**what is the average rate at which electrical energy is converted to thermal energy in the resistor?** This is a topic that many people are looking for. **star-trek-voyager.net** is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, ** star-trek-voyager.net ** would like to introduce to you **Power Dissipation In Resistors, Diodes, and LEDs**. Following along are instructions in the video below:

This video. Were gonna talk about how to find the power dissipated in each each resistor in this example. We have two resistors in series.

We can call them and r2 now in order to calculate the power dissipated by each resistor. We could use any one of the three formulas power is equal to voltage. Times current.

Its also equal to the square of the current flowing in the resistor times. The resistance and its also equal to the square of the voltage across the resistor divided by the resistance. Now the first thing we need to do is calculate the total resistance of the series circuit in the series circuit rt is just the sum of each resistor.

So the total resistance is going to be r1 is 20. So its gonna be 20 plus. 10.

Which is 30 ohms next. We need to calculate the current flowing in the circuit. Now we could use ohms law.

Its going to be the voltage of the battery divided by the total resistance thats 60 volts divided by 30 ohms. So we have a current of 2 amps flowing in this circuit. Now because theres only one path for the current to flow.

The current flowing through r1 and r2 is the same its 2 amps. Now that we have the current lets use this formula to calculate the power dissipated by each resistor. So lets start with the first resistor.

Its gonna be the current flowing through it squared times r1. So its 2 squared times. 20.

Ohms. 2. Squared is 4 times.

20. Thats 80. So thats the power dissipated by the first resistor to calculate the power dissipated by the second resistor.

Its going to be the square of the current times r2. So its 2 squared times 10. So thats 4 times.

10. That means that the second means is dissipated 40 watts. So lets think about what that means power is the rate at which energy is delivered to a device or to something a power of 1 watt.

Means. That one joule of energy is delivered each second so in 10 seconds. 10.

Joules of energy is being delivered. So are to this resistor is converting 40 joules of electrical energy into heat. Every.

Second r1 is converting 80 joules of electrical energy. Into heat. Every second so.

In 10 seconds. It converts 800 joules of electrical energy. Into thermal energy.

So thats the basic idea. Behind power dissipation. Now lets calculate the power delivered by the battery to do that were just going to multiply the voltage by the current because we dont know what the resistance of the battery is so the voltage of the battery of 60.

The current flowing from it is 2 amps. So 60 times 2. Reps is 120 notice that the power delivered by the battery is equal to the sum of the power dissipated by the two resistors this is in line with the law of conservation of energy.

The energy that the battery delivers to the circuit should equal what the elements in a circuit are consuming. So the two resistors are using up 120 watts. The battery is delivering 120 watts.

Those two values should be the same if not something is wrong. Now lets try another example at this time. We have two resistors connected in parallel calculate the power dissipated by each resistor.

And also the power delivered by the battery feel free to pause. The video. If you want to try this so first we need to calculate the current flowing through each resistor.

Because that its not the same whenever you have two resistors connected in parallel the voltage across is the same so the voltage across r. 1. And r.

2. Is the same as the voltage of the battery. Its 20 volts.

Now we could use ohms law to calculate the current flowing through each resistor. So to calculate i run its going to be the voltage across it divided by r 1. So its 20 volts divided by 5 ohms.

Which is 4. So we have a current of 4 amps. Flowing through r.

1.

Now to calculate i 2. Its going to be the voltage. Divided by r 2.

So its 20 volts divided by 10 ohms. Which is 2 amps. So thats the current flowing through r.

2. The current that is leaving the battery is the sum of these two currents. So the battery is delivering.

6 amps of current to the circuit. Now lets calculate the power dissipated by each resistor so starting with the first one. Its gonna be i squared or i 1 squared.

Times r. 1. This time i 1 and i 2 are different i 1 is 4 r.

1. Is 5 4 times. 5 is 20 times another 4.

Thats 80. So thats much power. The first resistor is dissipated for the second one this is gonna be i to square times r.

2. So thats 2 squared times 10 ohms. 2 squared is 4 times 10.

So thats 40 watts. Now the power delivered by the battery is going to be the voltage times. The current the voltage of the battery is 20 volts.

The current flowing from the battery. 6 amps. 20 times.

6. Is 120 watts. So once again the power delivered by the battery is equal to the sum of the power dissipated by the resistors number.

3. A 220 ohm resistor has a maximum power rating of point 5 watts. What is the maximum voltage that should be applied to this resistor now when you buy a resistor.

Theres two things that youll know or two impedes information that youll be given that is the value of the resistor. The resistance and the maximum power that it can handle so once you have that information when youre designing a circuit. You can determine what is the maximum voltage that you can safely apply to the resistor or the maximum current that should be flowing through this resistor and this example problem will help us to do that so all we know is that r is equal to 220 ohms and p is 05.

