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Q.

A given resistor cannot carry currents exceeding 20 A, without exceeding its maximum power dissipation ratings. By forced air cooling suppose that we increase the rate at which heat can be carried by a factor of 2. Now the maximum current that the resistor can carry is

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a

10 A

b

202A

c

302A

d

40 A

answer is B.

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Detailed Solution

Heat produced µ I2, initial heat produced = k (20)2, final heat produced = 2k (20)2. If final current is I', then                        kI'2=2 × k × (20)2 or I' =202A
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