1.

A resistor develops 300 J of thermal energy in 15s, when a current of 2A is passed through it. If the current increases to 3A, the energy developed in 10s is J.

Answer»

H = i2Rt

300 = 22 x R x 15

⇒ R = 300/60 = 5Ω

Now, for i = 3A, t = 10s, R = 5Ω

H = 32 x 5 x 10 = 450 J



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