Voltage drop question?

[kwired]

Of course it's a 240V range. What else would it be?

9.6 kW heater?

 

[Square D]

In this forum?!

I didn't realize there would be this much action in the thread either! But it has me fired up, I have taken this test (LLE) twice, and am not going to fail the third time, so I am going to resume my studies from the ground up. Thanks

 

[Square D]

Of course it's a 240V range. What else would it be?

The question stated "the appliance has a bad splice on "one of the phase conductors" so yeah probably a 240 range.

 

[ptonsparky]

Ok, this is a practice question I came across on a PSI exam. Does anyone know how they came up with this number? I apologize for my ignorance but I'm here to learn something. An appliance draws 40 amperes when used on a circuit. A bad splice is discovered on the line at the appliance on one of the phase conductors. The splice has a measured resistance of 0.5 ohms. How much power is lost at the splice? Answer: 800 watts

Here in lies my problem. I assumed several things.
I thought 240 volt. Voltage is not mentioned.
I assumed phase was single, but that isn't mentioned either.
Type of appliance is irrelevant, as everything I needed to know was given.

 

[kwired]

Here in lies my problem. I assumed several things.
I thought 240 volt. Voltage is not mentioned.
I assumed phase was single, but that isn't mentioned either.
Type of appliance is irrelevant, as everything I needed to know was given.

The one thing not really clear is whether the appliance is rated for 40 amps or is currently drawing 40 amps with that non intended resistance in the circuit.

If it were pure resistance load rated for 40 amps then you added another .5 ohms in series actual amps drawn drops to about 36.9.

 

[ptonsparky]

The one thing not really clear is whether the appliance is rated for 40 amps or is currently drawing 40 amps with that non intended resistance in the circuit.

If it were pure resistance load rated for 40 amps then you added another .5 ohms in series actual amps drawn drops to about 36.9.

That was my results but it didn't mesh with the answer.

 

[kwired]

That was my results but it didn't mesh with the answer.

Yes if there is multiple choice answers one can probably figure out what way they should go with it, as long as both possible answers are not in the selection list.

 

[LarryFine]

The one thing not really clear is whether the appliance is rated for 40 amps or is currently drawing 40 amps with that non intended resistance in the circuit.

If it were pure resistance load rated for 40 amps then you added another .5 ohms in series actual amps drawn drops to about 36.9.

In my opinion, that's real-world detail that is overthinking for a test question.

 

[kwired]

In my opinion, that's real-world detail that is overthinking for a test question.

Longer you have been in real world the more you think real world scenarios though

I likely question the questions more now than I would have questioned the same things I was at entry levels.

Other thing is you generally don't need to Ace the exam to pass, so as long as there aren't too many tricky questions it shouldn't be a problem.

 

[tallgirl]

Assuming it was multiple choice, the set of answers should lead to the single correct answer unless someone is just being mean.

As other said, P = I^2 R. Both I and R were, given, and truly volts aren't needed. In that situation, line voltage doesn't matter (it really doesn't) since what you need to calculate P is Vdrop, not Vline. In this instance, what you get for V in V = I * R is Vdrop. In this case, 40 amps * 0.5 ohms yields Vdrop = 20 volts. Now do P = I * V and you get 40 amps * 20 volts of voltage drop is 800 watts -- the same as P = I^2 * R.

Remember that Ohm's Law, taken as a whole, just needs two of P, I, R and V to derive a third value. You just need to know which "third value" you're deriving. In this case, it's the voltage drop across the series resistance. You can then take those three values and derive the fourth if you've remembered only 3/4th of the giant pile of equations.

For example, for a purely resistive load, you now have Vline = 220 (assuming Vline was originally 240). If the load was 40A * 240V = 9.6kW ("assume a perfect conductor"), you now have 40A * 220V = 8.8kW. You can also calculate R for the load since R = V / I -- 240 / 40 = 6 ohms. Now go back and calculate P = I ^ 2 * R and you get 40 * 40 * 6 = 9.6kW.

Such fun. I should have gotten that BSEE instead of wasting my life doing other things.

 

[Square D]

All these replies are helpful. The ideas is this thread are why I made the post. I was trying to plug "real world" numbers in to solve a vague question. The first couple of response posts made me think about Ohms law wheel ALOT.

 

[tallgirl]

My biggest Ohm's Law wheel realization was that I didn't have to remember the entire thing. Which for 12 or 13 year old me was a daunting task.

 

[don_resqcapt19]

Assuming it was multiple choice, the set of answers should lead to the single correct answer unless someone is just being mean.

