# Resistors with resistances of 150 Ohm and 90 Ohm are connected in series and included in the network

Resistors with resistances of 150 Ohm and 90 Ohm are connected in series and included in the network. How much heat was released in the first resistor if 18kJ of heat was released on the second?

R1 = 150 ohms.

R2 = 90 ohms.

Q2 = 18 kJ = 18000 J.

Q1 -?

According to the Joule-Lenz law, the amount of heat Q that is released in the conductor through which the electric current flows is determined by the formula: Q = I ^ 2 * R * t, where I is the current in the conductor, R is the resistance of the conductor, t is the time passage of current.

Q1 = I1 ^ 2 * R1 * t.

Q2 = I2 ^ 2 * R2 * t.

When the conductors are connected in series, the current strength in them is the same: I1 = I2.

I1 ^ 2 = I2 ^ 2 = Q2 / R2 * t.

The amount of heat energy Q1 that is released in the first resistor can be found by the formula: Q1 = Q2 * R1 * t / R2 * t = Q2 * R1 / R2.

Q1 = 18000 J * 150 Ohm / 90 Ohm = 30000 J = 30 kJ.

Answer: Q1 = 30,000 J of thermal energy will be released in the first resistor.

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