Watts. So to calculate the voltage. We could use this formula p is equal to v.

Squared over r multiplying. Both sides by r will give us the voltage. So v.

Squared is equal to the power x r. Taking the square root of both sides. We get v.

So the maximum voltage that should be applied to this resistor is the square root of the maximum power rated. Times the. Resistance so thats gonna be the square root of 05.

Times. 220 half of 220 is 110. So we need to take the square root of 110 and the square root of 121 10.

Rather is ten point four nine volts and thats a rounded answer. So thats the maximum voltage. That should be applied if youre applying 12 volts to this resistor.

Its gonna overheat and you could damage your component. Now lets move on to part b. Lets calculate the maximum current that should be flowing through this resistor.

So. Were gonna use this formula p is equal to i squared times r. So lets solve for i in order to solve for i will needs to divide both sides by r.

So we get that i square is equal to p. Over r next. We need to take the square root of both sides.

So. The maximum current is gonna be equal to the square root of the power i. Mean the maximum power divided.

By. R so we. Know p.

Is. 05. R is.

220.

05. Divided by 220. Thats point zero zero.

Two to seven. And if we take the square root of that its gonna be point zero four seven six seven which im gonna round to eight and this is in amps now if we multiply that by a thousand theres a thousand milliamps per one amp of current. This is gonna be forty seven point.

I take this back. There should be point zero four seven six seven which rounds two point zero four seven seven. So.

This is gonna be forty seven point seven milliamps. So that is the maximum current that should be applied or rather that should be flowing through that resistor so now you know how to design a circuit. If you know that a maximum power rating of a resistor and the value of the resistance.

You now know how to determine the maximum voltage. That should be applied to resistor and the maximum current that should be flown in that resistor. Lets move on to number four a six volt battery is in series with a forward bias.

Silicon diode and a 100 ohm resistor calculate the power dissipated by the diode. Assume the voltage drop of the silicon diode is 06. Volts.

So this is our well call this d. 1. So we have a 100 ohm resistor and a diode with a forward voltage drop of 06.

Volts. And we have a 6 volt battery in order to calculate the power dissipated by the diode. We need to know the current flowing through it so how can we calculate the current flowing in this circuit.

What we need to do is determine the current flowing through the resistor but we could do that if we can determine the voltage across the resistor now. What you need to understand is this the voltage drop of the diode and the voltage drop of the resistor must equal the voltage of the battery. So to calculate the voltage drop across the resistor is going to be the voltage of the battery.

Minus. The voltage drop of the diode. So its gonna be five point four volts according to kirchhoffs voltage law.

The sum of the voltages around a closed loop must add a zero so this is positive six. This is going to be negative point six and negative five point four these are negative because they consume energy from the circuit. So they have a voltage drop.

This is a voltage rise because it delivers energy to the circuit. But the sum of the voltages must add to zero in a closed loop. So thats the voltage across the resistor.

Its five point four volts so to calculate the current flowing through the resistor. Its going to be the voltage across it divided by the resistance based on ohms law. So the voltage is 5 point.

4 volts. The resistance is a hundred arms five point four divided by a hundred is plate zero five four amps. So that is the amount of current that is flowing in this circuit.

Now lets calculate the power dissipated by the silicon diode. So its going to be the voltage across the diode times two current flowing through it we know the voltage across it is point 6 volts. The current flowing through it is the same as the current flowing through the resistor because these two elements are in series with each other so its going to be point six times point zero five four and us the power dissipated by the diode is point zero three two four watts and if you multiply that by a thousand youll get 30 24.

Millions. So thats the power dissipated by the diode. Thats how you can calculate.

It number. 5. A 9 volt.

Battery is in series with a 680. Ohm. Resistor.

And the green. Led with a forward voltage drop of 2 volts calculate the power dissipated by the led the process for solving this problem is the same as number 4. The only difference is we have a light emitting diode as opposed to a regular silicon diode now.

We know the voltage drop of this diode is 2 volts. That means that the voltage drop of the resistor is 7. Because these two have to add up tonight.

So now we can calculate the current flowing in the circuit. Which is going to be the voltage across the resistor divided by the resistance. So that seven volts divided by the 680 ohm resistor.

So 7. Divided by 680 thats equal to a current of point zero 1 zero to. 9.

Amps. Now that we have the current. We can calculate the power dissipated by the led so power is voltage.

Times current. Theres a voltage of 2 volts across the led and a current of point zero one zero to 9. Amps flowing through it so the power dissipated is going to be point zero two zero five eight.

If you rhonda. Its like five nine. This is five eight eight multiplying that by a thousand we could say this is approximately twenty point six million watts.

So thats the power dissipated by the green led .

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