Series vs Parallel Circuits: Worked 24 V Study Examples

A circuit diagram can look complicated while containing only two useful relationships: components that share one current path, and branches connected across the same two points. Learning to recognize those relationships is the first step in solving series and parallel circuit questions.

All examples below are paper exercises using an ideal 24 V DC source, ideal fixed resistors and wires with zero resistance. They are not wiring instructions or a claim that any real 24 V circuit is safe to handle.

If rearranging V = IR is still unfamiliar, review the Ohm's law lesson first. This article uses that relationship to compare complete networks.

Read connections before numbers

Series resistors carry the same current; their resistances add, and their voltage drops add to the applied voltage. Parallel branches share the same two connection points and therefore the same voltage. Their currents add at the junction, while their equivalent resistance follows the reciprocal relationship. OpenStax: Resistors in series and parallel.

On paper, give the two ends of a network names such as A and B. Then identify the path between them. Two resistors drawn beside each other are not necessarily parallel; inspect their connections.

Example 1: one path, two voltage drops

Imagine this single path between the source terminals:

+24 V → 8 Ω resistor → 16 Ω resistor → 0 V

First add the resistances:

8 + 16 = 24 Ω

The source current is:

24 V ÷ 24 Ω = 1 A

That same 1 A passes through each resistor. Calculate each voltage drop separately:

The check is 8 + 16 = 24 V. Relative to the 0 V return, the point between the resistors is at 16 V: the first resistor has accounted for an 8 V drop from the +24 V terminal.

The current is not split in this example. Writing 0.5 A through each resistor would confuse sharing a source voltage with sharing current between branches.

Example 2: two paths across the source

Now draw points A and B with two separate paths between them:

• Upper path: A → 12 Ω → B

• Lower path: A → 24 Ω → B

Connect A to the ideal +24 V terminal and B to the 0 V terminal in the drawing. Each branch spans the full 24 V.

Calculate the branches first:

The source provides 2 + 1 = 3 A. The equivalent resistance is therefore:

24 V ÷ 3 A = 8 Ω

You can verify it independently: 1/12 + 1/24 = 3/24 = 1/8, giving 8 Ω after taking the reciprocal.

The equivalent resistance is smaller than either branch resistance. Adding 12 and 24 to get 36 Ω would describe a different, series network.

Example 3: work from a smaller network outward

Consider a 4 Ω resistor in series with a parallel pair of 12 Ω and 6 Ω. The entire combination connects across the same ideal 24 V source.

Start with the parallel pair:

1/12 + 1/6 = 3/12, so its equivalent resistance is 4 Ω.

The full network now reduces to 4 Ω + 4 Ω = 8 Ω. Source current is 24 ÷ 8 = 3 A.

The first, series resistor drops 3 × 4 = 12 V. That leaves 12 V across the parallel pair, giving:

• 12 Ω branch: 12 ÷ 12 = 1 A

• 6 Ω branch: 12 ÷ 6 = 2 A

Check both relationships: branch currents add to the 3 A source current, and the two main voltage drops add to 24 V.

The key decision was using 12 V for the branch calculations. Putting 24 V across each branch would ignore the voltage drop in the preceding series resistor.

Three questions to finish

1. Two 6 Ω resistors are in series across 24 V. Find the current and each voltage drop.

2. Two 12 Ω branches are in parallel across 24 V. Find each branch current and the source current.

3. In example 3, what current passes through the first 4 Ω resistor?

Answers: 1. 2 A; 12 V each. 2. 2 A per branch; 4 A total. 3. 3 A, before the current divides between the branches.

Redraw one example from memory during your next study-plan session. Explain why each voltage or current belongs to that particular part of the network, then continue through the G3 module overview.