07 · 22 min
Electricity
Current, voltage, resistance, power, charge, and series and parallel circuits.
V = IR
The idea
Current is the rate of flow of charge, I = Q / t, measured in amperes. One ampere is one coulomb per second. Voltage is the energy transferred per coulomb. Resistance is how hard the component makes it to push that current: V = IR.
Power is the rate of energy transfer, P = IV. With Ohm’s law you can also write P = I²R or P = V²/R. A 1.15 kW kettle on a 230 V supply draws 5 A.
In a series circuit the current is the same all the way round and the voltages add. Resistances add. In a parallel circuit the voltage across each branch is the same, the currents add, and the combined resistance is smaller than the smallest branch: 1/R = 1/R1 + 1/R2.
| Series | Parallel | |
|---|---|---|
| Current | Same through each component | Splits between branches |
| Voltage | Shares, and the shares add to the cell | Same across each branch |
| Resistance | R = R1 + R2 | 1/R = 1/R1 + 1/R2 |
Formulas
Ohm’s law
V = IR
For a resistor at constant temperature, V and I are proportional. R = V/I defines resistance even when they are not.
- V
- potential difference
- V
- I
- current
- A
- R
- resistance
- Ω
Electrical power
P = IV
Also P = I²R and P = V²/R. A kilowatt is 1 000 W.
- P
- power
- W
- I
- current
- A
- V
- potential difference
- V
Charge and current
Q = It
Charge in coulombs, current in amperes, time in seconds.
- Q
- charge
- C
- I
- current
- A
- t
- time
- s
Series resistors
R = R1 + R2
One path for the current. Add more resistors the same way.
- R
- combined resistance
- Ω
- R1
- first resistance
- Ω
- R2
- second resistance
- Ω
Two parallel resistors
1/R = 1/R1 + 1/R2
Do not forget to invert at the end. Two equal resistors in parallel halve.
- R
- combined resistance
- Ω
- R1
- first resistance
- Ω
- R2
- second resistance
- Ω
Worked examples
Current and charge from a cell
A 9.0 V cell is connected across a 180 Ω resistor. Find the current, and the charge that passes in 2.0 minutes.
- V
- 9.0 V
- R
- 180 Ω
- Time
- 2.0 minutes
Try it on paper first. Then open the working one step at a time.
Two resistors in series
A 6.0 Ω resistor and a 3.0 Ω resistor are in series across 12 V. Find the current and the voltage across the 6.0 Ω resistor.
- R1
- 6.0 Ω
- R2
- 3.0 Ω
- Supply
- 12 V
Try it on paper first. Then open the working one step at a time.
Checks
A kettle is labelled 1.15 kW and runs on 230 V. What current does it draw, in amperes?
A 4.0 Ω resistor and an 8.0 Ω resistor are in series across 6.0 V. What is the current, in amperes?
Two 10 Ω resistors are connected in parallel. What is their combined resistance, in ohms?
In a series circuit with one cell and two lamps, the current
Easy places to lose a mark
- Adding parallel resistors as if they were in series.
- Forgetting to convert kW to W, or minutes to seconds in Q = It.
- Saying current is used up around a circuit. Energy is transferred; the current returns.
Practical
Circuit bench
Predict the current, then close the switch and read the meters.