Series and Parallel Circuits Lesson Plan with a Virtual Lab
Updated 2026-10-02
This series and parallel circuits lesson plan uses a free virtual lab so every pair can build, break and measure circuits in one class period, with meters that never run flat. Students find out why the current is the same everywhere in a series loop, why parallel branches each get the full battery voltage, and why adding a branch makes the battery work harder. You get goals, setup, predictions, the readings students should get, a five-question set for the class link, an extension and differentiation. Every number comes from the simulation's own circuit solver.
If your class still needs V = IR itself, teach the Ohm's law lesson plan first. This lesson builds on it.
Lesson at a glance
- Level: grades 8–10 (ages 13–16), physical science or introductory physics.
- Time: one 55–60 minute period, plus an optional 20-minute extension.
- Prior knowledge: current, voltage and resistance, V = IR, milliamperes.
- Format: pairs on laptops or tablets, or the whole class with one projector.
- Simulations: DC circuit construction kit. Extension: Ohm's law – two resistors in series.
Learning goals
By the end of the lesson, students can:
- State that the current is the same at every point of a series circuit.
- State that the voltages across series components add up to the battery voltage.
- State that parallel branches each get the full battery voltage, and that the branch currents add up to the battery current.
- Calculate the total resistance of two resistors in series (R₁ + R₂) and in parallel (1/R = 1/R₁ + 1/R₂).
- Explain why bulbs in parallel glow brighter than the same bulbs in series.
What the simulation does
The kit is a free-form circuit board with two screens.
- Intro: drag wires, batteries, resistors, bulbs and switches from the box onto the board. Every battery is 9 V and every bulb and resistor is 10 Ω. Tap a component to see its voltage U, current I and power P. A voltmeter and an ammeter sit in the tray below the box.
- Lab: tap a component and set its value with a slider: batteries 0–120 V, bulbs and resistors 0–120 Ω, in steps of 0.5. Advanced sets the battery's internal resistance.
The Starting circuit setting loads a ready-made circuit on both screens: "Battery, switch and bulb", "Two bulbs in series", "Two bulbs in parallel", or "Resistor in series with two parallel bulbs". That saves building time.
A few details worth knowing before class:
- Readouts use three significant figures. Currents below 1 A appear in milliamperes, for example 450 mA, and larger ones in amperes, for example 1.80 A.
- Brightness follows power. Bulbs glow brighter as P = I²R grows.
- The voltmeter has a sign. It shows the red probe's potential minus the black one's. If a reading is negative, swap the probes.
Materials and setup before class
Materials: one device per pair (or a projector), the data tables below on paper, and a calculator.
Setup (10 minutes, once):
- Open the simulation and set Starting circuit to "Two bulbs in series" in the Starting values panel under it.
- Click Share and create a link for each class, for example "Science · Period 2". The link pins this starting circuit.
- Optional: on the Questions tab, enter the question set below and attach it to the link. The two predictions lock the simulation until each student commits. See the Predict–Observe–Explain guide for why this works.
- Post the link, or open it in present mode and show the QR code.
Students change the starting circuit themselves later, in the same panel. Changing it rebuilds both screens and returns to the Intro tab.
Lesson sequence
1. Hook and predictions (8 minutes)
Project the simulation with Starting circuit set to "Battery, switch and bulb". Tap the bulb: it reads U = 9.00 V and I = 900 mA. Tap the switch to open and close it while the class watches the electrons stop and start.
Then ask for two predictions, on paper or on the link:
- P1. "Two identical bulbs go on the same battery, first in series, then in parallel. When do they glow brighter?"
- P2. "Two bulbs are in parallel. You add a third bulb as another branch. Does the current through the battery go up, go down or stay the same?"
Many students expect the battery to give out a fixed current, so they predict the same brightness for P1 and "goes down" for P2. Don't correct anyone yet.
2. Series: one path, shared voltage (12 minutes)
Pairs open the link ("Two bulbs in series"). They tap each bulb, read the readout line "Current through Battery 1", then drag the ammeter onto three different wires. Finally they place both voltmeter probes across each bulb, then across both bulbs together.
| Measurement | Expected reading |
|---|---|
| Current through the battery | 450 mA |
| Current in bulb 1 and bulb 2 | 450 mA each |
| Ammeter on any wire | 450 mA |
| Voltage across each bulb | 4.50 V |
| Voltage across both bulbs | 9.00 V |
| Power of each bulb | 2.02 W |
Ask: "Is the current used up by the first bulb?" No: 450 mA before, between and after the bulbs. The voltage is what gets shared: 4.50 V + 4.50 V = 9.00 V.
