In a large building, there are 15 bulbs of 40W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be
Text Solution
Verified by ExpertsThe correct answer is:
C
Total power (P) = (15 × 40) + (5 × 100) + (5 × 80) + (1 × 1000) = 2500W
P = VI
⇒ I = 
A
= 
= 11.3 A
Minimum capacity should be 12 A
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A galvanometer having a coil resistance of 60 Ω Ω shows full scale deflection when a current of 1.0…
A student measures the terminal potential difference (V) of a cell (of emf ε and internal resistanc…
See the electrical circuit shown in this figure. Which of the following equations is a correct equa…
A thermocouple of negligible resistance produces an e.m.f. of 40 µV/°C in the linear range of tempe…
In the circuit shown in the figure, if potential at point A is taken to be zero the potential at po…
A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The…