Combination Circuits

The video lesson for this topic is coming soon.

Essential Question

How do we solve a circuit that has series and parallel parts?

Most real circuits mix series and parallel parts. You already know all the rules you need. By the end of this lesson, you should be able to find the total resistance of a combination circuit and then the current and voltage for every resistor in it.

Series and Parallel Together

Some parts in series
Some parts in parallel
Find the groups first

A combination circuit: a battery on the left, resistor R1 on the top wire in series, and then resistors R2 and R3 on two parallel branches. A dashed box surrounds R2 and R3 and is labeled parallel group. R1 is labeled series.

A combination circuit has some parts in series and some in parallel. Here, all of the current passes through R₁, so R₁ is in series with the rest of the circuit. Then the current splits between R₂ and R₃, which form a parallel group. The first job is always to spot these groups.

Step 1: Simplify Parallel Groups

Find each parallel group
Replace it with one resistor
Use the 1/R rule

A parallel pair of a 12 ohm resistor and a 4.0 ohm resistor, with an arrow pointing to a single 3.0 ohm resistor. A card shows 1 over R23 = 1/12 + 1/4.0 = 4/12, so R23 = 3.0 ohms.

Start with the parallel group, because it has to be replaced before anything can be added in series. For R₂ = 12 Ω and R₃ = 4.0 Ω in parallel: 1/R₂₃ = 1/12 + 1/4.0 = 1/12 + 3/12 = 4/12, so R₂₃ = 3.0 Ω. The two branches act exactly like a single 3.0 Ω resistor.

Step 2: Add the Series Parts

Now it’s a simple loop
R_T = R₁ + R₂₃
Then I = V/R_T

The simplified circuit: a 10 V battery, R1 = 2.0 ohms on the top wire, and R23 = 3.0 ohms on the right side, all in one loop. Below: R total = 2.0 + 3.0 = 5.0 ohms, and I = 10 V divided by 5.0 ohms = 2.0 A.

With the parallel group replaced by R₂₃ = 3.0 Ω, the circuit is a single loop, so the series rule applies: R_T = R₁ + R₂₃ = 2.0 Ω + 3.0 Ω = 5.0 Ω. Now use Ohm’s law on the whole circuit: I = V/R_T = 10 V ÷ 5.0 Ω = 2.0 A. This is the total current from the battery.

Step 3: Work Backward

Series part: use the total I
Parallel group shares one V
Use V = IR for each part

The original circuit labeled with results: total current 2.0 A; R1 = 2.0 ohms has 4.0 V across it; R2 = 12 ohms and R3 = 4.0 ohms each have 6.0 V across them, with 0.50 A through R2 and 1.5 A through R3.

Now return to the original circuit. All 2.0 A flows through R₁, so V₁ = IR₁ = (2.0 A)(2.0 Ω) = 4.0 V. The parallel group gets the rest of the battery voltage: V₂₃ = IR₂₃ = (2.0 A)(3.0 Ω) = 6.0 V, and both branches share that same 6.0 V. Then each branch gets its own current: I₂ = 6.0 V ÷ 12 Ω = 0.50 A and I₃ = 6.0 V ÷ 4.0 Ω = 1.5 A.

Check with the Rules

Junction rule: in = out
Loop rule: drops = battery
Both must check out

Two check cards. Junction rule: 0.50 A + 1.5 A = 2.0 A, check. Loop rule: 4.0 V + 6.0 V = 10 V, check.

Two rules let you check your work. The junction rule (conservation of charge) says the current into a junction equals the current out: 0.50 A + 1.5 A = 2.0 A. The loop rule (conservation of energy) says the voltage drops around any loop add up to the battery voltage: going through R₁ and either branch, 4.0 V + 6.0 V = 10 V. If either check fails, look for a mistake.

A Solving Table

Track R, I, V for each part
Fill in what you know
Use V = IR on each row

A table with columns R in ohms, I in amperes, and V in volts. R1: 2.0, 2.0, 4.0. R2: 12, 0.50, 6.0. R3: 4.0, 1.5, 6.0. Total: 5.0, 2.0, 10.