Then drag one bulb back into the box to remove it. The other bulb goes dark and the current drops to zero. One gap breaks the only path.
3. Parallel: full voltage, split current (12 minutes)
In the Starting values panel, pairs set Starting circuit to "Two bulbs in parallel" and repeat the measurements.
| Measurement | Expected reading |
|---|---|
| Current through the battery | 1.80 A |
| Current in each bulb | 900 mA |
| Ammeter on the wire next to the battery | 1.80 A |
| Ammeter on one branch | 900 mA |
| Voltage across each bulb | 9.00 V |
| Power of each bulb | 8.10 W |
Each bulb behaves as if it were alone on the battery: 9.00 V and 900 mA, the same as the single bulb in the hook. The branch currents add up: 900 mA + 900 mA = 1.80 A. Now remove one bulb. The other stays lit at 900 mA, and the battery current falls back to 900 mA.
Check P1. Parallel bulbs are brighter because each one uses 8.10 W, four times the 2.02 W of a series bulb. Check P2 the same way: a third branch draws another 900 mA, so the battery current rises to 2.70 A. Fast pairs can build that third branch from the box to see it.
4. Different resistances on the Lab screen (14 minutes)
Pairs open the Lab tab, where the same starting circuit waits. They tap a bulb, set its resistance with the slider and read every value again. For the parallel rows, they change Starting circuit and reopen the Lab tab.
| Circuit (9 V battery) | Battery current | Bulb 1 | Bulb 2 |
|---|---|---|---|
| Series, 10 Ω and 20 Ω | 300 mA | 3.00 V, 300 mA | 6.00 V, 300 mA |
| Parallel, 10 Ω and 20 Ω | 1.35 A | 9.00 V, 900 mA | 9.00 V, 450 mA |
| Series, 20 Ω and 40 Ω | 150 mA | 3.00 V, 150 mA | 6.00 V, 150 mA |
Ask pairs for two patterns. Good answers: "In series, the bigger resistance takes the bigger share of the voltage, in the same ratio" and "In parallel, the bigger resistance takes the smaller share of the current."
Then find the total resistance from the battery readings:
- Series: 9 V ÷ 0.300 A = 30 Ω, which is 10 Ω + 20 Ω.
- Parallel: 9 V ÷ 1.35 A ≈ 6.67 Ω, which matches 1/R = 1/10 + 1/20.
The key idea: a parallel branch adds a path, so the total resistance is smaller than the smallest branch. That is why the battery current rose in P2.
5. A mixed circuit (6 minutes)
Set Starting circuit to "Resistor in series with two parallel bulbs" and return to the Intro tab. Ask pairs to predict the readings before tapping:
- The two parallel bulbs act like 5 Ω, so the total is 10 Ω + 5 Ω = 15 Ω.
- Battery current: 9 V ÷ 15 Ω = 600 mA. The resistor carries all 600 mA and takes 6.00 V.
- Each bulb gets the remaining 3.00 V and carries 300 mA.
The simulation agrees, and the dim bulbs show that they now share the voltage with the resistor.
6. Exit check (5 minutes)
Use questions 3–5 of the set below. Then open View answers and show the Prediction and After columns for P1 and P2 side by side.
Question set for this lesson
Enter these on the simulation's Questions tab. Suggested Instructions for students: "Use the Starting circuit setting in the panel under the simulation. Tap a component to read its voltage and current."
1. Multiple choice · Before, as a prediction · Ask again after the simulation
- Question: "Two identical bulbs go on the same battery, first in series, then in parallel. When do they glow brighter?"
- Options: In series / In parallel (correct) / The same in both / It depends on which bulb is nearer the battery
- Explanation: "In parallel, each bulb gets the full 9 V, carries 900 mA and uses 8.10 W. In series, the bulbs share the 9 V, so each gets 4.50 V, the current is only 450 mA and each bulb uses about 2 W. Brightness follows power, so the parallel bulbs are brighter."
2. Multiple choice · Before, as a prediction · Ask again after the simulation
- Question: "Two bulbs are in parallel on a 9 V battery. You add a third identical bulb as another branch. The current through the battery…"
- Options: increases (correct) / decreases / stays the same / drops to zero
- Explanation: "Every branch gets the full 9 V and draws 900 mA, whatever the other branches do. Two branches draw 1.80 A and three draw 2.70 A. Each new branch is another path, so the total resistance goes down and the battery current goes up."
