Question

a circuit has a fuse of rating 5 ampere. what is the maximum number of 40 watt (200 volt) bulbs that can be safely used in the circuit.

Answer 1

here given is current =5A
power of each bulb is=40w
voltage is= 200v

let the no.of can be used be X
power =volt×current
40×x=200×5
40x=1000
x=1000/40
x=25
Ans.25 bulb can be used in the circuit...


Hope it's help you☺

Answer 2

Total no of bulbs be x

Now current in one bulb

P=VI

I= 40/200

I = 1/5 ampere

Now

Total power =x*40

X*40= vi

X*40= 200/5

X = 200/200

X= 1