This note covers static charge, electric current, potential difference, resistance and electrical power. These quantities are the language of every circuit question, and the equations link them together so you can find any one from the others.
Key ideas
Static charge. Insulating materials can be charged by rubbing. Electrons move from one surface to the other. The surface that gains electrons becomes negative and the surface that loses them becomes positive. Protons stay where they are.
- Like charges repel. Unlike charges attract.
- A charged object is surrounded by an electric field. A charge placed in the field feels a force.
- A charged insulator can pick up small uncharged scraps of paper. The scraps are attracted because charges in them are pulled or pushed slightly, so the nearer side has the opposite charge.
- Metals conduct charge away to the ground. Connecting a charged conductor to the ground (earthing) lets electrons flow and removes the charge.
Current. Electric current is the flow of charge. In a metal wire the moving charges are free electrons. By convention, conventional current is drawn from the positive terminal of a cell, around the circuit, to the negative terminal. The electrons actually move the opposite way.
A larger current means more charge passes a point every second. An ammeter measures current and is connected in series. A cell or battery provides a direct current (d.c.), which flows in one direction only.
Electromotive force and potential difference.
- The electromotive force (e.m.f.) of a source is the electrical work done by the source in driving one coulomb of charge around a complete circuit.
- The potential difference (p.d.) across a component is the work done by one coulomb of charge in passing through that component.
- Both are measured in volts (V). One volt is one joule per coulomb. A voltmeter measures p.d. and is connected in parallel with the component.
Resistance. Resistance is the opposition to current. For a given p.d., a larger resistance gives a smaller current. The unit is the ohm (Ω).
The resistance of a metal wire depends on its length and its cross-sectional area. Making a wire longer increases the resistance, because charges meet more obstacles. Making it thicker decreases the resistance, because there are more paths for charge. For extended work, the resistance is directly proportional to length and inversely proportional to cross-sectional area.
Current-voltage graphs.
- Fixed resistor at constant temperature: a straight line through the origin. Current is directly proportional to p.d. This is called ohmic behaviour.
- Filament lamp: the line curves and becomes less steep as p.d. rises. The filament gets hotter, so its resistance increases.
Electrical power and energy. Power is the energy transferred each second. In a resistor or lamp, electrical energy is transferred to thermal energy (and light for a lamp). The energy transferred in a time t is the power multiplied by t.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Q = It | Charge (C) = current (A) × time (s). Use when a current flows for a known time. |
| V = W ÷ Q | Potential difference (V) = work done or energy transferred (J) ÷ charge (C). |
| R = V ÷ I | Resistance (Ω) = p.d. (V) ÷ current (A). Works for any component at any point on its graph. |
| P = IV | Power (W) = current (A) × p.d. (V). |
| E = IVt | Electrical energy (J) = current × p.d. × time (s). Also E = Pt. |
| e.m.f. | Energy supplied per coulomb by the source, in volts. |
| Resistance and size (extended) | R increases with length l and decreases with area A: R ∝ l ÷ A. |
Worked examples
Example 1 A phone charger delivers a steady current of 0.25 A for 3.0 minutes. How much charge flows?
Convert the time to seconds: t = 3.0 × 60 = 180 s.
Use Q = It = 0.25 × 180.
Answer: 45 C
Example 2 A torch lamp has a p.d. of 12 V across it and a current of 0.40 A through it. Find (a) its resistance, (b) its power, (c) the energy it transfers in 5.0 minutes.
(a) R = V/I = 12 ÷ 0.40 = 30 Ω.
(b) P = IV = 0.40 × 12 = 4.8 W.
(c) The time is 5.0 × 60 = 300 s. Then E = Pt = 4.8 × 300 = 1440 J.
Answer: (a) 30 Ω, (b) 4.8 W, (c) 1440 J (1.44 kJ)
Example 3 (extended) A copper wire has a resistance of 4.0 Ω. A second wire is made from the same copper. It is twice as long and has half the cross-sectional area. What is its resistance?
Doubling the length doubles the resistance: 4.0 × 2 = 8.0 Ω.
Halving the area doubles the resistance again: 8.0 × 2 = 16 Ω.
Answer: 16 Ω
Common errors
- Saying that current is "used up" by a lamp. The same current leaves a component as enters it. It is the energy that is transferred.
- Forgetting to convert minutes to seconds before using Q = It or E = IVt.
- Connecting a voltmeter in series, or an ammeter in parallel, when describing a circuit.
- Saying electrons flow from positive to negative. Conventional current does, but the electrons move the other way.
- Describing the filament lamp graph as showing "constant resistance". The curve means the resistance increases as the filament heats.
Exam tip: Write the equation first, then substitute, then give the answer with its unit. If a question gives a time in minutes or hours, convert it to seconds before substituting.
Check yourself
- A current of 2.4 A flows through a heater element for 50 s. How much charge passes through it?
Answer
Q = It = 2.4 × 50 = 120 C. - A battery transfers 270 J of energy to 30 C of charge as it passes through a motor. What is the p.d. across the motor?
Answer
V = W/Q = 270 ÷ 30 = 9.0 V. - A 1.5 kW heater is connected to the 230 V mains. What current does it take?
Answer
P = 1500 W. I = P/V = 1500 ÷ 230 = 6.52 A. - (Extended) A wire has a resistance of 10 Ω. A wire of the same material and thickness is half as long. What is its resistance?
Answer
Resistance is proportional to length, so halving the length halves the resistance: 5.0 Ω.