3. Number · After the simulation
- Question: "On the Lab screen, use Two bulbs in series. Set one bulb to 10 Ω and the other to 20 Ω. What is the voltage across the 20 Ω bulb?"
- Answer: 6.00, tolerance ± 0.05, unit V
- Explanation: "The total resistance is 30 Ω, so the current is 9 V ÷ 30 Ω = 300 mA everywhere. The 20 Ω bulb takes 0.300 A × 20 Ω = 6.00 V and the 10 Ω bulb takes 3.00 V. The larger resistance gets the larger share, and the two add up to 9 V."
4. Number · After the simulation
- Question: "Now use Two bulbs in parallel with the same values, 10 Ω and 20 Ω. What is the current through the battery?"
- Answer: 1.35, tolerance ± 0.02, unit A
- Explanation: "Both branches get the full 9 V. The 10 Ω branch carries 900 mA and the 20 Ω branch 450 mA. The battery supplies both: 0.90 A + 0.45 A = 1.35 A. That means the pair acts like a single 6.67 Ω resistor, smaller than either one."
5. Short answer · After the simulation
- Question: "Explain why two resistors in parallel have a smaller total resistance than either resistor alone."
- Accepted answers (optional): *path*
- Model answer: "A parallel branch gives the current another path. The battery pushes the same voltage through each path, so the currents add up and the total current is bigger than through one resistor alone. A bigger current for the same voltage means a smaller total resistance."
Questions 1 and 2 are the predictions, asked again after the simulation so students can compare. The formative assessment guide explains how to read the results across classes.
Extension: voltage sharing and a real battery
Two resistors in series (10 minutes). The series resistor simulation prints I = U/(R1 + R2), U1 and U2 for a source and two resistors. Students set the values in its Starting values panel. With the defaults, U = 12 V, R1 = 10 Ω and R2 = 20 Ω, it shows I = 0.400 A, U1 = 4.00 V and U2 = 8.00 V.
Challenge pairs to make U1 exactly 3.00 V with U = 9 V. One answer: R1 = 15 Ω and R2 = 30 Ω give I = 0.200 A, U1 = 3.00 V and U2 = 6.00 V. Any pair where R2 is twice R1 works, which is the voltage divider rule in action.
Internal resistance (10 minutes). On the kit's Lab screen, set Battery internal resistance under Advanced to 1 Ω. With one 10 Ω bulb, the current drops to 818 mA and the battery's terminal voltage to 8.18 V. With "Two bulbs in parallel", the battery supplies 1.50 A and each bulb only gets 7.50 V. Ask: "Why does adding a branch make the bulbs dimmer now?" The bigger current loses more voltage inside the battery itself.
Differentiation
Support:
- Give the tables with the measurement names filled in, so students only read and record.
- Skip step 4 and the mixed circuit; compare only equal bulbs in series and in parallel.
- Give sentence starters: "In a series circuit, the current is…" and "In a parallel circuit, each bulb gets…"
Stretch:
- Find the single resistor that could replace the mixed circuit (15 Ω), set it on the Lab screen, and confirm that the battery current is the same 600 mA.
- On the Lab screen, tick Real bulbs (resistance rises as the bulb heats up), set a 10 Ω and a 20 Ω bulb in series, and explain why the voltages are no longer exactly 3.00 V and 6.00 V.
English learners: the simulation is available in six languages. Create a second link in the student's language so the labels are familiar while the discussion stays in English.
Standards alignment
This lesson fits the electricity unit of middle school physical science, GCSE Physics or introductory physics. NGSS has no performance expectation dedicated to series and parallel circuits, so we don't claim one. The lesson does exercise Planning and Carrying Out Investigations and Analyzing and Interpreting Data.
For a full lab write-up, see how to create a virtual lab activity. For more ideas in this unit, see interactive physics lesson ideas, and for general routines, how to use interactive simulations in the classroom.
FAQ
Why do the series bulbs read 2.02 W and not 2.03 W?
The wires and the battery are given a tiny resistance so the solver never divides by zero. That makes the current a fraction below 450 mA, and the power rounds to 2.02 W. Current and voltage still read 450 mA and 4.50 V.
How do I stop students building short circuits?
You don't need to. A short circuit only sets a virtual wire on fire and shows a warning. Use it to explain fuses, which students can add on the Lab screen.
Can students do this on phones?
Yes. Students open the link without an account, and tapping and dragging work on touch screens. On small screens, stick to the starting circuits rather than building from the box.