Heat Q = 5 kJ was released in the conductor during the time t = 10 s with a uniform increase in current

Heat Q = 5 kJ was released in the conductor during the time t = 10 s with a uniform increase in current from I1 = 1 A to I2 = 2 A. Find the resistance R of the conductor.

Task data: t (time) = 10 s; I1 (initial current) = 1 A; I2 (increased current) = 2 A; Q (heat) = 5 kJ = 5 * 10 ^ 3 J.

1) The law of current change: I (t) = x * t + y;

a) Current I1 = 1 A: I (0) = x * 0 + y = 1, whence y = 1;

b) Current I2 = 2 A: I (10) = x * 10 + y = x * 10 + 1 = 2, whence x = 1/10 = 0.1.

c) Law: I (t) = 0.1t + 1.

2) Heat released: Q = 0 ^ 10∫ I ^ 2 * R * dt = R * 0 ^ 10∫ (0.1t + 1) 2 dt = R * 0 ^ 10∫ (0.01t ^ 2 + 0, 2t + 1) dt = R * (0.01 * t ^ 3/3 + 0.2t ^ 2/2 + t) | 0 ^ 10 = R * (t ^ 3/300 + 0.1t ^ 2 + t ) | 0 ^ 10 = 23.33 (3) R.

3) Conductor resistance: R = Q / 23.33 (3) = 5 * 10 ^ 3 / 23.33 (3) = 214.29 Ohm.

Answer: The resistance is 214.29 ohms.