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.
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.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.