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V=IRPUPIL APP
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1 · Type your first name
2 · Pick your level
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Build it up step by step
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A bit more challenge
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Challenge
Push yourself
Type your name and pick a level to start.
Quick-fire recall
1. Current is measured in
2. Voltage is measured in
3. Which component pushes the current round the circuit?
the wirethe cell / batterythe switch
4. 🔭 Today's lesson: if a wire makes it harder for current to flow, we say it has a high
🪜 Stuck? "Victor has Volts." Amps = current. Ohms (Ω) = resistance — the thing that slows current down.
1. Resistance is measured in
2. Which instrument measures current?
3. In a circuit diagram, a small rectangle is the symbol for a
cellresistorswitch
4. 🔭 Today's lesson: if the battery pushes harder (more voltage) and the resistance stays the same, the current
More push, same blockage → more flow.
1. An ammeter is connected in
2. A voltmeter is connected in
3. The symbol Ω stands for
4. 🔭 Today's lesson: the voltage doubles but the resistance stays the same. Predict the current:
it halvesit doublesit stays the same
🚀 Challenge habit: for every answer, finish the sentence in your head: "I know because…"
Read the circuit
Watch the orange dots — that's the current flowing.
1. Name the parts:
A = B = C =
2. 🔮 Predict: we put a SECOND bulb into the same loop. The first bulb gets
brighterdimmerstays the same
🪜 Symbol help: cell = the small block · switch = the gap in the wire · bulb = circle with an ✕ · ammeter = circle with an A. Two bulbs in one loop must share the push.
1. Name the parts:
B = C = D =
2. 🔮 Predict: we add a second bulb into the loop. The reading on D will
,
because the total goes up.
More components in one loop = more total resistance = less current.
1. Component D is an .
To measure the voltage across C you would connect a voltmeter in
.
2. 🔮 Predict: a second identical bulb is added into the loop, doubling the resistance. The current
,
and each bulb now shares the battery's .
🚀 Think like a scientist: use the words proportional and shared when you explain your prediction.
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Watch your teacher
Your teacher is teaching the I Do for V = I × R. The next task will open when it's time.
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✅ You can continue now — press Continue.
Use the triangle
📐 Need the triangle? Tap here
Cover the one you want to find.
1. V = 12 V, R = 4 Ω. Cover I → I = V ÷ R. I = 12 ÷ 4 = A
2. V = 9 V, I = 3 A. Cover R → R = V ÷ I. R = 9 ÷ 3 = Ω
🪜 The four steps: 1️⃣ Cover the letter you want. 2️⃣ Read what's left. 3️⃣ Do the maths. 4️⃣ Write the unit!
1. I = 3 A, R = 4 Ω. Find V. V = 3 × 4 = V
2. V = 20 V, R = 5 Ω. Find I. I = 20 ÷ 5 = A
Finding V → multiply (I × R). Finding I or R → divide V by the other one.
1. V = 24 V, R = 8 Ω. Find I. I = A
2. V = 240 V, I = 4 A. Find R. R = Ω
3. If V stays the same and R doubles, the current will .
🚀 Sense-check: same push, bigger blockage → the current must get smaller. Does your answer obey that?
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Watch your teacher
Your teacher is teaching the I Do for series & parallel circuits. The next task will open when it's time.
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✅ You can continue now — press Continue.
Apply the rules
Series: one loop — the current is the same everywhere, resistances add.
1. Series 2 Ω + 4 Ω. Total R = Ω
2. That circuit has a 12 V supply. Current I = 12 ÷ 6 = A
🪜 Two jobs: Series = ONE loop → ADD the resistances first. Then I = V ÷ your total.
1. Series 3 Ω + 5 Ω. Total R = Ω
2. 16 V supply on that circuit → current = A
3. In parallel, the voltage across each branch is .
Two steps: add the Ω, then divide the volts by your total.
1. Series 4 Ω + 6 Ω, 20 V supply. Total R = Ω, current = A
2. In that series circuit, the voltage across the 6 Ω resistor (V = I × R) = V
3. Why do parallel branches each get the full voltage?
🚀 Power move: voltage across ONE component = I × that component's resistance.
On your own
🪜 Your toolkit: V = I × R — cover the one you want. Series circuits: add the resistances first.
1. Complete the table (V = I × R)
| V | I | R |
|---|---|---|
| 2 | 5 | |
| 20 | 4 | |
| 18 | 3 |
2. Label the circuit:
A (power source) =
B (slows current) =
C (measures amps) =
3. Find the total resistance of this series circuit:
Total R = Ω
4. Series, ammeter reads 2 A at the battery — further round it reads
1. Complete the table (V = I × R)
| V | I | R |
|---|---|---|
| 3 | 6 | |
| 24 | 4 |
2. Use the diagram:
Total R = Ω, so current = A
Add the three resistors first, then I = V ÷ your total.
3. In a parallel circuit the battery current is .
1. A 230 V supply drives 5 A. Resistance = Ω
2. Two 6 Ω resistors in series on 24 V. Total R = Ω, current = A, voltage across ONE resistor = V
Equal resistors share the voltage equally → 24 ÷ 2.
3. Look at the parallel circuit:
Both bulbs shine at full brightness because .
Last few — show what stuck
🪜 I = V ÷ R · R = V ÷ I
1. V = 10 V, R = 2 Ω. I = 10 ÷ 2 = A
2. In a series circuit, the current is
the same everywhereshared between branchesused up as it goes
3. In a parallel circuit, the voltage across each branch is
shared and adds upthe same across each branchalways zero
1. V = 18 V, I = 3 A. R = Ω
2. Two equal bulbs in series share 12 V — each gets V
3. Total resistance of 4 Ω + 6 Ω in series = Ω
1. 12 V supply, total resistance 8 Ω. Current = A
2. Why is the current the same all the way round a series circuit?
3. Two resistors in parallel across 6 V — voltage across each = V