Parallel RL circuits and phase angles.... - Page 1
Forum section: electronics / beginners

author: cvriv
author_rank: Frequent Contributor
timestamp: March 27, 2016, 07:24:00 pm
content: Ok. This is driving me nuts! The book I'm learning from isn't thorough enough for me. I know that IL lags IR by 90deg and IR is in phase with VS. Fine. What are the phase angles of XL and R though? The book talks about susceptance(B) which is the reciprocal of reactance. Is says that BL lags G, the reciprocal of resistance, by 90deg, so BL is -90deg. That means that 1/XL = -90deg, which means that XL = 90deg.Soooo, if I had to draw of a diagram of the relationship of XL and R of a parallel RL circuit... XL = 90deg and R = 0deg? Or 1/XL = -90deg and 1/R = 0deg?Thanks in advance.

author: cvriv
author_rank: Frequent Contributor
timestamp: March 27, 2016, 07:49:36 pm
content: Of a series RL circuit, the book says that IT =IL = IR, which makes sense, but what about their phase angles? Do they share the same phase angle?Does the same apply for VS = VL=VR of a parallel circuit? Do they share the same phase angle?

author: cvriv
author_rank: Frequent Contributor
timestamp: March 27, 2016, 09:13:08 pm
content: LOL.

author: Neganur
author_rank: Unknown
timestamp: March 28, 2016, 01:09:21 am
content: This topic is a bit iffy if you learn it from scratch without proper structure. What's your learning background? Judging from your posting history you're at uni?In my experience it is a good idea to be confident in using complex numbers and having a proper introduction to the names (and their meaning) of complex electric variables.Let me start by saying that your book is using magnitudes when it refers to XL. (i.e. they refer to the absolute value of jXL). And that I am assuming your book is using a visual approach since you mention diagrams with angles (probably phasors).Susceptance is not simply the reciprocal of reactance, this is only for pure reactances (i.e. no ohmic resistance involved). It's not necessarily wrong to say B = -1/X, but it is important to remember where the minus actually comes from and that R was equal to zero in this case. I'm sure that the book mentions this before they start explaining the other stuff.Maybe some intuition will help to grasp the concepts first:1. series circuits have the same current in all components2. parallel circuits have the same voltage across them3. voltage lags current in phase for the capacitor, current lags voltage in phase for the inductorThe first two should be clear.The third I always liked to remember by thinking of how a capacitor lets current flow before the voltage builds up while it charges (and then becomes an open circuit when fully charged). For the inductor it's the opposite. Bet there's a lot of other ways to remember it.ad 1) Series circuits, same current. It's the same current in both phase and magnitude.ad 2) Parrallel circuit, same voltage. It is the same magnitude and phase.Now you ask, what about the resistance/reactance?Reactance (impedance and all those other names) depend on both current and voltage (ohms law) and since current and voltage are not in phase for a capacitor or inductor it means that the reactance isn't in phase either (it either lags or leads the pure ohmic resistance by -90 or +90 degrees).The angle between pure ohmic resistance and reactance is also either -90 or +90 degrees, just like between the voltage and current over the inductor or capacitor.To draw the relationship of XL and R in a parallel circuit, you start with the common variable - the voltage across them: Us ("U supply" or whatever). This voltage is your normal position.* Draw an arrow that represents this voltage in any direction or angle you want, it doesn't matter.* Next, the current in the resistor will be some magnitude in the same direction as the normal posistion. Draw an arrow with the same starting point as the common voltage into the same direction as the voltage (zero degrees between them).* Further, the current through the inductor will lag this voltage, so draw an arrow with some other magnitude from the same starting point as the current through the resistor at an angle of 90 degrees lagging behind the voltage (the system is turning counter-clockwise).You can now geometrically add the two currents IL and IR to get the total current Is ("I supply").Since admittance Y is equal to 1/Z and Z = Us/Is, you know that G = 1/R = IR/Us for a pure resistive element and BL = 1/XL = IL/Us for a purly reactive element, so go ahead and divide those current arrows you drew above by the voltage across them. If you do that, all that happens is that you change the name of the arrows from Is to Is/Us = Y, from IR to IR/Us = G and from IL to IL/Us = B. The angles between the arrows didn't change (only the lenghts but it doesn't matter for the drawing)The picture you now have is a nice representation of the phasors of your system. The 90 degree angles are representing the j and the length of the arrows are the magnitudes of the real and imaginary part of the complex parallel admittance.

