Watts from ohms one line
- Thread starter Rjv0858
- Start date
- Status
- Location
- Tennessee NEC:2017
- Occupation
- Semi-Retired Electrician
You don't have to get Ugly about it!
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
No. Imagine or draw the circuit with all switches closed instead of open as shown.
(As shown, with the switches open, it's 35 ohms.)
- Location
- Tennessee NEC:2017
- Occupation
- Semi-Retired Electrician
Nope
- Location
- Tennessee NEC:2017
- Occupation
- Semi-Retired Electrician
Use the formula in the "pie" chart for power, I²R. You know the current is 10A, if the resistance was 50ohms, the answer would be 5000W. 5000 was not one of the choices for the answer.
BillyMac59
Member
- Location
- Wasaga Beach, Ontario
- Occupation
- Industrial Electrician
With all switches closed, SW3 effectively shorts the 5 and 20 ohm resistors. The only resistance in the circuit is therefore the two parallel 10 ohm resistors. The quick calculation for parallel resistors is their product over their sum. It simplifies further when the resistors have the same ohmic value. The total resistance is the resistor value divided by two. Here, total circuit resistance is 5 ohms. Using Ohm's Law Power = I X I X R =500 watts. Basic DC circuit analysis (from 35 years ago I might add)
So this is the best explanation I have gotten, do you know of a video or anything I can watch that teaches this. People have to understand my bigger problem is I can't find anything that teaches this. The book says the answer is 500 watts.
4x4dually
Senior Member
- Location
- Stillwater, OK
- Occupation
- Electrical Engineer/ Ex-Electrician
If all switches are closed, current will only flow through the two 10 ohms paths. SW3 shorts out all other paths.
Shortcut for equal parallel resistances paths is to divide in half.
V=I*R
Voltage across both 10 ohms resistors (5 ohms combined) is 10A * 5 Ohms = 50.
P=VI
Power dropped across both 10 ohm resistors is 50V * 10A = 500W
Long cut for parallel resistance: Rp = 1/R1 + 1/R2 + ............. +1/Rn
Shortcut for equal parallel resistances paths is to divide in half.
V=I*R
Voltage across both 10 ohms resistors (5 ohms combined) is 10A * 5 Ohms = 50.
P=VI
Power dropped across both 10 ohm resistors is 50V * 10A = 500W
Long cut for parallel resistance: Rp = 1/R1 + 1/R2 + ............. +1/Rn
Doesn't resistance usually show a load?
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
I am always amazed at how poorly electrician apprentices are taught the really basic stuff about electricity. This should have been something the op figured out in his head in about ten seconds.
The other issue that just makes me crazy is how few electricians can actually read electrical drawings and understand them.
The other issue that just makes me crazy is how few electricians can actually read electrical drawings and understand them.
The problem is a test of your knowledge of Ohm’s law. Google search will provide plenty of instructional information. The switches in the problem are really just a distraction. A basic, methodical way to solve the problem is to re-draw the circuit without all of the elements that are made irrelevant when the switches are closed. Then apply Ohm’s law to the resulting, simplified circuit.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Oh, well. So much for having the OP figure it out for himself.
LolOh, well. So much for having the OP figure it out for himself.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
I was hoping you would see that the only components that would contribute to the question as asked were the two 10-ohm resistors in parallel. Then you would have the two variables, 5 ohms and 10 amps, and use them to determine that the formula to use would be:
E = I x R = 5 x 10 = 50 volts
Then, to find power, you would use the two (one new, one already known) variables, 50 volts and 10 amps, to determine you need to use:
P = E x I = 50 x 10 = 500 watts
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
The whole purpose of that exercise, once you get the resistance, is to figure out what value you're looking for (P in this case), what values you know (I and R), and what formula you need to use to get there.
I prefer the two-step method, using E = I x R and P = E x I, over using the chart that uses squares and square roots, because I can remember the two simple formulae and use them in my head without a calculator.
But, this pie chart does show how to determine the value you're seeking, which will be one of the four values in the center circle, and then knowing which two values you have will tell you which formula to use.
I prefer the two-step method, using E = I x R and P = E x I, over using the chart that uses squares and square roots, because I can remember the two simple formulae and use them in my head without a calculator.
But, this pie chart does show how to determine the value you're seeking, which will be one of the four values in the center circle, and then knowing which two values you have will tell you which formula to use.
The only main issue I was having was getting the total ohms. I thought all resistors represented a load like a light bulb.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Basically, yes. Bulbs in parallel pass more current than one will, while bulbs in series pass less current than one will.
Resistors in series end up with a total greater than the largest one, because the current must pass through them all.
Resistors in parallel end up with a total lower than the smallest one, because the current has multiple pathways.
BillyMac59
Member
- Location
- Wasaga Beach, Ontario
- Occupation
- Industrial Electrician
Back to DC circuit analysis 101:
In a purely series circuit, the total resistance equals the sum of the individual resistances. In a purely parallel circuit, the RECIPROCAL of the total resistance is equal to the sum of the RECIPROCAL of the individual resistors. In the most basic parallel circuit of 2 resistors, the reduces to the product over the sums. Check with Prof. Google for basic DC circuit analysis for beginners.
In a purely series circuit, the total resistance equals the sum of the individual resistances. In a purely parallel circuit, the RECIPROCAL of the total resistance is equal to the sum of the RECIPROCAL of the individual resistors. In the most basic parallel circuit of 2 resistors, the reduces to the product over the sums. Check with Prof. Google for basic DC circuit analysis for beginners.
I'm not sure I understand what you are saying, but yes, resistors are normally a load. The exceptions are
1. When they are shorted out. (S2 shorts the 2 20ohm resistors. S3 shorts out 3 resistors).
2. When they are open and current cant flow through them. (S1 opens one 10 ohm, S4 opens one 20 ohm).
The formula for parallel resistors is 1/Rtotal = 1/R1 + 1/R2 + 1/R3....
This simplifies to to Rtotal = 1/ ( 1/R1 + 1/R2.....)
For equal resistors in parallel, the resistance is just divided by the numbers of resistors. (2) 10 ohm in parallel are 5 ohms. (3) 10 ohms in parallel are 3.33 ohms. 4 in parallel are 2.5 ohms.
Try the equation above with some of these values and see if you can get the same answers. Then you'll have some confidence doing both the calculations, and using the shortcuts.
- Status