as an error. It only means the current flows opposite to the direction you guessed.
• Assuming a battery's terminal voltage always equals its emf. With current flowing, $V = \varepsilon - Ir$.
• Putting a voltmeter in series or an ammeter in parallel. An ammeter in parallel (near-zero resistance) short-circuits the component.