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Q. If two electric bulbs have 40 W and 60 W rating at 220 V, then the ratio of their resistances will be

BVP MedicalBVP Medical 2007

Solution:

Key Idea: The rate at which electrical energy is dissipated into other forms of energy is called electric power P. Electrical power $ ={{\varepsilon }_{0}}\,\,E_{rms}^{2} $ (V = iR) $ =8.85\times {{10}^{-12}}\times {{(720)}^{2}} $ $ =4.58\times {{10}^{-6}}J/{{m}^{3}} $ where Vis potential difference, R is resistance. Given, $ {{10}^{-10}}s $ $ f=\frac{v}{v-{{v}_{s}}}f $ $ {{v}_{s}} $ $ v=320m/s,\,{{v}_{s}}=4m/s,\,f=240Hz $ $ f=\frac{320}{320-4}\times 240 $ $ =243Hz $ Note: Higher the wattage of a bulb smaller is the resistance of its filament that is thicker is the filament.