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P=I^2R formula confusion from book

DickFreed

Aug 6, 2021
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Sorry, couldn’t figure out how to make “I” to the second power as a superscript. I was looking through this old textbook and came across this short explanation on the power formula, which I circled in first pic, which explains that the power can be reduced by lowering the resistance. That makes sense to me because of basic math. However, the next page threw me off with the illustration showing the resistance of each bulb having less resistance, resulting in more power. The scribbles around the pic was just me trying to make sense of the conflicting info. I’m having difficulty understanding what seems to be conflicting examples.
https://imgur.com/a/nntS3yJ
 

Nanren888

Nov 8, 2015
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Not sure what you are missing.
The text says by lowering either the current or the resistance.

For the simple relationship of P changes linearly with R, what is the implication for I?
In the first equation, examine what the voltage must do if the current I is to remain constant as you change R.
.
You have drawn on that P = V^2 / R which gives you, for fixed voltage, power increasing as 1/R.
This is available by rearrangement of V=IR to make I the subject and substitution into the first equation.
.In the second case, work out the current.
 

Harald Kapp

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I’m having difficulty understanding what seems to be conflicting examples.
The conflict is that using P = I2 × R with varying resistances you assume I = constant whereas in the example of the lightbulbs V = constant .
Given V and R you have I = V / R and with P = I × V this results in P = V2 / R as @Nanren888 showed. You now easily see that with V = constant the power decreases with an increase in R.
 

hevans1944

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The scribbles around the pic was just me trying to make sense of the conflicting info.
It isn't conflicting, and your assumption that the light bulbs are supplied with 120 volts is incorrect, but irrelevant. Using the equation that P x R = V2, and substituting the values indicated for power and resistance, yields a value for V (after extracting the square root of the product of power and resistance) that is close to 110 volts instead of the 120 volts you used to calculate the current in each lamp. In any event, it is assumed that the voltage applied to household lamps is constant, so using the relationship that P = V2 / R it is easily seen that reducing the value of R will have the effect of increasing the value of P.

All of the relationships between V, I, R, and P for a particular circuit are algebraic derivatives of equations expressing Ohm's Law (V = I x R) and power (P = V x I), as @Harald Kapp mentioned in post #3 and @bertus illustrated in post #4. You just need to be careful when, where, and how you apply those equations.

Learn about Kirchoff's voltage law (algebraic sum of voltage around a loop is always zero) and current law (algebraic sum of currents into and out of a circuit node is always zero). These two laws, along with Ohm's Law, will allow you to analyze more complicated situations, such as circuits involving bridged resistances and two or more voltage and/or current sources. You should also be familiar with how to solve simultaneous equations and the superposition theorem, both of which are useful for analyzing circuits.
 
Last edited:

ramussons

Jun 10, 2014
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Sorry, couldn’t figure out how to make “I” to the second power as a superscript. I was looking through this old textbook and came across this short explanation on the power formula, which I circled in first pic, which explains that the power can be reduced by lowering the resistance. That makes sense to me because of basic math. However, the next page threw me off with the illustration showing the resistance of each bulb having less resistance, resulting in more power. The scribbles around the pic was just me trying to make sense of the conflicting info. I’m having difficulty understanding what seems to be conflicting examples.
https://imgur.com/a/nntS3yJ
The first example refers to Power in terms of Current and Resistance. I^2 * R gives the relationship. Just note that Power is Proportional to Square of Current.
In the next page, the examples are about bulbs with various powers and their resistances. The main factor here is not the Current, but the Voltage. Bulbs are rated for a given Power and Voltage. So, when the Resistance drops, the Current increases.
If the Resistance Halves, the Current doubles. And since the Power is proportional to the Square of the Current, the Power increases even though the Resistance reduces.
 

davenn

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Sep 5, 2009
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Sorry, couldn’t figure out how to make “I” to the second power as a superscript.

P=I2 R

look along the line of tools when you are creating a post and towards the right end you will see a sub and sup for subscript

sub sup.JPG


Wm or V2

when you type your text Wm or V2 as in my examples ... highlight the m or 2 and select sub or sup and you will see "open bracket sub close bracket m open bracket /sub close bracket"
that you want to appear around the character you highlighted
 

DickFreed

Aug 6, 2021
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P=I2 R

look along the line of tools when you are creating a post and towards the right end you will see a sub and sup for subscript

View attachment 52981


Wm or V2

when you type your text Wm or V2 as in my examples ... highlight the m or 2 and select sub or sup and you will see "open bracket sub close bracket m open bracket /sub close bracket"
that you want to appear around the character you highlighted

appreciate it
 
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