Question

an electrical bulb is rated as 60 W-220 V .what is its resistance and safe limit of current through it.

Answer 1

$\huge\mathbb\red{ANSWER}$

R=805.86 ohm

$\huge\mathbb\red{EXPLANATION}$

WE HAVE,

P =60WATT. &. V =220 volt.

P = V×I

60=220×I

I=60/220

I= 0.273 ampere

NOW,

V =I×R

220=0.273×R

R=220/0.273

R=805.86 ohm.

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