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Question 22: An electric heater which is connected to a 220 V supply line has two resistance coils A and B of 24 Ω resistance each. These coils can be used separately (one at a time), in series or in parallel. Calculate the current drawn when : (a) only one coil A is used. (b) coils A and B are used in series. (c) coils A and B are used in parallel. Lakhmir Singh Physics Class 10 |
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Answer» a) 9.16ampb) 4.58 ampc) 18.33 ampExplanation:An electric heater is connected to a supply LINE that has a voltage of 220V.This supply line has TWO resistance coils A and B.Resistance of A is 24ohm and B is also 24ohm.So, we can conclude that resistance of A and B are equal.Now, we know that the amount of voltage PASSED = amount of current drawn × Resistance. V=IRa) So, the amount of current drawn when A coil is used is : I= V/R I= 220/24 I= 9.16 ampb) When coil A and B are being used in series the amount of current drawn is : 220/48[ Since A+B = 24+24=48] Therefore I = 4.58 ampc) When coil A and B are in parallel series, TOTAL resistance would be 1/24 + 1/24 = 2/24 = 1/12Therefore amount of current drawn is : 220/12 = 18.33ampTherefore I = 18.33 amp |
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