Power Dissipation in Resistors: P31.52

In summary, your homework statement is that the voltage across the parallel combination of the 10 and 20 Ohm resistors is 21.9V.
  • #1
Phoenixtears
83
0

Homework Statement


Consider the circuit in Figure P31.52, in which R = 10 . The 10 resistor is dissipating 48 W of power. How much power are the other two resistors dissipating?
(Image Attached)

5 resistor ________

20 resistor ___24___


Homework Equations



VI=RP

The Attempt at a Solution



I began with the equation VI=RP. Then (I'm not even sure if I can do this!) I assumed the voltage to be constant throughout the resistors (which sounds extremely inaccurate). I was able to solve the equation for power: P= I/R. Then I used the given numbers (10 and 48W) and solved for the current: 480 amps. Placing the 'I' back into the equation I solved using the 20=R. I got the second part right with '24', but am confused why this process doesn't work for a resistor of 5.

Thank you so much in advance!

~Phoenix
 

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  • #2
Your equation is wrong. P=IV
 
  • #3
JaWiB said:
Your equation is wrong. P=IV

There are multiple equations I was given in class, two are: P=IV and V=IR. But how does P=IV help me if it doesn't include resistance. I can't seem to figure out a way to include the two equations for it to work for the first part of the problem. I do realize that I can substitute I out to make P=V^2/R, but I don't see how that helps me either.

Sorry if this is obvious, I'm just so confused! I've been looking at the problem for too long probably.

Thanks again!
 
  • #4
You are given the power of that resistor, which gives you the voltage across it. That gives you the current through the bottom two resistors, which gives you the current through the top resistor and it's voltage and current.

Post your work and solution based on that set of hints...
 
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  • #5
Thanks for the hints.

This is where I'm at now:

I used P=V^2/R to get the voltage of the bottom resistor: 21.91.

Then, using that number and the equation P=IV I found the current: 2.19

That current is constant throughout the system. I then used the equation V=IR to find the voltage of the top: 10.95.

Then I went back to the first equation I used, P=V^2/R, to find the power of the top resistor: 23.98.

However, that's not the answer. Am I applying a concept incorrectly?

Thanks again for all the help, this is all really helpful for me.

~Phoenix
 
  • #6
Phoenixtears said:
Thanks for the hints.

This is where I'm at now:

I used P=V^2/R to get the voltage of the bottom resistor: 21.91.

Then, using that number and the equation P=IV I found the current: 2.19

That current is constant throughout the system. I then used the equation V=IR to find the voltage of the top: 10.95.

Then I went back to the first equation I used, P=V^2/R, to find the power of the top resistor: 23.98.

However, that's not the answer. Am I applying a concept incorrectly?

Thanks again for all the help, this is all really helpful for me.

~Phoenix

Your voltage is correct -- the 21.9V is the voltage across the parallel combination of the 10 and 20 Ohm resistors. The current through the parallel combination of the 10 and 20 Ohm resistors is not just 21.9V/20 Ohms ... You need to use the parallel combiination resistance...
 
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FAQ: Power Dissipation in Resistors: P31.52

What is power dissipation in resistors?

Power dissipation in resistors refers to the amount of power that is converted into heat when an electric current flows through a resistor. This is due to the resistance of the resistor, which causes a drop in voltage and converts electrical energy into thermal energy.

Why is power dissipation important in resistors?

Power dissipation is important in resistors because it determines the maximum amount of power that a resistor can handle without overheating. This is crucial in electronic circuits as exceeding the power dissipation limit can damage the resistor and potentially cause the circuit to fail.

How is power dissipation calculated in resistors?

Power dissipation in resistors can be calculated using Ohm's Law, which states that power (P) is equal to the current (I) squared multiplied by the resistance (R). This can be represented by the formula P = I^2R. Additionally, the power rating of a resistor is usually specified by the manufacturer.

What factors affect power dissipation in resistors?

The main factors that affect power dissipation in resistors are the amount of current flowing through the resistor and the resistance value of the resistor. The higher the current or resistance, the greater the power dissipation. Other factors that can affect power dissipation include ambient temperature, the material and size of the resistor, and the design of the circuit.

How can power dissipation be reduced in resistors?

Power dissipation in resistors can be reduced by using resistors with a higher power rating, increasing the size of the resistor, or by using multiple resistors in parallel to distribute the power dissipation. Additionally, using more efficient materials and designing the circuit to minimize current flow can also help reduce power dissipation in resistors.

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