10 GCSE Physics questions about the video doorbell, from Foundation to Higher, each with a full worked solution.
A video doorbell runs on a rechargeable battery. A passive infrared (PIR) sensor spots the infrared given off by a warm person and wakes the camera up. The video is sent to your phone using radio waves (Wi-Fi), and at night infrared LEDs light up the doorstep so the camera can still see.
Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Topics: Electricity · Waves
📌 KEY DEFINITIONS – learn them word for word
Charge flow: the flow of electrical charge round a circuit. Current is the rate of flow of charge.
Radio waves: transverse electromagnetic waves with the longest wavelength. They are used for communication, such as Wi-Fi and TV.
Infrared radiation: emitted by all objects. The hotter the object, the more infrared radiation it emits in a given time.
Equations you’ll use
| Quantity | Symbol | Unit |
|---|---|---|
| charge flow | Q | coulombs (C) |
| current | I | amps (A) |
| time | t | seconds (s) |
| energy transferred | E | joules (J) |
| potential difference | V | volts (V) |
| power | P | watts (W) |
| wave speed | v | metres per second (m/s) |
| frequency | f | hertz (Hz) |
| wavelength | λ | metres (m) |
Use FIFA for every calculation
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Foundation: aim for Q1–8. Higher: aim for Q4–10.
Questions
Q1 (F) The doorbell draws a current of 0.05 A for 60 s while it records a visitor. Calculate the charge that flows. (Q = I t)
Show solution
F: Q = I t
I: Q = 0.05 × 60
F: Q is already the subject
A: 3 C
Q2 (F) The battery has a potential difference of 3.6 V. Calculate the energy transferred when 3 C of charge flows through the doorbell. (E = Q V)
Show solution
F: E = Q V
I: E = 3 × 3.6
F: E is already the subject
A: 10.8 J
Q3 (F) The doorbell sends video to the Wi-Fi router using radio waves with a frequency of 2.4 GHz. Radio waves travel at 3.0 × 108 m/s. Calculate the wavelength.
Show solution
Convert first: 2.4 GHz = 2.4 × 109 Hz
F: v = f λ
I: 3.0 × 108 = 2.4 × 109 × λ
F: λ = 3.0 × 108 ÷ 2.4 × 109
A: 0.125 m (12.5 cm)
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Q4 (F) The motion sensor is a passive infrared (PIR) sensor. Explain how it can detect a person walking up to the door. [3 marks]
Show answer
• All objects emit infrared radiation.
• A person is warmer than the doorstep and the air around it, so they emit more infrared.
• The sensor detects the change in infrared reaching it as the person moves, and switches the camera on.
Q5 (F/H) When fully charged, the battery can supply a current of 0.25 A for 24 hours before it runs flat. Calculate the charge it stores, in coulombs.
Show solution
Convert first: 24 hours = 24 × 60 × 60 = 86 400 s
F: Q = I t
I: Q = 0.25 × 86 400
F: Q is already the subject
A: 21 600 C
Q6 (F/H) On standby the doorbell draws an average current of 2.5 mA. Using your answer to Q5, calculate how many days the battery lasts.
Show solution
Convert first: 2.5 mA = 0.0025 A
F: Q = I t
I: 21 600 = 0.0025 × t
F: t = 21 600 ÷ 0.0025 = 8 640 000 s
A: 8 640 000 ÷ (24 × 3600) = 100 days
Q7 (F/H) The battery stores 21 600 C at a potential difference of 3.6 V. Calculate the energy it stores in kilojoules.
Show solution
F: E = Q V
I: E = 21 600 × 3.6
F: E is already the subject
A: 77 760 J ≈ 78 kJ
Q8 (F/H) When someone opens the app to watch a live video, the doorbell draws 0.40 A from the 3.6 V battery. Calculate the energy transferred during a 2-minute live view.
Show solution
Step 1 – power: P = V I = 3.6 × 0.40 = 1.44 W
Convert: 2 minutes = 120 s
F: E = P t
I: E = 1.44 × 120
F: E is already the subject
A: 173 J (about 170 J)
Q9 (H) At night, infrared LEDs light up the doorstep. The infrared has a wavelength of 850 nm. Calculate its frequency. (speed of light = 3.0 × 108 m/s)
Show solution
Convert first: 850 nm = 850 × 10−9 m = 8.5 × 10−7 m
F: v = f λ
I: 3.0 × 108 = f × 8.5 × 10−7
F: f = 3.0 × 108 ÷ 8.5 × 10−7
A: 3.5 × 1014 Hz
Q10 (H) Explain why the doorbell uses radio waves, not visible light, to send video to a router inside the house, and why the signal can be weak if the router is far away behind several brick walls. [4 marks]
Show answer
• Radio waves are transmitted through (pass through) walls, but visible light is absorbed or reflected by them.
• So radio waves can reach the router without a clear line of sight.
• Some of the radio wave energy is absorbed by each wall it passes through …
• … and the waves spread out over a larger area with distance, so less energy reaches the router.
⚠️ Where students lose marks
- Leaving time in hours when using Q = I t: multiply hours by 3600 to get seconds, so the charge comes out in coulombs.
- Using GHz or nm without converting to Hz and metres before using v = f λ.
- Saying the PIR sensor “sends out” infrared. It is passive: it only detects the infrared a person emits.
- Mixing up E = Q V (charge) and E = P t (power) – check which quantities the question gives you.
More devices: all 40 everyday devices · Electricity · Waves
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