Question

Above has voltage rating of 220 volt and power rating of 40 watt what is the resistance that can be connected with the bulb so that it will glow with normal brightness if a voltage source of EMF 330 volt is available?? Please answer fast I have my exam tmw

Answer 1

Answer:

1512.5

Explanation:

v1=220

p1=40

p=v^2/R

R=v^2/p = 1210

p=v^2/R

R=330^2/40

2722.5

Net = 2722.5-1210 = 1512.5