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Electrician Exam Free practice test

Series and parallel circuits

Updated · Written for the 2023 NEC

Short answer

In a series circuit the current is the same everywhere, the voltage divides among the loads, and total resistance is the sum of the resistances. In a parallel circuit the voltage is the same across every branch, the branch currents add up to the total, and total resistance is always less than the smallest branch. A combination circuit is solved by reducing the parallel groups to single resistances, then treating the result as a series circuit.

Why series and parallel circuits matter on the exam

These questions test whether you can apply Ohm's law to several loads at once: total resistance, total current, one resistor's voltage, or one branch's current. The same rules explain real wiring: receptacles and lights on a branch circuit are in parallel, while a switch and the load it controls are in series.

The rules side by side

QuantitySeries circuitParallel circuit
CurrentSame through every loadBranch currents add to the total
VoltageDrops add up to the source voltageSame across every branch
Total resistanceSum of all resistancesLess than the smallest branch
Adding a loadTotal resistance goes up, current goes downTotal resistance goes down, current goes up
One load opensEverything stopsOther branches keep working

These rules come straight from Kirchhoff's laws. The voltage law explains why series drops add up to the source. The current law explains why parallel branch currents add up to the total.

Total resistance formulas and shortcuts

For a series circuit, add the resistances.

Series: RT = R1 + R2 + R3 + ...

For a parallel circuit, add the reciprocals and take the reciprocal of the sum.

Parallel: RT = 1 ÷ (1/R1 + 1/R2 + 1/R3 + ...)

Three shortcuts save time on the exam:

  • Two resistors (product over sum): RT = (R1 × R2) ÷ (R1 + R2). For 10 and 15 ohms, 150 ÷ 25 = 6 ohms.
  • Equal resistors: divide one resistor's value by how many there are. Three 30-ohm resistors in parallel give 30 ÷ 3 = 10 ohms.
  • Total current method: if you know the voltage, find each branch current, add them, and divide the voltage by the total current. This avoids reciprocals entirely.
Product over sum: RT = (R1 × R2) ÷ (R1 + R2)
Equal resistors: RT = R ÷ N

Check: parallel total resistance must be smaller than the smallest branch.

Voltage division and current division

In a series circuit, each resistor takes a share of the source voltage in proportion to its resistance. You can find any drop without first finding the current.

Voltage divider: Ex = ET × Rx ÷ RT

In a two-branch parallel circuit, current divides in the opposite way: the smaller resistance carries more current. Each branch gets the total current times the other branch's resistance, divided by the sum.

Current divider: I1 = IT × R2 ÷ (R1 + R2)

Worked examples

Example 1: series circuit

Resistors of 12, 18 and 30 ohms are in series across 240 V.

  1. Total resistance: 12 + 18 + 30 = 60 ohms.
  2. Current: 240 ÷ 60 = 4 A, the same through all three.
  3. Drops: 4 × 12 = 48 V, 4 × 18 = 72 V, 4 × 30 = 120 V.
  4. Check: 48 + 72 + 120 = 240 V.

With the voltage divider, the 30-ohm drop is 240 × 30 ÷ 60 = 120 V, the same answer in one step.

Example 2: parallel circuit

Loads of 12, 24 and 40 ohms are in parallel across 120 V.

  1. Branch currents: 120 ÷ 12 = 10 A, 120 ÷ 24 = 5 A, 120 ÷ 40 = 3 A.
  2. Total current: 10 + 5 + 3 = 18 A.
  3. Total resistance: 120 ÷ 18 = 6.67 ohms.

The reciprocal method agrees: 1/12 + 1/24 + 1/40 = 0.15, and 1 ÷ 0.15 = 6.67 ohms. That is less than the smallest branch (12 ohms), as it should be.

Example 3: combination circuit

A 6-ohm resistor is in series with a parallel pair of 10 ohms and 15 ohms, across 240 V.

  1. Reduce the parallel pair: (10 × 15) ÷ (10 + 15) = 6 ohms.
  2. Total resistance: 6 + 6 = 12 ohms.
  3. Total current: 240 ÷ 12 = 20 A.
  4. Drop across the series 6-ohm resistor: 20 × 6 = 120 V.
  5. Voltage across the parallel pair: 240 − 120 = 120 V.
  6. Branch currents: 120 ÷ 10 = 12 A and 120 ÷ 15 = 8 A.
  7. Check: 12 + 8 = 20 A, which matches the total current.

The current divider gives the same split: the 10-ohm branch carries 20 × 15 ÷ 25 = 12 A and the 15-ohm branch carries 20 × 10 ÷ 25 = 8 A.

