Kaplan afternoon exam problem # 38

[mull982]

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.

 

[mull982]

Attached is the problem that I am referring to.

KMBT25020101018084107.pdf

 

[Benee]

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.

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

 

[mull982]

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.

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

Attached is the solution that was given.

KMBT25020101018114850.pdf

 

[Benee]

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.

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<75 instead 1.2247<75.

 

[mull982]

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.

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<75 instead 1.2247<75.

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?

 

[Benee]

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.

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<75 instead 1.2247<75.

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?

This is what I got but the solution is not match with any choice in the solution

Vnc = VnB + VBC

Vnc = (12.25 <75)* (-10j) + 173.2<-90) = 236.62<-60

 

[DK PE]

This is what I got but the solution is not match with any choice in the solutionVnc = VnB + VBC

Vnc = (12.25 <75)* (-10j) + 173.2<-90) = 236.62<-60

I punched it out quickly but I agree with your 236V <-60.

 

[cableguy]

I got 236<-60 as well, but did it a bit differently.

I set it up as a loop circuit with a source of 100<0 on the left, source of 100<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<0 minus IR.

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

 

[Flyer_PE]

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.

 

[cableguy]

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.

 

[Flyer_PE]

^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.

 

[mull982]

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.

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<75 instead 1.2247<75.

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?

This is what I got but the solution is not match with any choice in the solution

Vnc = VnB + VBC

Vnc = (12.25 <75)* (-10j) + 173.2<-90) = 236.62<-60

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?

 

[mull982]

Another part I dont get is that their final answer for Vcn= 176@-34deg 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???

 

[mull982]

[No message]

 

[cableguy]

Another part I dont get is that their final answer for Vcn= 176@-34deg 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.

 

[mull982]

Another part I dont get is that their final answer for Vcn= 176@-34deg 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.

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?

 

[mull982]

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

 

[DK PE]

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?

 

[cableguy]

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.