iGCSE Physics (0625) 4.3.1 Circuit diagrams and circuit components Paper 4 -Exam Style Questions- New Syllabus
Question
(i) Calculate the current in the 800 Ω lamp.
(ii) Calculate the energy used by the 800 Ω lamp in 5.0 hours. Give your answer in kWh.
(iii) Calculate the combined resistance of the two lamps in this circuit.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.3.1 — Circuit diagrams and circuit components (Part (a))
• Topic 4.2.4 — Resistance (Part (b)(i))
• Topic 4.2.5 — Electrical energy and electrical power (Part (b)(ii))
• Topic 4.3.2 — Series and parallel circuits (Part (b)(iii))
▶️ Answer/Explanation
Correct Answer: A circuit diagram with an a.c. source symbol (circle with a sine wave) connected to two lamp symbols (circle with an X) arranged in parallel branches.
Detailed solution: To draw the diagram, start with the standard symbol for an a.c. power supply. From the supply, the wires must split into two separate branches, each containing one lamp, before joining back together to return to the source. This parallel arrangement ensures that both lamps receive the full supply voltage of 230 V.
Correct Answer: 0.29 A (or 0.2875 A)
Detailed solution: In a parallel circuit, the potential difference across each branch is equal to the source voltage. Using Ohm’s Law, I= R V , we substitute the given values: I= 800 Ω 230 V . This calculation gives 0.2875 A, which is rounded to 0.29 A for two significant figures as per standard examination practice.
Correct Answer: 0.33 kWh
Detailed solution: Energy in kWh is calculated by multiplying Power in kW by Time in hours. First, find power: P=V×I=230 V×0.2875 A=66.125 W. Convert this to kilowatts by dividing by 1000 to get 0.066125 kW. Finally, E=P×t=0.066125 kW×5.0 h≈0.33 kWh.
Correct Answer: 480 Ω
Detailed solution: For two resistors in parallel, the combined resistance R p is calculated using the formula R p 1 = R 1 1 + R 2 1 or the product-over-sum method: R p = R 1 +R 2 R 1 ×R 2 . Substituting the values: R p = 1200+800 1200×800 = 2000 960000 . This simplifies to exactly 480 Ω. Note that the total resistance is less than the smallest individual resistance.