1 00:00:03,819 --> 00:00:09,099 Electrical energy and power. The power sources supply electrical power to the circuits. 2 00:00:09,980 --> 00:00:13,779 The load devices consume this power and convert it to other types of energy. 3 00:00:14,599 --> 00:00:20,140 In the international system, energy is measured in joules of J. Electrical power measures the 4 00:00:20,140 --> 00:00:24,760 energy that is consumed by a load device or delivered by a power source over a certain 5 00:00:24,760 --> 00:00:29,300 period. In the international system, it is measured in watts, W. 6 00:00:29,300 --> 00:00:36,640 Power is calculated as the result of multiplying the voltage of each component by the current that flows through it. 7 00:00:37,799 --> 00:00:40,500 The unit of energy often used is the kilowatt-hour. 8 00:00:41,759 --> 00:00:48,229 The Joule effect. 9 00:00:49,189 --> 00:00:56,270 In fact, all of the elements in a circuit, not just the load devices, consume a small amount of unwanted electrical energy. 10 00:00:57,070 --> 00:01:01,670 This is a phenomenon known as the Joule effect, and it is due to the resistance of these components. 11 00:01:02,549 --> 00:01:06,870 In our calculations in this unit, we will consider the resistance of conductors 12 00:01:06,870 --> 00:01:13,510 and power sources to be negligible. In this exercise you have to calculate the 13 00:01:13,510 --> 00:01:20,230 current that flows through a 1 watt motor powered by a 5 volt cell. The data is the power 14 00:01:21,109 --> 00:01:27,590 and the voltage and you do not know the intensity. To calculate it, go through the formula that is 15 00:01:27,590 --> 00:01:33,069 the word for power is the voltage by the intensity and clarify the equation 16 00:01:33,069 --> 00:01:38,150 clarify the I the intensity as you have ordered that you don't have any problems 17 00:01:38,150 --> 00:01:48,200 0.2 is the result remember not given the results in the form of fractions in this 18 00:01:48,200 --> 00:01:52,819 exercise we have to calculate the energy that consumes is consumed by a heater 19 00:01:52,819 --> 00:01:58,459 with a power of two kilowatts that runs for three hours first of all the data 20 00:01:58,459 --> 00:02:04,359 we have the power that is in kilowatts so we put it in watts that is just to 21 00:02:04,359 --> 00:02:11,960 multiply by 1000 and the time is given to us in three hours we have to put it 22 00:02:11,960 --> 00:02:22,139 in a unit of the international system that hours remember one hour has three 23 00:02:22,139 --> 00:02:29,460 three thousand and six hundred seconds so you have in total three hours ten thousand and eight 24 00:02:29,460 --> 00:02:37,560 hundred seconds and the thing with that we don't know is the energy then we have the formula that 25 00:02:37,560 --> 00:02:47,400 says that the energy is the power by the time so as we have the units of international system to 26 00:02:47,400 --> 00:02:51,759 because they are calculate we have to calculate it in joules that is a unit of 27 00:02:51,759 --> 00:02:57,060 the energy in the international system we have to use watts and seconds just 28 00:02:57,060 --> 00:03:04,800 making this operation of course you use the calculator you can solve it to 29 00:03:04,800 --> 00:03:12,840 calculate it in in kilowatts per hour it's easier because you have industry in 30 00:03:12,840 --> 00:03:16,960 the instructions they give you the power in kilowatts and the time in our 31 00:03:16,960 --> 00:03:20,960 So you just only have to multiply it and you have the result.