D. Physics: Principles and Applications of Electricity · Electricity: Charges, Circuits and EnergySeries Circuits: Current, Potential Difference and Resistance
D2.6 D2.5 A1.2About 75 minutes
Learning goals- I can construct and draw a series circuit and measure the current and potential differences in it.
- I can explain why the current is the same everywhere in a series circuit.
- I can show that the potential differences across the loads add up to the source's potential difference.
- I can calculate the total (equivalent) resistance of resistors in series: R_T = R₁ + R₂ + ...
- I can solve series-circuit problems using Ohm's law.
Words to know
series circuit
A circuit with only one path for the current. All the loads are connected end to end in a single loop.
total (equivalent) resistance (R_T)
The single resistance that could replace all the resistors in a circuit and draw the same current from the source.
Kirchhoff's voltage law
Around any closed loop, the potential difference gained in the source equals the total potential difference used by the loads.
Kirchhoff's current law
The total current flowing into any point in a circuit equals the total current flowing out. In a series circuit there are no branch points, so the current is the same everywhere.
voltage drop
The potential difference across a load: the energy each coulomb gives up as it passes through that load.
Remember from Grade 6
In Grade 6 you saw that in a series circuit there is only one path, so if one bulb is removed, all the bulbs go out. In Grade 9 you will measure and calculate exactly what happens to the current, the potential difference and the resistance.
Rule 1: the current is the same everywhere
In a series circuit, the charges have only one path. They cannot pile up or disappear anywhere, so the same amount of charge passes every point each second:
I_T = I₁ = I₂ = I₃ = ...
It does not matter where you put the ammeter: before the first resistor, between them, or after the last one, it reads the same.
Rule 2: the potential differences add up
Each coulomb gains energy in the battery and gives up all of it as it passes through the loads, sharing it out between them. So:
V_T = V₁ + V₂ + V₃ + ...
where V_T is the potential difference of the source. The larger resistor gets the larger share. This is Kirchhoff's voltage law.
Rule 3: resistances add up
Each extra resistor in the one path is one more obstacle, so the total resistance is the sum:
R_T = R₁ + R₂ + R₃ + ...
Adding more loads in series increases R_T, so the current from the source decreases, and lamps get dimmer.
Check with Ohm's law for the whole circuit: R_T = 100 Ω + 200 Ω = 300 Ω, and I = V_T ÷ R_T = 9.0 V ÷ 300 Ω = 0.030 A = 30 mA. That matches the ammeters!
Worked exampleExample 1: Solve a series circuit completely
A 12 V battery is connected in series with three resistors: R₁ = 2.0 Ω, R₂ = 4.0 Ω and R₃ = 6.0 Ω. Find the total resistance, the current, and the potential difference across each resistor.
- R_T = 2.0 + 4.0 + 6.0 = 12 Ω
- I = V_T ÷ R_T = 12 V ÷ 12 Ω = 1.0 A (the same through each resistor)
- V₁ = I × R₁ = 1.0 × 2.0 = 2.0 V; V₂ = 1.0 × 4.0 = 4.0 V; V₃ = 1.0 × 6.0 = 6.0 V
- Check: 2.0 + 4.0 + 6.0 = 12 V ✓ (Kirchhoff's voltage law)
R_T = 12 Ω, I = 1.0 A, and V₁ = 2.0 V, V₂ = 4.0 V, V₃ = 6.0 V.
Worked exampleExample 2: Find a missing value
A 9.0 V battery runs two lamps in series. A voltmeter reads 5.5 V across lamp 1, and the current is 0.25 A. Find the potential difference across lamp 2 and the resistance of each lamp.
- V₂ = V_T − V₁ = 9.0 − 5.5 = 3.5 V
- R₁ = V₁ ÷ I = 5.5 ÷ 0.25 = 22 Ω
- R₂ = V₂ ÷ I = 3.5 ÷ 0.25 = 14 Ω
- Check: R_T = 22 + 14 = 36 Ω, and I = 9.0 ÷ 36 = 0.25 A ✓
V₂ = 3.5 V, R₁ = 22 Ω, R₂ = 14 Ω.
