Question

12. A cooler of 1500W, 200V and a fan
of 500W. 20OV are to be used from a
house hold supply. The rating of fuse
to be used is *
O 2.5 A
O 5.0 A
O 7.5A
10 A​

Answer 1

Given,

• A cooler of 1500W , 200V and a fan 500W.
• 200V are to be used from a house hold supply

To Find:-

• The rating of fuse to be used is?

Calculations:-

Here we use the formula is P=p1 + P2

• 1500+500
• =2000W

Current drawn from the supply

I = $\frac{P}{V}$

= $\frac{2000}{200}$

I = 10 A

so,

The correct answer is option.D

Answer 2

[d] “10A” .

$\mathcal{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

• A cooler of 1500W, 200V and a fan of 500W, 200V are to be used from a house hold supply .

CALCULATION :-

✍️ We have know that,

$\rm{\purple{Power(P)\:=\:Current(I)\:\times{Voltage(V)}\:}}$

✍️ Current in cooler = $\rm{\dfrac{1500}{200}\:=\:7.5\:A\:}$

And

✍️ Current drawn by fan = $\rm{\dfrac{500}{200}\:=\:2.5\:A\:}$

✍️ So, the total current drawn is,

• 7.5A + 2.5A = 10A

$\therefore\:\rm{\blue{The\:rating\:of\:the\:fuse\:should\:be\:\:\underline\bold{10\:A}\:.}}$