Filament resistance of 20 W bulb and 40 W bulb, both rated to same voltage of operation is in the ratio 

Concept:

When the bulb is rated to voltage, then power consumed by the bulb is calculated as

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

When the bulb is rated to current, then power consumed by the bulb is calculated as

P = I² R

Where,

R is the resistance of the bulb

Calculation:

Given that 20W bulb and 40W bulb are rated to same voltage (V₁) of operation, then

P₁ = 20 W, P₂ = 40 W

Let the resistance of bulbs are R₁ & R_­2 respectively

Now, \({P_1} = \frac{{{V^2}}}{{{R_1}}} = 20\) 

\(\therefore {R_1} = \frac{{{V^2}}}{{20}}\;{\rm{\Omega }}\)

Now, \({P_2} = \frac{{{V^2}}}{R} = 40\)

\(\therefore {R_2} = \frac{{{V^2}}}{{40}}\)

Hence the ratio of filament resistance is

\(\frac{{{R_1}}}{{{R_2}}} = \frac{{{V^2}}}{{20}} \times \frac{{40}}{{{V^2}}}\)

R₁ : R₂ = 2 : 1