Worked exampleExample 3: Protecting an LED
An LED needs 2.0 V across it and a current of 20 mA. You want to run it from a 9.0 V battery. What resistor should be placed in series with it?
- The LED and resistor are in series, so they share the 9.0 V: V_resistor = 9.0 − 2.0 = 7.0 V.
- The current is the same through both: I = 20 mA = 0.020 A.
- R = V ÷ I = 7.0 V ÷ 0.020 A = 350 Ω.
A 350 Ω resistor in series. (Real resistors come in standard values; you would choose the next larger one, such as 390 Ω, so the current stays a little below 20 mA.)
| Quantity | Series rule | In words |
|---|
| Current | I_T = I₁ = I₂ = ... | The same everywhere |
| Potential difference | V_T = V₁ + V₂ + ... | Shared among the loads |
| Resistance | R_T = R₁ + R₂ + ... | Adds up; more loads → more resistance → less current |
Series circuits in real life
- Switches and fuses are always placed in series with the devices they control or protect.
- Cells in series add their voltages: four 1.5 V cells in series give 6.0 V.
- A big disadvantage: if one load fails (opens), the path is broken and everything stops.
Hands-on activitySeries circuit lab (virtual or real)
You need: the free PhET simulation "Circuit Construction Kit: DC", or a real kit (a holder for 2 AA cells, 2 or 3 small flashlight bulbs in holders, wires, and a multimeter).
- Build a series circuit with one bulb. Record the brightness, the current, and the potential difference across the bulb.
- Add a second bulb in series. Record the current at three places and the potential difference across each bulb and across the battery.
- Add a third bulb and repeat.
- Unscrew (or, in the simulation, remove) one bulb. Record what happens.
- Write a conclusion: what happens to the current, the brightness and the potential differences as bulbs are added?
What to look for: the current is the same at every point, it decreases as bulbs are added, and the bulbs get dimmer. The bulb voltages always add up to the battery voltage. Removing one bulb turns them all off.
Safety (grown-up note): Use only AA batteries. Never connect a wire straight from one battery terminal to the other. Small bulbs can get warm; let them cool before touching.
Just for funSome older strings of holiday lights were wired in series. When one bulb burned out, the whole string went dark, and someone had to test the bulbs one at a time to find the culprit. A true Canadian winter tradition: untangling lights and playing detective.
Your turn
1In a series circuit, the current through R₁ is 0.40 A. What is the current through R₂?
0.40 A. The current is the same everywhere in a series circuit.
2A 6.0 V battery is connected to two resistors in series. The potential difference across R₁ is 2.5 V. What is it across R₂?
3Find the total resistance of 15 Ω, 25 Ω and 10 Ω in series.
R_T = 15 + 25 + 10 = 50 Ω.
4A 9.0 V battery is connected to 30 Ω and 60 Ω in series. Find the current.
R_T = 90 Ω. I = 9.0 ÷ 90 = 0.10 A.
5For the circuit in the last question, find the potential difference across each resistor.
V₁ = 0.10 × 30 = 3.0 V; V₂ = 0.10 × 60 = 6.0 V. Check: 3.0 + 6.0 = 9.0 V ✓.
6What happens to the current in a series circuit when another lamp is added? Why?
The current decreases, because the total resistance increases while the battery's voltage stays the same.
7Why does removing one bulb from a series string turn off all of them?
There is only one path. Removing a bulb leaves a gap, so the circuit is open and no current flows anywhere.
More practice: warm-up
1How many paths for current does a series circuit have?
2Write the rule for total resistance in series.
3Three 1.5 V cells are connected in series. What is the total voltage?
4Two equal resistors are in series with a 12 V battery. What is the potential difference across each?
They share it equally: 6.0 V each.
5True or false: in a series circuit, the resistor nearest the battery gets the most current.
False. All resistors in series get the same current.
