Answer to Question #252542 in Electrical Engineering for Kehinde

Question

• 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

Answer 1

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