Example 4: power in a series circuit

Using Example 1, power in each resistor is I² × R: 4² × 12 = 192 W, 4² × 18 = 288 W and 4² × 30 = 480 W. The total is 960 W, which equals 240 V × 4 A. In series, the largest resistor uses the most power. In parallel, the smallest resistor draws the most current and uses the most power. See Watt's law for the power formulas.

Common exam mistakes

  • Adding resistances in parallel. Adding only works for series. In parallel, the total always drops below the smallest branch.
  • Forgetting the last reciprocal. After adding 1/R values you have 1/RT, not RT. Flip it.
  • Using product over sum on three resistors at once. It only works for two. Combine two, then combine that result with the third.
  • Giving the full source voltage to a series resistor. Each series load gets only its share.
  • Skipping the series drop in a combination circuit. The parallel group only sees what is left after the series resistor's drop.

You can check your answers with the Ohm's law calculator.

Practice questions

Question 1 · Electrical theoryResistors of 4, 6, and 12 ohms are connected in parallel. What is the total resistance?
Show answer

Answer: D. 2 Ω.

Use the reciprocal method: add the reciprocals of each branch and take the reciprocal of the sum. 1/4 + 1/6 + 1/12 = 0.5, and 1 ÷ 0.5 = 2 ohms.

  1. 1/RT = 1/4 + 1/6 + 1/12
  2. 1/RT = 0.25 + 0.1667 + 0.0833 = 0.5
  3. RT = 1 / 0.5 = 2 Ω
Question 2 · Electrical theoryA 10-ohm load and a 20-ohm load are connected in parallel across 120 volts. What is the total current from the source?
Show answer

Answer: C. 18 A.

Each parallel branch sees the full 120 V. The 10-ohm branch draws 12 A and the 20-ohm branch draws 6 A, so the source supplies 12 + 6 = 18 A.

  1. I1 = 120 / 10 = 12 A
  2. I2 = 120 / 20 = 6 A
  3. IT = 12 + 6 = 18 A
Question 3 · Electrical theoryWhat happens to the total resistance of a parallel circuit when another resistor is added in parallel?
Show answer

Answer: C. It decreases.

Each added branch gives current another path, so the source supplies more current at the same voltage. More current at the same voltage means lower total resistance.

Question 4 · Electrical theoryA 4-ohm resistor is in series with a parallel combination of a 6-ohm and a 12-ohm resistor. The circuit is connected to 120 volts. What is the total current?
Show answer

Answer: A. 15 A.

First combine the parallel pair: (6 × 12) ÷ (6 + 12) = 4 ohms. Add the series 4 ohms for 8 ohms total, so current is 120 ÷ 8 = 15 A.

  1. R(parallel) = (6 × 12) / (6 + 12) = 4 Ω
  2. RT = 4 + 4 = 8 Ω
  3. IT = 120 / 8 = 15 A
Question 5 · Electrical theoryA 4-ohm resistor is in series with a parallel combination of a 6-ohm and a 12-ohm resistor across 120 volts. How much current flows through the 6-ohm resistor?
Show answer

Answer: C. 10 A.

The circuit's total resistance is 8 ohms, so 15 A flows and the series 4-ohm resistor drops 60 V. The remaining 60 V appears across the parallel pair, so the 6-ohm branch carries 60 ÷ 6 = 10 A.

  1. RT = 4 + (6 × 12)/(6 + 12) = 8 Ω; IT = 15 A
  2. E(parallel) = 120 − (15 × 4) = 60 V
  3. I(6 Ω) = 60 / 6 = 10 A

Frequently asked questions

What is the main difference between series and parallel circuits?

A series circuit has one path, so the current is the same everywhere and the voltage divides. A parallel circuit has several paths across the same two points, so the voltage is the same on every branch and the current divides.

Why does adding a resistor in parallel lower the total resistance?

Each new branch gives current another path. At the same voltage the source supplies more total current, and more current at the same voltage means lower total resistance.

Are household outlets wired in series or parallel?

Receptacles and lights on a branch circuit are connected in parallel, so each one gets the full circuit voltage and one device failing does not shut off the others.

How do you solve a combination circuit?

Reduce each parallel group to one equivalent resistance, add the series parts to get the total, and find total current. Then work back out to find the drop across each series part and the current in each branch.

What is a quick way to find two parallel resistors?

Multiply the two values and divide by their sum. For 6 and 3 ohms, 18 ÷ 9 = 2 ohms. For equal resistors, divide one value by the number of resistors.

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