题型概述：电动势、内阻与电路Question type: EMF, internal resistance & circuits
Top GoldSection 1 每年约有 10 分落在直流电路上。反复出现的套路有两个：一是"两种状态 → 两个方程 → 解出 ε 和 r"（同一个电池接两种外电路，各给一个电流）；二是最大功率传输（外电阻等于内阻时外电路功率最大）。另外常考用基尔霍夫定律和"接地 = 0 V"求某点电势。物理只需要三件事，下面逐一讲清。About 10 marks of Section 1 each year are on DC circuits. Two patterns recur: "two states → two equations → solve for ε and r" (the same cell drives two different external circuits and you are told the current each time), and maximum power transfer (the external power peaks when the load equals the internal resistance). Questions on Kirchhoff's laws and potentials relative to earth (0 V) also appear. Only three ideas are needed; each is explained below.
1电动势、内阻与端电压EMF, internal resistance and terminal p.d.
电动势 ε 是电池对每库仑电荷做的功（J/C = V）。真实电池内部也有电阻 r，电流 I 流过时要在电池内部"损失"一部分电压 Ir（常叫 lost volts）。所以接在电池两端的电压表读到的端电压 V 比 ε 小，只有断路（I = 0）时才等于 ε。整条回路的能量守恒写成 ε = I(R + r)。The emf ε is the work the cell does on each coulomb of charge (J/C = V). A real cell also has an internal resistance r, so when a current I flows, a p.d. Ir is "lost" inside the cell (the lost volts). A voltmeter across the terminals therefore reads the terminal p.d. V, which is less than ε and equals ε only on open circuit (I = 0). Energy conservation round the whole circuit gives ε = I(R + r).
V = ε − Ir = IR ⇔ ε = I(R + r) Imax = ε / r (短路short circuit, R = 0)ε 和 r 是两个未知数，所以只要题目给出同一个电池在两种外电阻下的电流（或电压），就能列两个方程解出它们。ε and r are two unknowns, so whenever a question gives the current (or p.d.) for the same cell with two different external resistances, you can write two equations and solve for both.
2最大功率传输：R = rMaximum power transfer: R = r
外电阻 R 上的功率 P = I²R。R 很小时电流大但 R 小，功率小；R 很大时电流几乎为零，功率也小。中间必有最大值：对 R 求导令其为零，得到 R = r。此时外电路和内阻上分到的功率相同，效率只有 50%。The power in the external resistor is P = I²R. When R is tiny the current is large but R is small, so P is small; when R is huge the current is almost zero, so P is small again. There must be a maximum in between: setting dP/dR = 0 gives R = r. At that point the load and the internal resistance take equal power, so the efficiency is only 50%.
P(R) = ε²R / (R + r)² ⇒ Pmax = ε² / 4r at R = r, η = R/(R + r) = 50%P–R 曲线（ε = 9 V，r = 3 Ω），可拖动P–R curve (ε = 9 V, r = 3 Ω); drag it
拖动滑块：R 经过 3 Ω（= r）时 P 达到峰值 6.75 W，效率恰好 50%。R 再增大，效率上升但功率下降。Move the slider: as R passes 3 Ω (= r) the power peaks at 6.75 W with efficiency exactly 50%. Beyond that the efficiency rises but the power falls.
3沿回路走一圈：电势与接地Walking round a loop: potential and earth
电势只有相对值，接地点规定为 0 V。从某点出发沿回路走：顺着电流穿过电阻，电势下降 IR；从电池负极穿到正极，电势升高 ε（再在内阻上降 Ir）。回到起点时电势必须回到原值，这就是基尔霍夫电压定律。电路只有一个接地点时，没有电流流进大地，它只决定"零点"在哪里。Only differences of potential are physical; an earthed point is defined to be 0 V. Walk round the loop: going with the current through a resistor the potential falls by IR; going through a cell from − to + it rises by ε (and falls Ir across the internal resistance). You must arrive back at the starting value, which is Kirchhoff's voltage law. With a single earth connection no current flows to earth; the earth only fixes where zero is.
Σ ε = Σ IR (任一闭合回路round any closed loop) VX − VY = IR (电流从 X 经 R 流到 Ycurrent flows X → R → Y)判断正负的办法：电流总是从高电势流向低电势（在电阻里）。所以电流流入接地点的那一侧是正电势，流出接地点的那一侧是负电势。Sign rule: in a resistor, current always flows from high to low potential. So the side from which current flows into the earth point is positive, and the side it flows on to is negative.
| 子题型Sub-type | 识别特征How to spot it | 核心步骤Core step | 真题Past paper |
|---|---|---|---|
| 两种状态求 ε、rTwo states → ε and r | 同一电池，外电路改变（并联一个电阻、串联↔并联），给两个电流Same cell, external circuit changed (resistor added in parallel, series ↔ parallel), two currents given | 每种状态写 ε = I(R外 + r)，联立Write ε = I(Rext + r) for each state; solve together | 2017 Q1(h) · 2012 Q1(j) |
| 最大功率传输Maximum power transfer | "power … is a maximum"、"maximum current"、可变电阻"power … is a maximum", "maximum current", a variable resistor | 负载 = 从负载两端看进去的等效内阻Load = resistance of the rest of the circuit seen from the load | 2017 Q1(i) · 2023 Q1(p) |
| 电势 / 接地Potentials / earth | "earth"、"zero potential"、求 A、B 点电势"earth", "zero potential", potential at points A, B | 先定电流方向，再从 0 V 点沿回路累加Fix the current direction, then add up p.d.s from the 0 V point | 2018 Q1(i) |