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2.1 Ohm's law

The current through a resistor is the voltage across it divided by its resistance. That is Ohm's law: I = V / R. Double the voltage and the current doubles; double the resistance and the current halves.

One law, three ways round

In 1827 Georg Ohm found that the current through a metal wire rises in step with the voltage across it. Push twice as hard, and twice as much current flows. The resistance is what links the two:

V = I x R
I = V / R
R = V / I

All three say the same thing. Use whichever gives the one you do not know. With V in volts and R in ohms, I comes out in amps.

Worked examples

9 V across 1 kΩ: I = V / R = 9 mA
5 V and 20 mA: R = V / I = 250 Ω
2 mA through 4.7 kΩ: V = I x R = 9.4 V

Prefixes make these easy. Volts divided by kilohms gives milliamps, and milliamps times kilohms gives volts, so the thousands cancel. In the last example, 2 × 4.7 = 9.4, and the answer is in volts.

Here is a whole circuit:

circuit ohm: 6 V / 1.2 kΩ = 5 mA

2.2 Power and energy

Power is how fast energy is used, in watts: P = V × I. A resistor turns all of its power into heat, so its power rating must be bigger than the power it takes.

Power is volts times amps

Volume 01 said a volt is a joule for each coulomb, and an amp is a coulomb each second. Multiply them, and you get joules each second. That is power, measured in watts (W): one watt is one joule per second.

P = V x I

Ohm's law gives two more forms, for when you know the resistance instead:

P = I x I x R
P = V x V / R

12 V at 2 A: P = V x I = 24 W
9 V across 1 kΩ: P = V x V / R = 81 mW
20 mA through 220 Ω: P = I x I x R = 88 mW

Choosing a resistor's power rating

A resistor that takes more power than its rating overheats. A good habit is to choose a rating at least twice the power it will take.

circuit hot: current 50 mA; R1 takes 250 mW

250 mW is right at the limit of a quarter-watt resistor, so it would run hot. Choose a half-watt one, or better, a one-watt one.

Energy is power times time

Energy is power multiplied by how long it lasts. Joules are watts times seconds. Electricity bills use a bigger unit, the kilowatt hour (kWh): a thousand watts for an hour.

a 10 W lamp for 5 hours: 10 W x 5 h = 50 Wh = 180 kJ
1 kWh = 3.6 MJ

2.3 Resistors in series and the voltage divider

In series, parts share one current, and their resistances add. The supply voltage is shared between them in proportion to their resistances. That idea is the voltage divider.

Resistances in series add

Parts in series sit one after another, in a single path. The same current must pass through each of them, so each one adds its own resistance:

R = R1 + R2 + R3

circuit series: total 6 kΩ; current 2 mA
R1 has 2 V across it, R2 has 4 V, R3 has 6 V
2 V + 4 V + 6 V = 12 V
the joins are at 12 V, 10 V and 6 V

Each resistor's share is its resistance times the current. So the biggest resistor gets the biggest share of the voltage, and the shares always add up to the supply.

The voltage divider

Two resistors in series make a voltage divider: a way to get a smaller voltage from a bigger one. The output is taken from the join between them:

Vout = Vin x R2 / (R1 + R2)

circuit divider: Vout = 9 V x 10 kΩ / (10 kΩ + 10 kΩ) = 4.5 V

Unequal resistors split it unequally. From 5 V:

Vout = 5 V x 15 kΩ / (10 kΩ + 15 kΩ) = 3 V

A potentiometer is a voltage divider you can adjust. Its sliding contact splits one resistor into two, and the output rises as it slides towards the top:

a 10 kΩ potentiometer on 5 V, set 30 percent of the way up: 1.5 V

2.4 Resistors in parallel and the current divider

In parallel, parts share one voltage, and their currents add. The combined resistance is less than the smallest one, because each branch is one more path for the current.

Same voltage, currents add

Parts in parallel are connected side by side, with both ends joined. Each branch has the whole voltage across it, and draws its own current. The supply provides the total:

circuit parallel: A1 reads 3 mA; A2 reads 2 mA; A3 reads 1 mA
2 mA + 1 mA = 3 mA
together the two resistors act like 6 V / 3 mA = 2 kΩ

Combining resistances in parallel

The combined resistance comes from adding the upside-down values:

1 / R = 1 / R1 + 1 / R2 + 1 / R3

For just two resistors, a shorter form is the product over the sum:

R = R1 x R2 / (R1 + R2)

3 kΩ and 6 kΩ in parallel: 3 x 6 / (3 + 6) = 2 kΩ
10 kΩ and 10 kΩ in parallel: 5 kΩ
1 kΩ and 1 MΩ in parallel: 999 Ω

Two equal resistors give half of one. And a big resistor beside a small one hardly changes it: the small one carries almost all the current.

