CBSE Class 10 Answered
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Asked by riyamaths | 30 May, 2018, 07:14: PM
Expert Answer
Specifications of device-1 :-
Power P = 44 W; operating voltage V = 220 V;
operating current I = P/V = 44/220 = 0.2A
Resistance = P/I2 = 44×5×5 = 1100 Ω
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Specifications of device-2:-
Power P = 11 W; operating voltage V = 220 V;
operating current I = P/V = 11/220 = 0.05A
Resistance = P/I2 = 11×20×20 = 4400 Ω
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when two devices are connected in series, combined resistance is = 1100+4400 = 5500 Ω
current drawn by the devices connected in series under operating voltage 440 V = 440/5500 = 0.08A
fuse connected to device-2 will likely to blow, because 0.08 A is greater than its operating current 0.05 A
Answered by Thiyagarajan K | 31 May, 2018, 01:53: PM
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