More practice: on your own
1A 24 V supply is connected to resistors of 4.0 Ω, 8.0 Ω and 12 Ω in series. Find R_T, I, and the voltage across each resistor.
R_T = 24 Ω. I = 24 ÷ 24 = 1.0 A. V = 4.0 V, 8.0 V, 12 V (sum 24 V ✓).
2Two lamps are in series with a 12 V battery. The current is 0.50 A and lamp 1 has a resistance of 10 Ω. Find the resistance of lamp 2.
R_T = 12 ÷ 0.50 = 24 Ω. R₂ = 24 − 10 = 14 Ω.
3An LED needs 3.0 V at 15 mA. What resistor is needed in series to run it from a 6.0 V battery?
V_R = 6.0 − 3.0 = 3.0 V. R = 3.0 ÷ 0.015 = 200 Ω.
4A 100 Ω and an unknown resistor are in series with 9.0 V. The voltmeter across the 100 Ω resistor reads 3.0 V. Find the unknown resistance.
I = 3.0 ÷ 100 = 0.030 A. V across the unknown = 9.0 − 3.0 = 6.0 V. R = 6.0 ÷ 0.030 = 200 Ω.
5A student says, "The first bulb in a series circuit uses up some current, so the second bulb gets less." Explain the error.
Current is not used up. The same charges flow through both bulbs, so the current is the same. What the bulbs use is energy, shown by the potential difference across each one.
Challenge
1Three identical lamps are connected in series to a 9.0 V battery and the current is 0.30 A. One lamp is replaced with a wire (shorted out). Find the new current and the new potential difference across each remaining lamp. (Assume the lamps' resistance does not change.)
Each lamp: R_T = 9.0 ÷ 0.30 = 30 Ω, so each is 10 Ω. With 2 lamps: R_T = 20 Ω, I = 9.0 ÷ 20 = 0.45 A. Each lamp gets 0.45 × 10 = 4.5 V. The remaining lamps get brighter.
2You have a 12 V battery and want exactly 3.0 V across a 60 Ω device. What resistor should go in series with it?
I = 3.0 ÷ 60 = 0.050 A. The other resistor must take 12 − 3.0 = 9.0 V. R = 9.0 ÷ 0.050 = 180 Ω. (Check: R_T = 240 Ω, I = 12 ÷ 240 = 0.050 A ✓.) This is called a voltage divider.
3Design an investigation to test the hypothesis: "Adding resistors in series increases total resistance by the value of each resistor." Include what you measure and how you would analyse the data.
Use a fixed battery voltage. Measure the current with 1, 2, 3 and 4 identical resistors (for example 100 Ω each) in series, and the battery voltage. Calculate R_T = V ÷ I for each. If the hypothesis is right, R_T will be about 100, 200, 300 and 400 Ω, and a graph of R_T against the number of resistors will be a straight line with slope 100 Ω. Control: the same battery, wires and temperature.
Quiz
1. In a series circuit, the current:
2. Resistors of 5 Ω, 10 Ω and 15 Ω are in series. R_T =
3. A 12 V battery runs two resistors in series. One has 8.0 V across it. The other has:
4. What happens when a third identical lamp is added in series to two lamps?
5. A 6.0 V battery is connected to 10 Ω and 20 Ω in series. The current is:
6. Which statement is Kirchhoff's voltage law for a series circuit?
7. A 9.0 V battery, a 100 Ω resistor and a 200 Ω resistor are in series. The potential difference across the 200 Ω resistor is:
Big idea: Expectation D2.6 asks students to construct series and parallel circuits and compare current, potential difference and resistance in them. This lesson covers series circuits; the next covers parallel. Calculations also practise D2.5.
Common difficulties:
- "Current gets used up." It does not; energy does. The current is the same everywhere in series.
- Forgetting to check that the voltages add up to the source voltage: this is the best self-check.
- Subscripts. R_T means total resistance; R₁, R₂ are the individual resistors.
How to help: Ask your teen to solve a circuit completely (R_T, I, each V) and then check with V₁ + V₂ = V_T. The PhET simulation is a safe way to build circuits at home.