GATE EE 2014 Set 1 — Question 39

NAT+2 / -0MediumR, L, C & M ElementsNetwork Elements & SourcesElectric Circuits

Electric Circuits → Network Elements & Sources → R, L, C & M Elements

Last updated 5 September 2026

Question

40\text{ W}, 240\text{V}

R_{ON} = \frac{V^2}{P} = \frac{240^2}{40} = \frac{57600}{40} = 1440\ \Omega

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R_{ON} = \frac{V^2}{P} = \frac{240^2}{40} = \frac{57600}{40} = 1440\ \Omega

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