Kaplan afternoon exam problem # 38

[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?

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 <-60. Note that also means Vcn (which is what they state in their solution but calculate with wrong current and subscripts) = 236<120. Rename the center point y or something as a previous poster suggested if it will help.

 

[mull982]

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?

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.

 

[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?

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?

YES

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.

Let's call the mid point of resistor/cap a new point y and assume Va of source is 100<0 which fixes the sources.

Now using data from earlier posts and notes, Vab = 173<30.

and i = 12.2 < 75

so V yb: that is voltage of midpoint= y w.r.t. point b is 122< -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<-90 so just substitute above and you're done and you get and get 236<-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.

 

[mull982]

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?

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?

YES

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.

Let's call the mid point of resistor/cap a new point y and assume Va of source is 100<0 which fixes the sources.

Now using data from earlier posts and notes, Vab = 173<30.

and i = 12.2 < 75

so V yb: that is voltage of midpoint= y w.r.t. point b is 122< -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<-90 so just substitute above and you're done and you get and get 236<-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.

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.

 

[VectrenEng]

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.

 

[cableguy]

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.

 

[DK PE]

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?

 

[mull982]

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?

Aaaahhhaa! The light bulb just went on!

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