The current divider

A current splits between parallel branches, and the smaller resistance takes the bigger share. For two branches:

I1 = I x R2 / (R1 + R2)
3 mA x 6 kΩ / (3 kΩ + 6 kΩ) = 2 mA

Note the swap: the current in R1 uses R2 on top. The other branch gets the rest.

A divider with a load

Now the voltage divider from Module 3 can be finished. Anything connected to its output - a load - sits in parallel with R2, and lowers it:

circuit divider-load: R2 and RL together: 5 kΩ; Vout falls from 4.5 V to 3 V

This is exactly the meter loading of Volume 01. There, a 10 MΩ meter sat in parallel with a 1 MΩ resistor:

1 MΩ and 10 MΩ in parallel: 909 kΩ
9 V x 909 kΩ / (1 MΩ + 909 kΩ) = 4.29 V

So a divider only gives its formula's voltage when its load is much bigger than its resistors.

Series and parallel together

Most circuits mix the two. Work from the inside out: combine each group, then combine the groups.

R2 and R3 in parallel: 1 kΩ
add R1 in series: 1 kΩ + 1 kΩ = 2 kΩ
current from the battery: 9 V / 2 kΩ = 4.5 mA
R1 has 4.5 V across it; the pair has 4.5 V
R2 and R3 carry 2.25 mA each
circuit mixed: solved, and it agrees

2.5 Kirchhoff's laws

Two rules solve any circuit. Kirchhoff's current law: the currents into a join equal the currents out. Kirchhoff's voltage law: round any loop, the voltage rises equal the voltage drops.

The current law

Charge cannot pile up at a join, so whatever flows in must flow out. In the parallel circuit, 3 mA flowed into the join, and 2 mA + 1 mA flowed out. Rivers work the same way: where one splits in two, the water in the two branches adds up to the water in the river.

The voltage law

Go round any loop and add up the voltages. The battery raises the voltage; each resistor lowers it. By the time you are back where you started, you must be at the same voltage again. So the rises equal the drops. In the series circuit, 12 V rose in the battery, and 2 V + 4 V + 6 V fell in the resistors.

Two supplies

These two rules solve circuits that series and parallel cannot, such as one with two batteries:

Call the unknown voltage at the join V. With kilohms and milliamps, the current law at the join says:

current in through R1: (12 - V) / 2
current in through R2: (6 - V) / 1
current out through R3: V / 1
(12 - V) / 2 + (6 - V) / 1 = V / 1
multiply by 2: 12 - V + 12 - 2V = 2V
so 24 = 5V, and V = 4.8 V

With V known, every current follows from Ohm's law, and both laws can be checked:

circuit two-supplies: V = 4.8 V
R1 carries 3.6 mA; R2 carries 1.2 mA; R3 carries 4.8 mA
current law: 3.6 mA + 1.2 mA = 4.8 mA
voltage law round B1, R1 and R3: 12 V = 7.2 V + 4.8 V
voltage law round B2, R2 and R3: 6 V = 1.2 V + 4.8 V

1. Name the unknown voltages, one for each join, with ground as 0 V.
2. Write the current law at each join, with each current from Ohm's law.
3. Solve for the voltages, then work out every current.
4. Check with the voltage law round a loop or two.

This method, called nodal analysis, is how circuit simulators work. The solver behind this course does exactly this for every circuit it draws.

What you learned

• Ohm's law: V = I × R, so I = V / R and R = V / I. Volts over kilohms gives milliamps.
• Power is P = V × I, or I × I × R, or V × V / R. Choose a resistor rated at twice its power.
• Energy is power times time; 1 kWh is 3.6 MJ.
• In series, one current flows, resistances add, and the voltages share out the supply.
• A voltage divider gives Vout = Vin × R2 / (R1 + R2) - until a load draws current from it.
• In parallel, one voltage is shared, currents add, and the combined resistance is less than the smallest.
• The current law: currents in equal currents out. The voltage law: rises equal drops round a loop.

Key words from this volume

Every word below has a plain-English entry in the glossary.

• Ohm's law
• Power (electrical)
• Watt (W)
• In series
• Voltage divider
• Potentiometer
• In parallel
• Load
• Kirchhoff's current law
• Kirchhoff's voltage law
• Nodal analysis

Practice

12 V across 470 Ω: current 25.5 mA, power 306 mW

1 kΩ + 2 kΩ = 3 kΩ; current 3 mA; 3 V across 1 kΩ and 6 V across 2 kΩ

1 / R = 1 / 1 kΩ + 1 / 1 kΩ + 1 / 2 kΩ, so R = 400 Ω

Vout = 5 V x 3.3 kΩ / (1.8 kΩ + 3.3 kΩ) = 3.24 V

Interview corner

Volume 03 adds parts that store energy: capacitors, which make circuits that take time to change, and inductors.