# Kaplan afternoon exam problem # 38



## mull982 (Oct 17, 2010)

In the Kaplan sample afternoon exam problem #38 I understand the solution they give however I'm curios if thres another way to find thid solution.

They used the Iab current dropped across the 10ohm resistor to determine the voltage dropped across the resistor and then use this voltage drop subtracted from Vac to find the Vcn voltage.

Can you find the same solution by calculating the voltage drop across the -j10ohm capacitor and then subtract this voltage to find Vcn? To do this would you have to use Iba current phasor instead of Iab? Also would you use Vbc or Vcb to subtract the voltage across -j10 capactor from?

I'll try to scan and post the problem tomorrow.


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## mull982 (Oct 18, 2010)

Attached is the problem that I am referring to.

KMBT25020101018084107.pdf


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## Benee (Oct 18, 2010)

mull982 said:


> In the Kaplan sample afternoon exam problem #38 I understand the solution they give however I'm curios if thres another way to find thid solution.
> They used the Iab current dropped across the 10ohm resistor to determine the voltage dropped across the resistor and then use this voltage drop subtracted from Vac to find the Vcn voltage.
> 
> Can you find the same solution by calculating the voltage drop across the -j10ohm capacitor and then subtract this voltage to find Vcn? To do this would you have to use Iba current phasor instead of Iab? Also would you use Vbc or Vcb to subtract the voltage across -j10 capactor from?
> ...



I worked it out and I have the VnC is 236.62&lt;-60 Volt. which is not match to none of the choice in the question. Can you post the solution ?


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## mull982 (Oct 18, 2010)

Benee said:


> mull982 said:
> 
> 
> > In the Kaplan sample afternoon exam problem #38 I understand the solution they give however I'm curios if thres another way to find thid solution.
> ...


Attached is the solution that was given.

KMBT25020101018114850.pdf


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## Benee (Oct 18, 2010)

mull982 said:


> In the Kaplan sample afternoon exam problem #38 I understand the solution they give however I'm curios if thres another way to find thid solution.
> They used the Iab current dropped across the 10ohm resistor to determine the voltage dropped across the resistor and then use this voltage drop subtracted from Vac to find the Vcn voltage.
> 
> Can you find the same solution by calculating the voltage drop across the -j10ohm capacitor and then subtract this voltage to find Vcn? To do this would you have to use Iba current phasor instead of Iab? Also would you use Vbc or Vcb to subtract the voltage across -j10 capactor from?
> ...



Yes, you can do either way and you will get the same answer, by the way, the solution is have an error, The current Iab is 12.247&lt;75 instead 1.2247&lt;75.


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## mull982 (Oct 18, 2010)

Benee said:


> mull982 said:
> 
> 
> > In the Kaplan sample afternoon exam problem #38 I understand the solution they give however I'm curios if thres another way to find thid solution.
> ...



Thanks! Can you briefly show me how to do it the other way. I cant seem to arrive at the same answer doing it the other way. I'm thinking I would have to use Vbc instead of Vca and use the current Iba instead of Iab?


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## Benee (Oct 18, 2010)

mull982 said:


> Benee said:
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This is what I got but the solution is not match with any choice in the solution

Vnc = VnB + VBC

Vnc = (12.25 &lt;75)* (-10j) + 173.2&lt;-90) = 236.62&lt;-60


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## DK PE (Oct 18, 2010)

Benee said:


> This is what I got but the solution is not match with any choice in the solutionVnc = VnB + VBC
> 
> Vnc = (12.25 &lt;75)* (-10j) + 173.2&lt;-90) = 236.62&lt;-60


I punched it out quickly but I agree with your 236V &lt;-60.


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## cableguy (Oct 19, 2010)

I got 236&lt;-60 as well, but did it a bit differently.

I set it up as a loop circuit with a source of 100&lt;0 on the left, source of 100&lt;240 on the right, and 10-j10 in the middle between them. Solve for I.

Then the voltage to neutral between R and X is 100&lt;0 minus IR.

Then the voltage at the middle, referenced to C, is Vmiddle (136&lt;-60) minus 100&lt;120.


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## Flyer_PE (Oct 19, 2010)

I looked at their solution. They botched the current by a factor of 10.

