D.C. circuits

A breadboard lets you build and test practical circuits without soldering.

e.m.f. and p.d.

The electromotive force 电动势 (e.m.f.) $\varepsilon$ of a source is the energy 能量 given to each unit of charge by the source as it drives the charge around a full circuit. Unit: volt.

The potential difference 电势差 (p.d.) across a component is the energy changed from electrical to other forms by each unit of charge as it passes through that component.

Both are in volts; they differ in energy direction:

• e.m.f. — energy put into the circuit by the source (chemical → electrical in a battery, mechanical → electrical in a generator).
• p.d. — energy taken out of the electrical form (electrical → thermal in a resistor, → light in a lamp, → kinetic in a motor).

In the lab you often build a circuit on a breadboard 面包板 (a board with rows of holes that connect components without soldering) and measure currents and p.d.s with a multimeter 万用表.

A real circuit on a breadboard, being measured with a multimeter

Internal resistance

A real source has some internal resistance 内阻 $r$ — usually the resistance of the electrolyte 电解质 in a cell 电池, or the wire windings in a generator. When current $I$ flows, an internal p.d. of $Ir$ is "lost" inside the source, so the terminal p.d. 端电压 across the outside circuit is

So:

• no current (open circuit 开路, $I = 0$): the terminal p.d. equals the e.m.f.
• larger current: the terminal p.d. falls.
• short circuit 短路 ($R_{\text{external}} \to 0$): $I = \varepsilon / r$, a large current, with all the energy turned to heat inside the source.

To measure $r$, change the outside resistance and plot $V_{\text{terminal}}$ against $I$: the line has $y$-intercept $\varepsilon$ and gradient $-r$.

Worked example. A cell of e.m.f. $1.5\ \text{V}$ and internal resistance $0.50\ \Omega$ is connected to a $2.5\ \Omega$ resistor. Find the current and the terminal p.d.

The e.m.f. drives the current through both resistances: $I = \dfrac{\varepsilon}{R + r} = \dfrac{1.5}{2.5 + 0.50} = 0.50\ \text{A}$. Then

Circuit for measuring the e.m.f. and internal resistance of a cell

Terminal p.d. against current — the intercept is the e.m.f. and the gradient is minus the internal resistance

The power 功率 given to the outside load is $P_{\text{ext}} = (\varepsilon - Ir) I$; the power lost inside is $P_{\text{int}} = I^{2} r$; the total power from the source is $\varepsilon I$.

Circuit symbols

You must recognise and draw the standard symbols in the syllabus: cell, battery, switch, resistor, variable resistor, ammeter 电流表, voltmeter 电压表, lamp, diode (and LED 发光二极管), capacitor 电容器, inductor, thermistor, light-dependent resistor, fuse 保险丝, earth, junction. An ideal ammeter has zero resistance 电阻 and goes in series 串联. An ideal voltmeter has infinite resistance and goes in parallel 并联.

The standard circuit symbols you need to recognise and draw

First law (junction rule)

At any junction 节点, the total current flowing in equals the total current flowing out. This follows from conservation of charge 电荷守恒 — charge cannot build up at a point in a steady circuit, so charge in per second equals charge out per second.

For a junction with three wires: $I_{1} = I_{2} + I_{3}$ if currents 2 and 3 flow out and current 1 flows in.

Current divides at a junction in a parallel circuit (3 A in equals 2 A plus 1 A)

Second law (loop rule)

Around any closed loop 回路, the total e.m.f. equals the total p.d. across the components in that loop. This follows from conservation of energy 能量守恒: as a unit of charge goes once round a loop, the energy it gains from sources equals the energy it gives up to components.

Pick a direction round the loop. Take an e.m.f. as positive when the loop direction goes from − to + of the source, and a p.d. as positive when the loop direction is the conventional current direction through the resistor.

Combining resistors

Resistors in series carry the same current; the total p.d. is the sum:

so $R_{\text{series}} = R_{1} + R_{2} + \ldots$.

Two resistors in series and their single equivalent resistor

Resistors in parallel have the same p.d.; the total current is the sum:

so $\dfrac{1}{R_{\text{parallel}}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \ldots$.

Two resistors in parallel and their single equivalent resistor

Two equal resistors $R$ in parallel give $R/2$; $N$ equal ones give $R/N$. A parallel combination is always smaller than any of its resistors; a series combination is always larger.

Worked example. A $4.0\ \Omega$ resistor and a $12\ \Omega$ resistor are connected in parallel. Find their combined resistance.

Solving a circuit

1. Label every current with a symbol and a chosen direction.
2. Use Kirchhoff's first law 基尔霍夫第一定律 at each junction to link the currents.
3. Use Kirchhoff's second law 基尔霍夫第二定律 around each loop to get equations in the p.d.s.
4. Use $V = IR$ for each resistor.
5. Solve the equations together.

For symmetric resistor networks, use the symmetry to spot branches with equal currents — the branch with the most current gives the most power ($P = I^{2}R$).

A potential divider 分压器 is two (or more) resistors in series across a source. The p.d. across each resistor is in direct proportion to its resistance:

The output (tapped between $R_{1}$ and $R_{2}$) can be set to any voltage 电压 between $0$ and $V_{\text{in}}$ by choosing the resistances. A rheostat 变阻器 (a slider on a uniform-resistance wire) gives a smoothly variable divider.

Worked example. A $6.0\ \text{V}$ supply is connected across a $2.0\ \text{k}\Omega$ resistor in series with a $4.0\ \text{k}\Omega$ resistor. Find the output voltage tapped across the $4.0\ \text{k}\Omega$ resistor.

A potential divider — the p.d. splits between R1 and R2 in proportion to their resistances

Sensor circuits

Replace one fixed resistor with a sensor 传感器 whose resistance changes with a physical quantity:

• thermistor 热敏电阻 (NTC): $R$ falls as temperature rises. In a divider, the output voltage changes with temperature in a fixed direction.
• light-dependent resistor 光敏电阻 (LDR): $R$ falls as light intensity 光强 rises, giving a brightness-dependent output.

Connect the output to a transistor 晶体管 base or a comparator 比较器 to switch a load on or off when the temperature or light passes a threshold 阈值.

A thermistor in a potential divider gives an output voltage that changes with temperature

Potentiometer and the null method

A potentiometer 电位差计 is a uniform resistance wire of length $L_{0}$ with a sliding contact (jockey 滑动触头). The resistance per unit length is uniform, so the p.d. from one end to the jockey is proportional to the length:

To compare two e.m.f.s (an unknown cell against a standard cell), connect each in turn with the jockey through a galvanometer 检流计. Slide the jockey until the galvanometer reads zero (a null — no current flows through the cell being measured, because the potentiometer's voltage there exactly opposes the cell's e.m.f.). The two balance lengths are in the ratio of the e.m.f.s:

This is a null method 零点法: you find the balance (zero current) instead of measuring a current's value. Its advantage is that at balance the unknown cell gives no current, so its internal resistance does not affect the result.

A potentiometer comparing two cell e.m.f.s by the null method

• Apply Kirchhoff's laws: current into a junction $=$ current out (charge conserved); $\sum \text{e.m.f.} = \sum \text{p.d.}$ round a loop (energy conserved).
• Combine resistors: series $R = R_1 + R_2$; parallel $1/R = 1/R_1 + 1/R_2$.
• A potential divider splits voltage in the ratio of the resistances.
• Include internal resistance: $\text{e.m.f.} = I(R + r)$ — the "lost volts" are $Ir$.