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Your question isn't very clear. What exactly is the problem? The red segment is a reactance and the green segment is resistance. Where the two segments meet, there is an angle--θ2.
In an impedance diagram (in phasor form) of resistance/reactance, inductive reactance is shown as a positive angle, capacitive reactance is a negative angle, and resistance is a straight line along the origin. The green segment, I believe, is representing a resistance (hence the 'R' in I1Re). I may be wrong, but I believe that's how this problem is set up.
I don't know if what you say is correct or not; I'm still not able to grasp that what leads the author to conclude that the I1X1 makes an angle equal to phi_2.
All to do with properties of triangles rather than electrical engineering.
Look at my re-hashed version of your diagram.
There is always 180 degrees in a triangle.
Consider the trangle ADC, if angle CAD is Phi2 and angle ADC is a right angle, then angle ACD is (90 - Phi2).
Now consider the triangle ABC, if angle ACB is a right angle (90deg), we already know that angle ACD is (90 - Phi2), then angle BCD must be (90 - (0 - Phi2)) which is equal to Phi2.
All to do with properties of triangles rather than electrical engineering.
Look at my re-hashed version of your diagram.
There is always 180 degrees in a triangle.
Consider the trangle ADC, if angle CAD is Phi2 and angle ADC is a right angle, then angle ACD is (90 - Phi2).
Now consider the triangle ABC, if angle ACB is a right angle (90deg), we already know that angle ACD is (90 - Phi2), then angle BCD must be (90 - (0 - Phi2)) which is equal to Phi2.
Thanks a lot, Jim. And DerStrom thanks to you too for being there. Perhaps, I didn't think of the problem in terms of geometry of triangles rather I was too preoccupied with reactance, impedance etc. and that's why didn't phrased the question correctly.
There is a common well known theorem in plane geometry, which states that if two sides of an angle are perpendicular to two sides of another angle, left side to left side and right side to right side, then the two angles are equal. If they are perpendicular left side to right side and right side to left side, then the angles are supplementary.
Ze is an impedance, made up of a resistance and a reactance. We'll start with the first part: Ze cos(θe - θ2) (the 'e's and '2' are subscript, by the way). There is a rule in trigonometry that says cos (x - y) is equal to ((cos x * cos y) + (sin x * sin y)). Therefore, Ze cos(θe - θ2) is equal to Ze (cos θe * cos θ2 + sin θe * sin θ2). When you split up the impedance, and think of sin and cos on a real-imaginary plane (where x is real and 'y' is imaginary (j)), the cos corresponds to the real value, and sin corresponds to j. Since resistance is real and reactance is imaginary, you'll end up with (Re cos θ2 + Xe sin θ2).
Do you follow? I realize that's probably pretty hard to read. I think if you read through it a few times though, you might get the gist of it.
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