Series Circuits

The video lesson for this topic is coming soon.

Essential Question

What happens when parts of a circuit share one path?

Real circuits have more than one part. When the parts are connected one after another in a single loop, they form a series circuit. By the end of this lesson, you should be able to find the total resistance, the current, and the voltage across each part of a series circuit.

One Path

Parts connected end to end
Charge has only one path
It flows through every part

A series circuit: a battery on the left connected in a single rectangular loop to three glowing bulbs, two on the top wire and one on the right side. Caption: one loop, one path.

In a series circuit, the parts are connected end to end in a single loop. Charge leaving the battery has only one path to follow, so it must pass through every part, one after another, before it returns to the battery.

Same Current Everywhere

Same current at every point
Charge is not used up
I = I₁ = I₂

A series circuit with a battery, two resistors R1 and R2, and three ammeters placed at different spots in the loop. All three ammeters read 2.0 A.

Because there is only one path, every coulomb that leaves the battery passes through every part. Charge is not used up along the way, so the current is the same everywhere in a series circuit. An ammeter placed anywhere in the loop gives the same reading.

Voltages Add Up

Each part uses some voltage
Drops add to the battery voltage
V = V₁ + V₂ + …

A series circuit with a 12 V battery, a 4.0 ohm resistor on the top wire, and an 8.0 ohm resistor on the right side. The voltage across the 4.0 ohm resistor is 4.0 V and across the 8.0 ohm resistor is 8.0 V. Caption: 4.0 V + 8.0 V = 12 V.

The battery gives each coulomb a certain amount of energy (12 J per coulomb for a 12 V battery). As the charge travels around the loop, it gives that energy away to the parts it passes through. By the time it returns to the battery, all of it has been used, so the voltage drops across the parts add up to the battery voltage. This is conservation of energy. The bigger resistor gets the bigger share: here the 8.0 Ω resistor drops 8.0 V and the 4.0 Ω resistor drops 4.0 V.

Total Resistance

Resistances in series add
R_T = R₁ + R₂ + …
More resistors, less current

Three resistors in a row, 2.0, 3.0, and 5.0 ohms. An arrow points down to a single equivalent resistor labeled R total = 10.0 ohms. Caption: 2.0 + 3.0 + 5.0 = 10.0 ohms.

In series, the charge has to push through every resistor, one after another, so the resistances simply add: R_T = R₁ + R₂ + R₃ + …. Three resistors of 2.0 Ω, 3.0 Ω, and 5.0 Ω act like one 10.0 Ω resistor. Adding another resistor in series always increases the total resistance, which decreases the current from the battery.

One Bulb Out, All Out

A break anywhere opens the loop
Current stops everywhere
Every bulb goes dark

The same series circuit with three bulbs, but the middle bulb on the top wire has a broken filament and is labeled burned out. All three bulbs are dark. Caption: one bulb out, all out.

When a bulb burns out, its filament breaks and leaves a gap. In a series circuit that gap opens the only loop, so the current stops everywhere and every bulb goes dark, even the ones that still work. Old strings of holiday lights were wired this way, which is why one bad bulb could knock out the whole string.

Solving Series Circuits

Work from the whole to the parts
Same I through every part
Check: drops add to V

Four steps in a flowchart. Step 1: R total = R1 + R2. Step 2: I = V divided by R total. Step 3: V1 = I R1 and V2 = I R2. Step 4: check that V1 + V2 = V.

To solve a series circuit, start with the whole circuit and work toward the parts. First add the resistances to get R_T. Then use Ohm’s law on the whole circuit, I = V/R_T, to get the current, which is the same through every part. Next use V = IR on each resistor to get its voltage drop. Finally, check that the drops add up to the battery voltage.

Worked Examples

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

Example 1: Total Resistance and Current

A 12 V battery is connected in series with a 2.0 Ω resistor and a 4.0 Ω resistor. Find the total resistance and the current.

A series circuit with a 12 V battery on the left, a 2.0 ohm resistor on the top wire, and a 4.0 ohm resistor on the right side.

Total resistance: R_T = R₁ + R₂ = 2.0 Ω + 4.0 Ω = 6.0 Ω
Current equation: I = V/R_T
Substitute: I = 12 V / 6.0 Ω
Answer: R_T = 6.0 Ω and I = 2.0 A, the same through both resistors

Example 2: Voltage Drops

An 18 V battery is in series with R₁ = 3.0 Ω and R₂ = 6.0 Ω. Find the voltage across each resistor.

A series circuit with an 18 V battery on the left, R1 = 3.0 ohms on the top wire, and R2 = 6.0 ohms on the right side.

Total resistance: R_T = 3.0 Ω + 6.0 Ω = 9.0 Ω
Current: I = V/R_T = 18 V / 9.0 Ω = 2.0 A, the same through both resistors
Each drop: V₁ = IR₁ = (2.0 A)(3.0 Ω) = 6.0 V, and V₂ = IR₂ = (2.0 A)(6.0 Ω) = 12 V
Answer: V₁ = 6.0 V and V₂ = 12 V. Check: 6.0 V + 12 V = 18 V.

Example 3: Missing Resistor

A 24 V battery pushes 1.5 A through two resistors in series. One resistor is 4.0 Ω. What is the other?

A series circuit with a 24 V battery, an ammeter reading 1.5 A on the bottom wire, a 4.0 ohm resistor on the top wire, and an unknown resistor on the right side labeled question mark.

Knowns: V = 24 V, I = 1.5 A, R₁ = 4.0 Ω, R₂ = ?
Total resistance: R_T = V/I = 24 V / 1.5 A = 16 Ω
Subtract: R_T = R₁ + R₂, so R₂ = R_T − R₁ = 16 Ω − 4.0 Ω
Answer: R₂ = 12 Ω

Example 4: Adding a Bulb

Two identical 4.0 Ω bulbs are in series with a 12 V battery. A third identical bulb is added in series. Find the current before and after. What happens to the brightness?

Before: R_T = 4.0 Ω + 4.0 Ω = 8.0 Ω, so I = 12 V / 8.0 Ω = 1.5 A
After: R_T = 4.0 Ω + 4.0 Ω + 4.0 Ω = 12 Ω, so I = 12 V / 12 Ω = 1.0 A
Power per bulb: before: P = I²R = (1.5 A)²(4.0 Ω) = 9.0 W. After: P = (1.0 A)²(4.0 Ω) = 4.0 W.
Answer: The current drops from 1.5 A to 1.0 A, and every bulb gets dimmer.

Big Ideas

One path: same current everywhere
Voltage drops add to the battery voltage
R_T = R₁ + R₂ + …
One bulb out, all out

Back to the essential question: in a series circuit, the parts share one path, so the same current flows through every part. The voltage drops add up to the battery voltage, the resistances add to give R_T, and if one part breaks, everything stops. Next up, we’ll give charges more than one path with parallel circuits.