A current of 0.5 A flows through a resistor when connected to a 40 V battery. How much energy is dispatched in 2 minutes?

The energy dispatched by an electrical device is given by the product of the power consumed by the device and the time it is operating. In this case, the device is a resistor, and its power consumption can be calculated using Ohm's Law, which states that the power (P) consumed by a resistor is equal to the product of its resistance (R) and the square of the current (I) passing through it: P = I^2 * R. The resistance of the resistor is not given, but we can calculate it using the information given in the problem. Ohm's Law can also be rearranged to find the resistance of a resistor: R = V / I, where V is the voltage applied across the resistor. In this case, the voltage is given as 40 V and the current is 0.5 A, so the resistance of the resistor is R = 40 V / 0.5 A = 80 Ω. Now that we know the resistance of the resistor, we can use Ohm's Law to find its power consumption: P = I^2 * R = 0.5 A^2 * 80 Ω = 20 W. The problem asks for the energy dispatched in 2 minutes. To find this, we need to convert the time from minutes to seconds, since power is measured in watts, which are joules per second. 2 minutes is equal to 120 seconds. The energy dispatched is the product of the power consumption and the time: E = P * t = 20 W * 120 s = 2400 J. Therefore, the answer is: 2400 J.