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Originally Posted by Smart \$
Resistors connected in series sum arithmetically. Apply Ohm's law to the result. Current is the same through all the resistors.

That's the most I can give you without providing the actual answer.
I'm not sure if I should add all the ohms of the resistors together and divide by the voltage which would be 50÷50 equals 1 or divide the voltage by the 10 which would be 50÷10 = 5 honestly not 100% sure this is why am trying to get help with this problem

2. Originally Posted by sclement
I'm not sure if I should add all the ohms of the resistors together and divide by the voltage which would be 50÷50 equals 1 or divide the voltage by the 10 which would be 50÷10 = 5 honestly not 100% sure this is why am trying to get help with this problem
Ohm's law: E=IxR

You have determined you have 50 volts and total circuit resistance of 50 ohms. 50V / 50 Ohms = 1 Amp.

Now figure how much voltage is across the 10 ohm resistor if it has 1 amp flowing through it.

Did that help?

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Originally Posted by kwired
Ohm's law: E=IxR

You have determined you have 50 volts and total circuit resistance of 50 ohms. 50V / 50 Ohms = 1 Amp.

Now figure how much voltage is across the 10 ohm resistor if it has 1 amp flowing through it.

Did that help?
10×1 = 10 so 10 volts! Thanks for the assistance I really appreciate it all

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## Voltage Drop

thanks to all the folks that have been helping with the different problems I've posted! here's another one that's got me scratching my head

Single phase 120 V, 16 amp single phase load is connected to a12 AWG THHN conductor. The equipment name plate requires no more than a 3% voltage drop on the branch. What is the maximum wire length from the panel where the voltage drop will not exceed 3% (k=12.9)

5. Originally Posted by sclement
thanks to all the folks that have been helping with the different problems I've posted! here's another one that's got me scratching my head

Single phase 120 V, 16 amp single phase load is connected to a12 AWG THHN conductor. The equipment name plate requires no more than a 3% voltage drop on the branch. What is the maximum wire length from the panel where the voltage drop will not exceed 3% (k=12.9)
Write out your VD calculation - fill in the known values, you will have to calculate what 3% is before filling that in. You should have all components of the formula except length - reorganize the equation to solve for that missing value.

I want you to try this before it gets shown to you.

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## Transpose equations

[QUOTE=sclement;1851686]thanks to all the folks that have been helping with the different problems I've posted! here's another one that's got me scratching my head

Single phase 120 V, 16 amp single phase load is connected to a12 AWG THHN conductor. The equipment name plate requires no more than a 3% voltage drop on the branch. What is the maximum wire length from the panel where the voltage drop will not exceed 3% (k=12.9)[/Qequipm

The questions you are asking are generally the same questions in which you are trying to find (×) write this formula on a piece of paper. Cma=2×k×i×l÷vd. I expained to you what the letters represented in the last question. With this formula you can answer all the questions asking for (l)length,(cma) circular mil or the wire size, vd, or (i)amp. Which ever variable they are looking for needs to be isolated by itself using the the formula. Transposing is important to know because this skill will be needed in basic ohms law questions. Hope this helps.

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Originally Posted by kwired
Write out your VD calculation - fill in the known values, you will have to calculate what 3% is before filling that in. You should have all components of the formula except length - reorganize the equation to solve for that missing value.

I want you to try this before it gets shown to you.
Here is my attempt I know something is out of order, and I'm not sure when to apply the 3% in the equation.

2 x 12.9 x 16 = 412.8
412.8÷6530=0.063215926
??

8. Originally Posted by sclement
Here is my attempt I know something is out of order, and I'm not sure when to apply the 3% in the equation.

2 x 12.9 x 16 = 412.8
412.8÷6530=0.063215926
??
You are missing Vd and Cmil in your equation
Vd = Nominal voltage × voltage drop percentage
Vd = 120V × 3%
'Plug' this Vd into the equation as a known value.
Cmil = 2 × K × I × L ÷ Vd

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Originally Posted by Smart \$
You are missing Vd and Cmil in your equation
Vd = Nominal voltage × voltage drop percentage
Vd = 120V × 3%
'Plug' this Vd into the equation as a known value.
Cmil = 2 × K × I × L ÷ Vd
I had circular Mills in there the 6530 and length is unknown I'm just not sure when to apply the 3% Factor and how to come up with length? Thanks!!

10. Originally Posted by sclement
I had circular Mills in there the 6530 and length is unknown I'm just not sure when to apply the 3% Factor and how to come up with length? Thanks!!
Yeh. You did as a second step. Learn to do it all in one step.

In the equation I wrote, take all knowns to one side of the equal sign; change preceding operator sign when you do: multiplication to division, division to multiplication.
Cmil ÷ 2 ÷ K ÷ I × Vd = L

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