Tuco tamps, onu taled how a doov and hu thar Gowot ao ose conmeced iw

1.	Tuco tamps, onu taled how a doov and hu thar Gowot ao ose conmeced iw posxodldl dechic suppluh. totaQampsonanaatt tor one OLd
Answer» We have given two lamps in such a way : Power of 1st lamp, P₁ = 40W , voltage of 1st lamp, V₁ = 220V power of 2nd lamp , P₂ = 60W , voltage of 2nd lamp ,V₂ = 220V we know, one things , Power = V²/R [ when potential difference is same then consider P = V²/R ] R = V²/P so, resistance of 1st lamp , R₁ = V₁²/P₁ = (220)²/60 = 4840/6 = 2420/3Ωresistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω energy consumed by lamps in one hour = Energy consumed by 1st lamps in one hour + energy consumed by 2nd lamps in one hour = P₁ × 1hour + P₂ × 1 hour =(P₁ + P₂) × 1 hour = (60W + 40W) × 1 hour = 100 Wh = 100/1000 KWh [ ∵ 1 KWh = 10²Wh ]= 0.1 KWh

Tuco tamps, onu taled how a doov and hu thar Gowot ao ose conmeced iw posxodldl dechic suppluh. totaQampsonanaatt tor one OLd

Answer»

We have given two lamps in such a way : Power of 1st lamp, P₁ = 40W , voltage of 1st lamp, V₁ = 220V power of 2nd lamp , P₂ = 60W , voltage of 2nd lamp ,V₂ = 220V

we know, one things , Power = V²/R [ when potential difference is same then consider P = V²/R ] R = V²/P so, resistance of 1st lamp , R₁ = V₁²/P₁ = (220)²/60 = 4840/6 = 2420/3Ωresistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω

energy consumed by lamps in one hour = Energy consumed by 1st lamps in one hour + energy consumed by 2nd lamps in one hour = P₁ × 1hour + P₂ × 1 hour =(P₁ + P₂) × 1 hour = (60W + 40W) × 1 hour = 100 Wh = 100/1000 KWh [ ∵ 1 KWh = 10²Wh ]= 0.1 KWh

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