Calculating Parallel Circuit

Bigbri0104

Member
Location
New Jersey
Occupation
Student
Hello Everyone,

I'm new to the forum. I'm currently taking electrical vocational classes at night. In our night class we were calculating electrical circuits using Ohms law and I wanted to find out, why voltage is squared when calculating the known load in a parallel circuit?
In the question the voltage and power were given. The question was taking out of Mike holts electrical theory book figure 9-21
Just curious.

Thanks
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Squaring the voltage has nothing to do with a parallel circuit. It has to do with Power.

Power is the product of voltage and current (P=V*I), and since I increases when V increases (I=V/R), V gets squared.

Since I= V/R, you can substitute V/R for I in the first equation. That gives you P = V* V* I which is V squared * I.

For an example, start with 10 Volts across a 10 ohm resistor. Calculate the Power (10V/10 ohms = 1 amp. Then 10 volts * 1 amp = 10 watts).

Then increase (mentally) the voltage to 20 volts. Now 20V/10 ohms = 2 amps. And 20V * 2 amps = 40 watts. So the power has gone up 4 x for a 2 x increase in voltage.

Try using V squared /R and I squared * R and convince yourself they all work out the same.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Its also good to learn the limitations of ohms law pretty quick. For example, motors don't have a nice constant "R" like resistors do. They are more a constant wattage loads. Whatever load they are forced to move pretty much sets the required wattage. Since P stays pretty constant even when V is changed, the current can actually go up as the voltage goes down.

Incandescant light bulbs are also more complex. Their resistance is almost 0 when they are cold, but it shoots up to 100 ohms or so when they are glowing red hot.

For now, just realize that there will be exceptions that you will learn in the future.
 
Incandescant light bulbs are also more complex. Their resistance is almost 0 when they are cold, but it shoots up to 100 ohms or so when they are glowing red hot.

I remember many moons ago testing the resistance of a light bulb, not knowing that filiment resistance changed significantly with temperature, and being quite perplexed! I remember checking my math like 5 times and using every meter I had to try and find the "mistake" 😂
 

Dsg319

Senior Member
Location
West Virginia
Occupation
(Green)Master Electrician
I remember many moons ago testing the resistance of a light bulb, not knowing that filiment resistance changed significantly with temperature, and being quite perplexed! I remember checking my math like 5 times and using every meter I had to try and find the "mistake" 😂
Your not alone lol
 

Carultch

Senior Member
Location
Massachusetts
I remember many moons ago testing the resistance of a light bulb, not knowing that filiment resistance changed significantly with temperature, and being quite perplexed! I remember checking my math like 5 times and using every meter I had to try and find the "mistake" 😂

That's why light bulbs commonly burn out when you first turn them on. The bulb draws the greatest current at its initial cold temperature, as it heats up for that fraction of a second. That's the time when the bulb is most likely to overheat, is when you first turn it on.
 

mikeames

Senior Member
Location
Germantown MD
Occupation
Teacher - Master Electrician - 2017 NEC
We have had this discussion. I have had the same experiences, and use it in class now. Don't forget "T" rating on switches.
 

Bigbri0104

Member
Location
New Jersey
Occupation
Student
Hey Mike I like the example of breaking down ohms law into other forms that you showed in your post. Are you able to take other variables in ohms law and break it down the way you did in the recent example? The visuals help

Thanks
 

mikeames

Senior Member
Location
Germantown MD
Occupation
Teacher - Master Electrician - 2017 NEC
Hey Mike I like the example of breaking down ohms law into other forms that you showed in your post. Are you able to take other variables in ohms law and break it down the way you did in the recent example? The visuals help

Thanks
Yes of course you can. Like Larry said the formula wheel gives them all to you. I cant remember that wheel because I don't use all those combinations very much. I remember Ohms law and Watts law. I know that cold. If I am even in a situation where I need to find something and the simple formula wheel does not work, I derive it on the spot, or look it up on the phone.

The point I was making is there are 12 different formulas related to watts law and ohms law and you questioned the square. Its nothing fancy just the basics broken down to a different level.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
I remember Ohms law and Watts law. I know that cold. If I am even in a situation where I need to find something and the simple formula wheel does not work, I derive it on the spot, or look it up on the phone.
Same here. I just use the two basic formulae. The square and square root steps save a calculation step, but don't change results.
 
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