173/30o/(10-j10) = 12.25/75

Their solution is showing 1.225/75 which would seem to indicate they should have set the impedance to 100-j100 rather than 10-j10.


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## cableguy (Oct 19, 2010)

Kaplan botched a whole bunch of their solutions. I'm starting a list as I'm reworking them. ld-025: Stupid stuff like giving us an efficiency of .85, and then using .80 in the calculations (Morning #4). Or forgetting to divide by 3 for the capacitance value (Morning #2). Or doing the 2x the subtraction 6.818-5.249 and getting 3.3138 as the answer (and that's marked as "a") (Afternoon 21). Their proofreading sucks.


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## Flyer_PE (Oct 19, 2010)

^If it makes you feel any better, I had an instructor in college that was always doing stuff like that on problems he created for homework. I had the material down cold in that class because I was literally working the problems backwards and forwards trying to figure out where the screw-up was. The process is painful but may very well serve you well when you take the exam.


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## mull982 (Oct 20, 2010)

Benee said:


> mull982 said:
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> > Benee said:
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To get Vnc dont you have to use VnB + VCB instead of VBC. VBC would be a vector pointed downwared and would not give you the corrosponding Vnc value. VCB would be a value pointing upward and would give you the corrosponding Vnc value according to the ABC phase rotation.

Do you agree?


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## mull982 (Oct 20, 2010)

Another part I dont get is that their final answer for Vcn= [email protected] does not match the VCN shown in the phasor diagram in the solution. The phasor for VCN they show in the solution looks to be at about -240deg, unless this is the phasor of the supply voltage?

I think I may be getting my phasors confused???


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## mull982 (Oct 20, 2010)

[No message]


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## cableguy (Oct 20, 2010)

mull982 said:


> Another part I dont get is that their final answer for Vcn= [email protected] does not match the VCN shown in the phasor diagram in the solution.


This is because they used the wrong current value. They used 1.2247 as the current magnitude, they should have used 12.247... 176 is just flat out wrong, you have to ignore it.

If you use the correct current value in their formula, you get 236 at angle -60 degrees...

And their phasor diagram is wrong. Just another wonky Kaplan problem... lol.


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## mull982 (Oct 20, 2010)

cableguy said:


> mull982 said:
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> > Another part I dont get is that their final answer for Vcn= [email protected] does not match the VCN shown in the phasor diagram in the solution.
> ...


O.k. heres the part I'm hung up on maybe someone can help.

I understand in their solution (although current is wrong) that they subtracted the Van voltage or Vr from Vac to come up with the Vcn voltage.

I understand that this problem can be solved a similar way using Vbc and the voltage drop across the -j10 cap however I dont see why when using their solution you subtract Vr or Van from Vac but when using the alternate solution shown above, you add Vbn to Vbc instead of subtracting Vbn from Vbc. In other words why is the solution Vcn=Vbn+Vbc instead of Vcn=Vb-Vbn as is used with the origonal solution?


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## mull982 (Oct 21, 2010)

Can someone please help me identify what I am doing wrong with understanding the phaosor notation above? It is driving me nuts


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## DK PE (Oct 21, 2010)

mull982 said:


> Can someone please help me identify what I am doing wrong with understanding the phaosor notation above? It is driving me nuts


Using double subscript notation you would get V nc = V nb + V bc, correct? One thing that may be tripping you up is remember the current (passive sign convention, so + sign) is entering the "n" side terminal of cap where the sign on the resistor is opposite so they add there. Does that help at all?


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## cableguy (Oct 21, 2010)

Looking a little more closely, I notice that the middle point of the Y is labeled "n", and they have another "N" floating around the spot where they labeled phase B.

Relabel that middle point as "y" or something instead of "n".

Then you see that they're really solving for Vcy.

This means that the phasor diagram they drew is "mostly" correct - they used capital N for V(CN) and V(BN), whereas V(Cn) is a completely different phasor.

The use of n and N make the solution fairly confusing... is that where it's going wrong for you?

Also, redraw it with 3 different voltage sources (100 angle 0, 100 angle 240, 100 angle 120), and an open circuit where the voltmeter is.


