Question

A fuse wire rated 15A is used in a house, when a heater of 2000 watt and N number of bulbs each 100W are connected in parallel with a supply of 220 V. Maximum of N ?

Answer 1

Hello Dear.

Here is the answer---

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Power of the Heater(P) = 2000 W
 Voltage of the Mains(V) = 220 V.

Using the Formula,

   P = V × I
⇒  I =  2000/220
⇒ I = 100/11 A

Thus, the Excess current which can be safely used = 15 - 100/11
                                                                                  = (165-100)/11
                                                                                  =  65/11 A.

Now,
Power of Each Bulbs = 100 W
Voltage = 220 V          
[Since, the Heater and Bulbs are connected in Parallel]

Thus, Current in each Bulb = P/V
                                            = 100/220
                                            = 5/11 A.


Now,

  Number of Bulbs(N)  = $\frac{Excess Current Available}{Current Drawn by each bulb}$
                                  = $\frac{65/11}{5/11}$
                                  = 13 Bulbs.


Thus,the number of Bulbs N s 13.



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Hope it helps.

Have a Nice day.

Answer 2

Number of bulbs = N
bulbs are connected in parallel with a supply of 220V .
power of each bulb = 100 W

for maximum number of bulbs
total Power dissipated = Power dissipated by all bulbs + power dissipated by heater
So,Pnet = (P₁ + P₂ + P₃ + ...... +Pₙ ) + Power of heaters
Vi = (100 + 100 + 100 + .... + 100) + 2000W

Given,
V = 220V and i = 15A
∴ 220 × 15 = 100N + 2000W
3300 - 2000 = 100N
1300 = 100N
N = 13

Hence, number of bulbs = 13