A table keeps a combination circuit organized. Make a row for each resistor and one for the total, with columns for R, I, and V. Fill in the given values first, then fill in the rest. Every row must follow V = IR, so if you know any two boxes in a row, you can find the third.

Worked Examples

Try each problem on your own first, then check your work against the solution.

Example 1: Total Resistance

An 18 V battery is connected to R₁ = 4.0 Ω in series with a parallel pair, R₂ = 6.0 Ω and R₃ = 3.0 Ω. Find the total resistance.

A combination circuit with an 18 V battery, R1 = 4.0 ohms on the top wire, and R2 = 6.0 ohms and R3 = 3.0 ohms on two parallel branches.

Parallel group: 1/R₂₃ = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0, so R₂₃ = 6.0/3 = 2.0 Ω
Series parts: R_T = R₁ + R₂₃
Substitute: R_T = 4.0 Ω + 2.0 Ω
Answer: R_T = 6.0 Ω

Example 2: Current and Voltages

An 18 V battery is connected to R₁ = 4.0 Ω in series with a parallel pair, R₂ = 6.0 Ω and R₃ = 3.0 Ω (the circuit from Example 1, where R_T = 6.0 Ω and R₂₃ = 2.0 Ω). Find the total current, the voltage across R₁, and the voltage across the parallel pair.

Total current: I = V/R_T = 18 V / 6.0 Ω = 3.0 A
Voltage across R1: all of the current passes through R₁, so V₁ = IR₁ = (3.0 A)(4.0 Ω) = 12 V
Voltage across the pair: V₂₃ = IR₂₃ = (3.0 A)(2.0 Ω) = 6.0 V
Answer: I = 3.0 A, V₁ = 12 V, V₂₃ = 6.0 V. Loop rule check: 12 V + 6.0 V = 18 V.

Example 3: Branch Currents

In the circuit from Examples 1 and 2 (18 V battery, R₁ = 4.0 Ω in series with R₂ = 6.0 Ω and R₃ = 3.0 Ω in parallel), the voltage across the parallel pair is 6.0 V and the total current is 3.0 A. Find the current through R₂ and through R₃.

Branch voltage: R₂ and R₃ are in parallel, so each has 6.0 V across it
Through R2: I₂ = V/R₂ = 6.0 V / 6.0 Ω = 1.0 A
Through R3: I₃ = V/R₃ = 6.0 V / 3.0 Ω = 2.0 A
Answer: I₂ = 1.0 A and I₃ = 2.0 A. Junction rule check: 1.0 A + 2.0 A = 3.0 A, the total current.

Example 4: Series Inside a Branch

A 12 V battery is connected to two branches in parallel. One branch has R₁ = 2.0 Ω and R₂ = 4.0 Ω in series. The other has R₃ = 3.0 Ω. Find R_T and the total current.

A 12 V battery connected to two parallel branches. The first branch has R1 = 2.0 ohms and R2 = 4.0 ohms in series, one above the other. The second branch has R3 = 3.0 ohms.

Series inside the branch: R₁₂ = R₁ + R₂ = 2.0 Ω + 4.0 Ω = 6.0 Ω
Parallel branches: 1/R_T = 1/6.0 + 1/3.0 = 3/6.0, so R_T = 2.0 Ω
Total current: I = V/R_T = 12 V / 2.0 Ω
Answer: R_T = 2.0 Ω and I = 6.0 A

Check: each branch gets 12 V, so the first branch carries 12 V / 6.0 Ω = 2.0 A and the second carries 12 V / 3.0 Ω = 4.0 A. Together that’s 6.0 A.

Big Ideas

Simplify groups, starting with the innermost
Add the series parts to get R_T
I = V/R_T, then work backward
Check: junction rule and loop rule

Back to the essential question: we simplify it step by step. Replace each parallel group with one equivalent resistor, add the series parts to get R_T, and find the total current with I = V/R_T. Then work backward with V = IR, the junction rule, and the loop rule to find the current and voltage for every part. This wraps up our unit on electric circuits.