TEJ3M Series, Parallel and Voltage Dividers: Notes and Practice

Key things to remember

1Series: current is shared, voltages add, resistances add. Total > largest resistor.

2Parallel: voltage is shared, currents add, reciprocals add. Total < smallest branch.

3Two in parallel:

R_{T} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}

4n equal resistors in parallel give

R / n

5Voltage divider:

V_{o u t} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}}

6Kirchhoff: voltages around a loop sum to zero; currents into a junction equal currents out.

There are only two ways to connect two components, and every circuit you will ever analyze is some combination of them. In series there is one path, so the same current must flow through everything. In parallel there is a junction, so the same voltage appears across every branch. Getting those two sentences straight makes the rest of this topic almost mechanical.

Fig 1One path versus a junction — and what that does to the totals.

Series circuits

Current is the same everywhere: $I_{T} = I_{1} = I_{2} = I_{3}$
Voltages add up to the supply: $V_{T} = V_{1} + V_{2} + V_{3}$ (Kirchhoff's voltage law)
Resistances add: $R_{T} = R_{1} + R_{2} + R_{3}$
The total is always LARGER than the largest single resistor.
Break the circuit anywhere and everything stops — which is why old Christmas lights all went out together.

Parallel circuits

Voltage is the same across every branch: $V_{T} = V_{1} = V_{2} = V_{3}$
Currents add up to the total: $I_{T} = I_{1} + I_{2} + I_{3}$ (Kirchhoff's current law)
Resistance combines reciprocally: $\frac{1}{R _{T}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$
The total is always SMALLER than the smallest single branch.
Remove one branch and the others carry on — which is why house wiring is parallel.

R_{T} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}

The product-over-sum shortcut — valid for exactly two resistors in parallel.

Check every answer against the rule

A series total must exceed the largest resistor. A parallel total must be below the smallest branch. If your parallel answer comes out larger than a branch, you added instead of using reciprocals, or you forgot the final reciprocal step — which is far and away the most common error in this topic.

Identical resistors in parallel

n identical resistors of value R in parallel give exactly

R / n

. Two 100 Ω resistors give 50 Ω; four give 25 Ω. This is worth knowing because it gives you an instant sanity check on any calculation.

Loading interactive…

Switch between series and parallel with the same resistors and compare.

Voltage dividers

Two resistors in series across a supply divide that voltage between them, in proportion to their resistance. Tapping the junction between them gives an output voltage that is a fixed fraction of the input. This tiny circuit is genuinely everywhere: it is how a potentiometer sets a volume, how a thermistor becomes a temperature reading, and how a 5 V signal is safely reduced to 3.3 V.

Fig 2The output is whatever fraction of the supply falls across the lower resistor.

V_{o u t} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}}

R₂ is the resistor between the output node and ground.

The sensor application is the important one. Replace R₂ with a component whose resistance changes with something physical — a light-dependent resistor, a thermistor, a force sensor — and the output voltage now varies with light, temperature or force. Feed that node into a microcontroller's analog input and you have turned a physical quantity into a number. Every analog sensor project in the next unit is built on this circuit.

Loading a divider

The formula assumes nothing draws current from the output node. Connect a real load with resistance comparable to R₂ and it sits in parallel with R₂, lowering the effective resistance and pulling the output below the calculated value. Microcontroller analog inputs draw almost nothing, so they are fine — but a divider cannot be used as a power supply.

Working through a combination circuit

Redraw the circuit clearly, if it is drawn awkwardly. Most mistakes are misreadings.
Find the innermost group that is purely series or purely parallel and reduce it to a single equivalent resistance.
Redraw with that group replaced, and repeat until only one resistance remains.
Use Ohm's law on the total to find the total current.
Work back outward: in a series section the current is known, so find each voltage; in a parallel section the voltage is known, so find each current.
Check: the individual voltages in any series loop must add to the supply voltage.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A 100 Ω, a 220 Ω and a 330 Ω resistor are in series across a 12 V supply. Find the total resistance, the current, and the voltage across each resistor.

Example 2

Two resistors, 470 Ω and 680 Ω, are connected in parallel across 9 V. Find the total resistance, each branch current, and the total current.

Example 3

Design a voltage divider to produce 3.3 V from a 5 V supply, using a 10 kΩ resistor as R₁. What value is R₂?

Where marks get lost

Adding reciprocals and forgetting the final flip —

1/ R_{T}

is not

R_{T}

Getting a parallel total larger than one of the branches and not noticing it is impossible.

Assuming voltage is shared equally between series resistors. It divides in proportion to resistance.

Using product-over-sum with three resistors. That shortcut is only valid for exactly two.

Forgetting that a real load connected to a divider's output sits in parallel with R₂ and changes the answer.

Mixing up which resistor is R₂ in the divider formula. R₂ is the one between the output node and ground.

Key things to remember

Worked examples

Where marks get lost

Flashcards 10

Practice