As other said, P = I^2 R. Both I and R were, given, and truly volts aren't needed. In that situation, line voltage doesn't matter (it really doesn't) since what you need to calculate P is Vdrop, not Vline. In this instance, what you get for V in V = I * R is Vdrop. In this case, 40 amps * 0.5 ohms yields Vdrop = 20 volts. Now do P = I * V and you get 40 amps * 20 volts of voltage drop is 800 watts -- the same as P = I^2 * R.

Remember that Ohm's Law, taken as a whole, just needs two of P, I, R and V to derive a third value. You just need to know which "third value" you're deriving. In this case, it's the voltage drop across the series resistance. You can then take those three values and derive the fourth if you've remembered only 3/4th of the giant pile of equations.

For example, for a purely resistive load, you now have Vline = 220 (assuming Vline was originally 240). If the load was 40A * 240V = 9.6kW ("assume a perfect conductor"), you now have 40A * 220V = 8.8kW. You can also calculate R for the load since R = V / I -- 240 / 40 = 6 ohms. Now go back and calculate P = I ^ 2 * R and you get 40 * 40 * 6 = 9.6kW.

Such fun. I should have gotten that BSEE instead of wasting my life doing other things.

I have seen many times, that where a simple math mistake can lead to an incorrect answer, that incorrect answer will be give in the 4 choices.

 

[tallgirl]

I have seen many times, that where a simple math mistake can lead to an incorrect answer, that incorrect answer will be give in the 4 choices.

Yes, that is a thing. With something like Ohm's Law that might involve swapping numerator and denominator in a equation. Unless it's a timed test where running out of time is part of the strategy for creating non-perfect scores, use that extra time to check your work. That's why I included the 2nd to last paragraph -- you can often check your work with some other equation you should also know by heart.

 

[kwired]

I have seen many times, that where a simple math mistake can lead to an incorrect answer, that incorrect answer will be give in the 4 choices.

Or maybe math was correct but if you skip a certain step in a process will lead to an answer they put in the possible selections.

 

[LadyDi]

The tricky part of calculations is understanding when to use the values listed outside the Formula Wheel and when to apply the specific rules and formulas regarding the type of circuit rules (i.e., series or parallel) to calculate the total values. Here’s a suggestion that might help.

View attachment 2560150
Step 1: Solve individual branch circuit components using the values in the outer Formula circle.

Step 2: Solve the total values using the specific type of circuit with those rules and formulas.

In this question, you’re dealing with an individual component and trying to discover Power (P) in watts. The test writer provided two values in the question. In the green-colored section, choose the formula based on the two values provided in the question.

• I= 40 amps
• R= 0.5 ohms

Therefore, choose the formula P= I (squared) X R

Next, substitute the known values.

• P= 0.50
• P= 800 watts

Now substitute the known values and solve the problem.

• P= 0.50
• P= 800 watts

 

[LadyDi]

Go to the free stuff on Mike Holt to get a cool Ohm's Law Poster.

Also, his is Formula wheel is not green-colored like the one I wanted to upload but could not because of the rules.

In the other noted case, (totals or type of circuit) use the specific rules and formulas that apply. For example, look at the rules and formulas for parallel circuits found on the attachment. I made this for my students. Use it if it helps you.

 

[LadyDi]

The tricky part of calculations is understanding when to use the values listed outside the Formula Wheel and when to apply the specific rules and formulas regarding the type of circuit rules (i.e., series or parallel) to calculate the total values. Here’s a suggestion that might help.

View attachment 2560150
Step 1: Solve individual branch circuit components using the values in the outer Formula circle.

Step 2: Solve the total values using the specific type of circuit with those rules and formulas.

In this question, you’re dealing with an individual component and trying to discover Power (P) in watts. The test writer provided two values in the question. In the green-colored section, choose the formula based on the two values provided in the question.

• I= 40 amps
• R= 0.5 ohms

Therefore, choose the formula P= I (squared) X R

Next, substitute the known values.

• P= 0.50
• P= 800 watts

Now substitute the known values and solve the problem.

• P= 0.50
• P= 800 watts

I=40 Squared R=0.50 40X40X0.50=800 LOL always check your stuff!

 

[LarryFine]

• I= 40 amps
• R= 0.5 ohms

Therefore, choose the formula P= I (squared) X R

The problem I have with that type of formula is that, on the rare occasions I need to use the formulae while working, it's too hard to remember, let alone use the ones with squares and square roots.

It's so much easier to remember and use the basics, especially since I usually do the math in my head. In my opinion, using E = I x R and P = E x I works so well and easily that it's worth the extra step.

 

[ptonsparky]

Eat In Restaurant
what do you eat there?
PIE