Electric boiler resistance R = 100 Ohm. The current in the circuit is I = 2 A.

Electric boiler resistance R = 100 Ohm. The current in the circuit is I = 2 A. What amount of energy will the boiler release during the time interval t = 2 minutes.

R = 100 ohms.

I = 2 A.

t = 2 min = 120 s.

Q -?

The amount of thermal energy Q that is released in a conductor when an electric current passes through it is determined by the Joule-Lenz law: Q = I * U * t,

Where I is the current in the circuit, U is the voltage in the conductor, t is the time of passage of the electric current.

We express the voltage U from Ohm’s law for a section of the circuit: I = U / R, where R is the resistance of the conductor.

U = I * R.

The formula for the Joule-Lenz law will take the form: Q = I * I * R * t = I ^ 2 * R * t.

Q = (2 A) ^ 2 * 100 Ohm * 120 s = 48000 J.

Answer: Q = 48000 J of thermal energy is released in the boiler.