This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Here are the solutions for questions A16 and A17: A16 Which fuse is most suitable for use in a 2000W electric heater connected to a 240V supply? Step 1: Calculate the operating current (I) of the electric heater using the formula P = VI, where P is power and V is voltage. I = (P)/(V) I = 2000 W240 V I = (200)/(24) A I = (25)/(3) A I ≈ 8.33 A Step 2: Choose a fuse rating that is slightly higher than the operating current to allow for normal operation but will blow if there is an overload. Comparing 8.33 A with the given options: A: 3 A (too low) B: 5 A (too low) C: 10 A (suitable, as it's the next standard rating above 8.33 A) D: 13 A (also above, but 10 A provides closer protection) The most suitable fuse is 10 A. The correct option is C. --- A17 What is the operating resistance of an electric lamp rated 60W, 240V? Step 1: Use the formula relating power (P), voltage (V), and resistance (R): P = (V^2)/(R). Rearrange the formula to solve for resistance: R = (V^2)/(P) Step 2: Substitute the given values into the formula. R = (240 V)^260 W R = (240 × 240)/(60) R = (57600)/(60) R = 960 Step 3: Compare the calculated resistance with the given options. A: 4 B: 300 C: 960 D: 14 400 The calculated resistance is 960 . The correct option is C.