iGCSE Physics (0625) 1.7.4 Power Paper 4 -Exam Style Questions- New Syllabus
Question
Fig. 9.1 shows a mobile phone (cell phone) being charged on a wireless charging plate

(a) When the charging plate is switched on, there is an alternating current (a.c.) in the primary coil. A secondary coil is in the mobile phone. Explain how a current is produced in the secondary coil.
(b) The maximum energy stored in the battery of the mobile phone is 0.012 kW h.
(i) Show that this maximum energy is \(4.3 × 10^4\) J
(ii) The charging plate in Fig. 9.1 has a useful output power of 15 W. The phone manufacturer claims that the battery can be charged to 50% capacity in less than 30 minutes. Show that this claim is true.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.5.1 — Electromagnetic induction (Part (a))
• Topic 1.7.4 — Power (Part (b))
• Topic 4.2.5 — Electrical energy and electrical power (Part (b))
▶️ Answer/Explanation
Correct Answer: (alternating current / a.c. in primary coil / plate produces) changing magnetic field (in primary coil)
secondary / phone coil cuts (this) magnetic field OR secondary / phone coil is in this / changing magnetic field
(changing magnetic field) causes induced current (in secondary coil)
Detailed solution: An alternating current in the primary coil generates a continuously changing magnetic field around it. This varying magnetic field links with the secondary coil inside the phone. According to Faraday’s law of electromagnetic induction, the changing magnetic flux through the secondary coil induces an electromotive force (e.m.f.), which drives a current in the secondary coil.
Correct Answer: 1 kW h = \(1000\times 60\times 60 J\) AND \(0.012\times 3.6\times 10^6\) \((= 4.3\times 10^4 J)\)
Detailed solution: One kilowatt-hour (kW h) represents the energy used by a 1 kW device operating for 1 hour. Converting this to joules involves multiplying the power in watts (1 kW = 1000 W) by the time in seconds (1 h = 60 × 60 = 3600 s), giving 1 kW h = \(3.6 \times 10^6\text{ J}\). Multiplying the given capacity (0.012 kW h) by this conversion factor yields \(0.012 \times 3.6 \times 10^6 = 43200\text{ J}\), which rounds to \(4.3 \times 10^4\text{ J}\).
Correct Answer: 50% charged = \((4.3\times 10^4 / 2 =) 2.15\times 10^4\) (J) OR 63% charged in 30 min
OR (50% charged in t =) 24 min
P = E / t OR (t =) E / P OR \(2.15\times 10^4 / 15\) (s)
OR energy provided in 30 min = \((15\times 30 \times 60=) 2.7\times 10^4\) (J)
\(2.7\times 10^4 > 2.15\times 10^4\) OR 63% > 50% OR 30 min > 24 min
Detailed solution: To verify the claim, calculate the time needed to reach 50% capacity. The energy required for 50% charge is half the total energy: \(E = \frac{4.3 \times 10^4\text{ J}}{2} = 2.15 \times 10^4\text{ J}\). Using the power equation \(t = \frac{E}{P}\), the time taken is \(t = \frac{2.15 \times 10^4}{15} \approx 1433\text{ s}\), which is approximately 24 minutes. Since 24 minutes is less than the claimed 30 minutes, the manufacturer’s statement is validated.
