Theory — Resistor Circuits

Back to Electronics hub Glossary

Series and parallel resistor networks, then voltage dividers — the circuits behind most sensor inputs. Try the examples, then test yourself.

Concept 1

Resistor networks (up to three resistors)

Key words

Series
components in one loop, joined end to end.
Parallel
components side by side on separate branches.
Total resistance (RT)
the one resistor that could replace the whole network.
Series
RT = R1 + R2 + R3
resistances add
Parallel
1RT = 1R1 + 1R2
do 1 ÷ each resistor, add them, then do 1 ÷ that total
A series pair and a parallel pair of resistors R₁R₂ Series Parallel
In series, just add the resistors. In parallel, do 1 ÷ each, add, then 1 ÷ the total.
Substitute & solve A 12 Ω resistor and a 4.0 Ω resistor are connected in parallel. Calculate the total resistance.
1RT=112 + 14.0
1RT=0.0833 + 0.25 = 0.3333common error: don't stop here
RT=1 ÷ 0.3333flip back at the end
RT=3.0 Ω
Multi-step A 6.0 Ω resistor is in series with a parallel pair of 12 Ω and 4.0 Ω. Calculate the total resistance of the network.
RP=3.0 Ωparallel pair (above)
RT=RP + 6.0now add the series part
RT=3.0 + 6.0 = 9.0 Ω
Multi-step → current & power The 9.0 Ω network above is connected across an 18 V supply. Calculate (a) the current drawn from the supply and (b) the power dissipated in the 6.0 Ω resistor.
V=IRTstart from V = IR
I=VRTrearrange for current
I=18 ÷ 9.0 = 2.0 A
P=I2 R6.0 Ω carries the full current
P=2.02 × 6.0square first
P=24 W
We do — fill the gaps A 20 Ω resistor and a 30 Ω resistor are connected in parallel. Find the total resistance.
1 Do 1 ÷ each: 1 ÷ 20 =  ·  1 ÷ 30 =
2 Add them: 1 ÷ RT =
3 Flip back: RT = 1 ÷
4 Answer: RT = Ω

Don't stop at step 2 — that's the most common mistake. You must flip back.

See the full answer
1 ÷ 20 = 0.05 1 ÷ 30 = 0.0333 1 ÷ R_T = 0.05 + 0.0333 = 0.0833 R_T = 1 ÷ 0.0833 R_T = 12 Ω

In a series circuit the total resistance is the of the resistors, so the total is always than any one of them. In parallel you add the (1 ÷ R), and the total is than the smallest resistor.

Practice 1

Three 6.0 Ω resistors are connected in parallel. Calculate the total resistance.

Answer
1/R_T = 1/6 + 1/6 + 1/6 = 3/6 = 0.5 R_T = 1 ÷ 0.5 = 2.0 Ω

✅ You can now…

  • add resistors in series.
  • find the total of a parallel pair (1 ÷ each, add, then flip back).
Concept 2

Voltage dividers

Key words

Voltage divider
two resistors that split the supply voltage between them.
Output (V2)
the voltage across the bottom resistor.
Sensor divider
a divider with an LDR or thermistor, so its output changes with light or heat.
Ratio form
V1V2 = R1R2
the voltage splits between the resistors — the bigger one gets more volts
Output form
V2 = R2R1 + R2 × Vs
output across the lower resistor R₂
  • R₁ = top, R₂ = bottom. The output V2 is the p.d. across the bottom resistor.
  • Replace R₁ or R₂ with an LDR or thermistor to make a sensor whose output voltage changes with light or temperature.
  • Pick the sensor for the quantity: a light trigger uses an LDR; a temperature trigger uses a thermistor.

The divider output usually feeds a 741 comparator, which switches an output on at a set light or heat level — that circuit is covered in Logic gates & ICs — control circuits. A divider can also feed a transistor switch.

Substitute & solve A divider has R₁ = 3.0 kΩ (top) and R₂ = 7.0 kΩ (bottom) across a 12 V supply. Calculate the output p.d. across R₂.
V2=R2R1 + R2 × Vs
V2=(7.0 ÷ (3.0 + 7.0)) × 12kΩ cancel — no convert
V2=0.70 × 12 = 8.4 V
We do — fill the gaps A divider across 10 V has R₁ = 6.0 kΩ (top) and R₂ = 4.0 kΩ (bottom). Find the output across R₂.
1 Formula: V2 = (R2 ÷ (R1 + R2)) × Vs
2 Numbers in: V2 = ( ÷ ( + )) × 10
3 Work it out: V2 = × 10
4 Answer: V2 = V

Both resistors are in kΩ, so they cancel — no need to convert.

See the full answer
V₂ = (4.0 ÷ (6.0 + 4.0)) × 10 V₂ = 0.40 × 10 V₂ = 4.0 V

A voltage divider has two resistors in . The output voltage is taken across the resistor (R₂). To sense light, replace a resistor with an ; to sense temperature, use a .

Voltage-divider & comparator

Set the supply, the two resistors (kΩ) and the comparator reference. The output is the p.d. across the bottom resistor R₂.

Practice 2

A divider across 5.0 V has R₁ = 4.0 kΩ (top) and an LDR as R₂ (bottom). In the dark the LDR is 16 kΩ. Calculate the output voltage in the dark.

Answer
V₂ = (R₂ / (R₁ + R₂)) × V_s V₂ = (16 / (4 + 16)) × 5.0 = (16/20) × 5.0 = 4.0 V

✅ You can now…

  • find the output voltage of a voltage divider.
  • make a sensor divider with an LDR or thermistor whose output changes with light or heat.
Interactive

Network Navigator challenge

Find the total resistance of ten series and parallel networks. Type each total in ohms (Ω). Score 8 / 10 to earn the 🔗 Network Navigator badge. Remember: series adds up; parallel uses 1/RT = 1/R1 + 1/R2.

Your networks scored

Answers within 2% are accepted, so sensible rounding is fine.

View my progress

Check

Check your understanding

Answer the multiple-choice questions, then mark yourself with the RAG self-check.

RAG self-check

Tap Red / Amber / Green for how confident you feel.