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Section-A1. (i) Powers of two electric bulb are 100 W and 200 WBoth are connected from mains of 220V. The ratio of resis-0tance of their filaments will be:10(a) 1:2(b) 2:1(a) 1.d) 4.1explain with solution |
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Answer» Given: P₁=100W, P₂=200W, and V=220V Answer: Power=Current×Potenial difference P=VI......(1) V=IR (Ohm's law) I=V/R....(2) Substituting (2) in (1); P=V²/R =>R=V²/P Risistance for 1ST bulb; R₁=(220)²/100 R₁=48400/100 Thus, R₁=484 Ohm......(3) Now, RESISTANCE for 2nd bulb; R₂=(220)²/200 and, R₂=242 Ohm......(4) Ratio of R₁and R₂: R₁:R₂=484/242 Therefore, R¹:R²=2:1Thus, OPTION (b) is the answer I hope it helps ^_^Please mark as Brainliest |
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