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## DK PE (Oct 21, 2010)

DK PE said:


> mull982 said:
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> 
> > Can someone please help me identify what I am doing wrong with understanding the phaosor notation above? It is driving me nuts
> ...


In addition to the the mistake on the current calculation, their subscript notation is wrong which may be what is misleading you. Take their solution, toss it away and see if by drawing it out you can get V nc = 236 &lt;-60. Note that also means Vcn (which is what they state in their solution but calculate with wrong current and subscripts) = 236&lt;120. Rename the center point y or something as a previous poster suggested if it will help.


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## mull982 (Oct 22, 2010)

DK PE said:


> mull982 said:
> 
> 
> > Can someone please help me identify what I am doing wrong with understanding the phaosor notation above? It is driving me nuts
> ...


O.k. that is starting to make a little more sense. So with double subscript notation the inner terms out of the group of 4 terms cancel (in this case the b's) and we are left with a vector consisting of the outside terms? Is this the same for subtracting vectors as well? This may be where my hangup is?

How can you tell which side of the cap is the + sign when you say that it is entering the "n" side? Can you briefly desctibe this current flow notation.

I think I will have it after these become clear. Thanks for all the help.


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## DK PE (Oct 22, 2010)

mull982 said:


> DK PE said:
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Let's call the mid point of resistor/cap a new point y and assume Va of source is 100&lt;0 which fixes the sources.

Now using data from earlier posts and notes, Vab = 173&lt;30.

and i = 12.2 &lt; 75

so V yb: that is voltage of midpoint= y w.r.t. point b is 122&lt; -15

note this passes snif test as current leads in the capacitor so i is at angle of +75 and voltage across cap is at -15 so 90 degree difference.

Using double subscript notation you would get V yc = V yb + V bc,

We know V bc = 173&lt;-90 so just substitute above and you're done and you get and get 236&lt;-60. Note this is equivalent to the former V nc

You don't really have to handle the + terminal of capacitor but using passive sign convention and given the direction we have chosen for current, the + terminal for the cap voltage is where the current enters which is the midpoint of C and R.


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## mull982 (Oct 22, 2010)

DK PE said:


> mull982 said:
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> > DK PE said:
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O.k. that makes sense. I see how this arrives at the solution using the voltage yb. The only part that isn't jiving with me is why we use Vbc for adding to vyb? I see that this makes sense in the equation using doulbe notation but is this the only reason we choos this? Because I just cant see it circuit wise when looking at the circuit.


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## VectrenEng (Oct 22, 2010)

Is everyone in agreement that the "N" next to "B" represents neutral or ground? If so, I am trying to think of an application on a wye configuration to ground one phase; it is somewhat common on delta networks.


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## cableguy (Oct 22, 2010)

VectrenEng said:


> Is everyone in agreement that the "N" next to "B" represents neutral or ground? If so, I am trying to think of an application on a wye configuration to ground one phase; it is somewhat common on delta networks.


I agree with this. Perhaps this is their way of telling us that the system voltages are solid and won't be moved around by an unbalanced 2- (or 3) phase load.


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## DK PE (Oct 25, 2010)

mull982 said:


> O.k. that makes sense. I see how this arrives at the solution using the voltage yb. The only part that isn't jiving with me is why we use Vbc for adding to vyb? I see that this makes sense in the equation using doulbe notation but is this the only reason we choos this? Because I just cant see it circuit wise when looking at the circuit.


I think the clearest circuit explanation I can think of is:

Think of the bottom part of circuit of node c, then V bc + V yb as two batteries in series in a flashlight. Just like two 1.5 v cells add to 3V, V bc and V yb add to V yc (the path up the right side which is actually a high impedance meter.... this help?


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## mull982 (Oct 25, 2010)

DK PE said:


> mull982 said:
> 
> 
> > O.k. that makes sense. I see how this arrives at the solution using the voltage yb. The only part that isn't jiving with me is why we use Vbc for adding to vyb? I see that this makes sense in the equation using doulbe notation but is this the only reason we choos this? Because I just cant see it circuit wise when looking at the circuit.
> ...


Aaaahhhaa! The light bulb just went on!

Great example this helped me see whats going on very clearly now. Thanks for the help!!!


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