# The circuit section consists of two parallel-connected conductors, the resistance of which is 40 and 60 ohms.

**The circuit section consists of two parallel-connected conductors, the resistance of which is 40 and 60 ohms. The voltage on the section of the circuit is 60V. What amount of heat will be released in each of the resistors in 1 min?**

R1 = 40 ohms.

R2 = 60 ohms.

U = 60 V.

t = 1 min = 60 s.

Q1 -?

Q2 -?

According to the Joule-Lenz law, the amount of heat Q that is released in a conductor is determined by the formula: Q = I2 * R * t, where I is the current in the conductor, R is the resistance of the conductor, t is the time of current passage through the conductor.

Q1 = I12 * R1 * t.

Q2 = I22 * R2 * t.

With a series connection of conductors: I = I1 = I2, R = R1 + R2.

We find the current strength I according to Ohm’s law for a section of the circuit: I = U / (R1 + R2).

I = 60 V / (40 Ohm + 60 Ohm) = 0.6 A.

Q1 = (0.6 A) ^ 2 * 40 Ohm * 60 s = 864 J.

Q2 = (0.6 A) ^ 2 * 60 Ohm * 60 s = 1296 J.

Answer: Q1 = 864 J, Q2 = 1296 J.