A house is fitted with `20` length of `60` watt each `10` fans consuming `0.5` ampere each an electric kettle of resistance `110 Omega`. If the energy is supplied at `120 V` and costs `150` paise `kWh`, calculate monthly bill for running these appliances for 6 hours a day (1 length = 30 days).
Text Solution
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Here, `n_(1) = 20` lamps, `P_(1) = 60W`, `n_(2) = 10` fans, `I_(2) = 0.5 A, n_(3) = 1` kettel, `R_(3) = 110 Omega` `V_(1) = V_(2) = V_(3) = 220 V, t_(1) = t_(2) = t_(3) = 6h//"day"` `Rate = 50` paise/kWh, Total cost = ? Electric energy consumed per day `n_(1) P_(1) t_(1) + n_(2)V_(2)I_(2)t_(2) + n_(3) xx (V_(3)^(2))/R_(3) xx t_(3)` `[ 20 xx 60 xx 6 +10 xx 220 xx 0.5 xx 6 +1 xx ((220)(^2))/110 xx 6 ] Wh` `= (7200+ 6600 + 2640)/1000 = 16.44 kWh` `:.` Total energy consumed in November month ` = 16.44 xx 30 = 493.2 kWh` `:.` Total bill ` = 493.2 xx 50/100 = 246.60` rupees.
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