CI-II-54

[rg1]

I think the answer to this question is wrong by CI. What do you say?

 

[Omer]

I got around 449 V for terminal phase voltage.

(512∠14)-(47∠-acos(.82))*(.6+1.3i)= 449

 

[rg1]

I got around 449 V for terminal phase voltage.

(512∠14)-(47∠-acos(.82))*(.6+1.3i)= 449

Thanks Omer. The solution starts with changing reference of current , that too is done wrong I think and gets answer 455<9.7. 

 

[Omer]

@rg1

after thinking for a while I think my answer is wrong..

the book answer could be right if they specify in the question the PF is at the Eg. that is to say current lags Eg by 34.9.

However, general convention for PF is always at the load if nothing mentioned, that is current lags the terminal voltage Vt by 34.9.

My answer is definitely wrong since

(512∠14)-(47∠-acos(.82))*(.6+1.3i)= 447∠11.58

in this case the PF is neither at the load nor the source.

actually the problem is unsolvable if PF is at the load...

what do you think?

 

[rg1]

@rg1

after thinking for a while I think my answer is wrong.. I did not pay attention so deeply

the book answer could be right if they specify in the question the PF is at the Eg. that is to say current lags Eg by 34.9. You are right.

However, general convention for PF is always at the load if nothing mentioned, that is current lags the terminal voltage Vt by 34.9. You are right.

My answer is definitely wrong since

(512∠14)-(47∠-acos(.82))*(.6+1.3i)= 447∠11.58

in this case the PF is neither at the load nor the source.

actually the problem is unsolvable if PF is at the load... yes and no. If I assume the Vt as the reference for all the angles i.e. Vt<0 then I can write the equation V+IZ=E and equate the magnitudes on both sides of the equality. The answer for V will not differ much from what you got but if I do not know the reference at all, yes the question is not solvable. I think the owner of the question is around and must be watching us!!

what do you think?

 

[Omer]

I mean, if your reference is Vt, the problem is not solvable for the given condition.

it could be solvable in other conditions. Much higher currents or less Eg.

 

[rg1]

I mean, if your reference is Vt, the problem is not solvable for the given condition. I am not able to follow. If reference is Vt then you have a reference phasor and then why it remains unsolvable, I could not get. If no reference is there then only we say it is unsolvable.

it could be solvable in other conditions. Much higher currents or less Eg. I could not follow what is conveyed here.

 

[Omer]

Using trigonometry there is no such triangle with these values specified. 

in my pervious solution I assume the reference is Vt@0

current I is [email protected] lagging Vt

when I solve

(512∠14)-(47∠-acos(.82))*(.6+1.3i)= 447∠11.58

VT in the answer is not at 0 degree, that mean angle between I and VT is now 34.9+11.58 which is a different PF

 

[rg1]

Using trigonometry there is no such triangle with these values specified. 

in my pervious solution I assume the reference is Vt@0

current I is [email protected] lagging Vt

when I solve

(512∠14)-(47∠-acos(.82))*(.6+1.3i)= 447∠11.58

VT in the answer is not at 0 degree, that mean angle between I and VT is now 34.9+11.58 which is a different PF.

@Omer you may be right. It is actually not Vt@o but some reference other than Vt. All angles are from that reference. Yea something wrong. However I propose to leave it here; I feel this question is eating away more time than benefits. We will take it up again after exam.

 

[Stephen2awesome]

I had this exact question. The power factor is only to find the angle BETWEEN the Voltage and the current, not Current angle itself. Find the angle, arcos(.82). So the current is lagging the V by angle 34.9. We are given the angle for Voltage of 14, so the current angle will be 14-34.9, which is -20.9 degrees.

Usually, Voltage we assume its at angle 0 so the PF is the currents angle. Maybe that's what the confusion is?

 

[Omer]

I had this exact question. The power factor is only to find the angle BETWEEN the Voltage and the current, not Current angle itself. Find the angle, arcos(.82). So the current is lagging the V by angle 34.9. We are given the angle for Voltage of 14, so the current angle will be 14-34.9, which is -20.9 degrees.

Usually, Voltage we assume its at angle 0 so the PF is the currents angle. Maybe that's what the confusion is?

what you are saying is absolutely right.

the only issue with this problem is that, the PF should be taken at the load not at the source.

the current should be 34.9 lagging the terminal voltage which have to be calculated.

if we assume the PF at the source then what you mention is 100% right.

 

[Stephen2awesome]

what you are saying is absolutely right.

the only issue with this problem is that, the PF should be taken at the load not at the source.

the current should be 34.9 lagging the terminal voltage which have to be calculated.

if we assume the PF at the source then what you mention is 100% right.

In this problem, it is stated that the load draws 47A with the pf= 0.82 lagging. I took that as taking the pf at the load.