author: cvriv
author_rank: Frequent Contributor
timestamp: March 28, 2016, 02:35:45 am
content: Thanks for the reply. I understand a lot of what you are saying. I think it will be best for me to just ask my professor in class on monday. I was hoping to clear this up before then. I think I have it straight, I just need someone to verify it. Thanks.

author: johm
author_rank: Contributor
timestamp: March 28, 2016, 04:04:51 pm
content: Here's a mnemotechnic rule to remember all the -ances (impedance(Z) resistance(R) reactance(X) admittance(Y) conductance(G) susceptance(B) inductance(L) capacitance (C)):GBY CLR ZX.Game Boy Color ZX (its processor).

author: broz
author_rank: Regular Contributor
timestamp: March 29, 2016, 03:31:03 pm
content: Quote from: Neganur on March 28, 2016, 01:09:21 amOne way I've always remembered which leads/lags which is "ELI the ICE man." Where E is referring to voltage, I to current, L to inductance, and C to capacitance.For "ELI:" since E appears before I, voltage leads current (or current lags voltage) in inductors/inductive loads.For "ICE:" since I appears before E, current leads voltage (or voltage lags current) in capacitors/capacitive loads.

author: AustinTxBob
author_rank: Contributor
timestamp: March 30, 2016, 05:51:36 pm
content: I'm taking the same class as you so let's see.Quote from: cvriv on March 27, 2016, 07:24:00 pmIf you are drawing a diagram of the impeadance, then Xl would be 90 and Xr would be 0.Quote from: cvriv on March 27, 2016, 07:49:36 pmYes they should. You can prove this to yourself. Work out a single parallel circuit with a 5A 0 degree supply with a 1k resistor and an inductor with Xl = 1K. You'll end up seeing that the resistor has a current of 3.5A 45 degrees and the inductor will have a current of 3.5A -45 degrees. Vs, Vr and Vl should all be 3535 at 45 degrees.Hope this helps.

author: Neganur
author_rank: Unknown
timestamp: March 30, 2016, 06:16:24 pm
content: The current in a resistor will be 0 degrees, in phase with the voltage over it. Always.The current through the inductor is not 45 degree either vs the voltage over it. It can never be 45 degrees.Were you maybe thinking of the total current through the circuit?

author: AustinTxBob
author_rank: Contributor
timestamp: March 30, 2016, 06:26:48 pm
content: Quote from: Neganur on March 30, 2016, 06:16:24 pmThe Ir is in phase with V; they're both at 45 degrees. I specified a current source of 5A at 0 degrees, not a voltage source.(I attached my math. Did I miss something?)

author: Neganur
author_rank: Unknown
timestamp: March 30, 2016, 07:00:40 pm
content: Nah, I just didn't read properly and it's indeed the total current.In my mind, I was also referencing to the common variable (the voltage) but you did specify the current as 0 degrees (and so of course the voltage is not 0 degrees)

author: AustinTxBob
author_rank: Contributor
timestamp: March 30, 2016, 07:09:02 pm
content: Quote from: Neganur on March 30, 2016, 07:00:40 pmIn that case, I might pass my test tomorrow.

author: Neganur
author_rank: Unknown
timestamp: March 30, 2016, 07:24:23 pm
content: Fingers crossed!

author: bson
author_rank: Unknown
timestamp: March 31, 2016, 03:44:30 am
content: Quote from: cvriv on March 27, 2016, 07:24:00 pmYes. Yes. They're both correct.\$\frac{1}{1+1j} = 0.5-0.5j\$The reciprocal of a complex number causes mirroring around the real axis. Aka conjugation. If you think about it this makes sense since the reactances \$X_C = 1/\omega C\$ and \$X_L = \omega L\$ are 180 degrees apart and cancel. (\$Z_C = -j X_C\$ and \$Z_L = j X_L\$ respectively)And this is an excellent excuse to try out MathJax.