If R 1 and R 2 are the resistances of the filaments of 200 W and 100 W electric bulbs operated at 220 V , then R 1 R 2 is

Open MARKS Web Download App

Search any question & find its solution

Question: Answered & Verified by Expert

If

R_{1}

and

R_{2}

are the resistances of the filaments of

200 W

and

100 W

electric bulbs operated at

220 V

, then

(\frac{R_{1}}{R_{2}})

is

PhysicsCurrent ElectricityNEET

Options:

A $2$
B $0.5$
C $4$
D $0.25$

Solution:

2604 Upvotes Verified Answer

The correct answer is:

0.5

Power of bulb,

P = \frac{V^{2}}{R}

∴ \frac{P_{2}}{P_{1}} = \frac{R_{1}}{R_{2}}

Or

\frac{R_{1}}{R_{2}} = \frac{100}{200} = 0.5

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.

Use on iOS or Desktop Download from Google Play