Answer to Question #252542 in Electrical Engineering for Kehinde

Question #252542
  • The electrical load in a small workshop consists of 14 lamps, each rated at 240volt, 60watt and 3 electric fire, each rated at 240, 1kw. What is the effective resistance of the load
1
Expert's answer
2021-10-18T03:45:33-0400

Lamp Ratings = 240V, 60W

Therefore P = 60 W, and V = 240 V

Power P is given by: "P = \\frac{V^2}{R}"

Hence Lamp Resistance "R = \\frac{V^2}{P} = \\frac{240 * 240}{60} = 960 Ohms"

Current drawn by each lamp = "I = \\frac{V}{R} = \\frac{240}{960} = 0.25 Amp."

Total current drawn by 14 lamps = 14 * 0.25 = 3.5 amp.


Electric Fire Ratings: 240 V, 1 KW

V = 240, P = 1KW = 1000 W

current drawn by each electric fire = "\\frac{P}{V} = \\frac{1000}{240} = 4.17 amp."


Therefore, total current drawn by 3 electric fires = 3 * 4.17 = 12.50 amp.


Total current drawn by 14 lamps and 3 electric fires = 3.5 + 12.50 = 16 amp.


Voltage = 240V

Current Load = 16 Amp.

Therefore, the effective resistive load = "\\frac{V}{I} = \\frac{240}{16} = 15 Ohms"


Final Answer:

The effective resistive load = 15 ohms



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