Steady-state analysis and impedance-admittance calculations

Zazoo

Joined Jul 27, 2011
114
oh wait wait!! I think we covered that right. V1 is 8V.
So:
I1 = E1/R1 = 8/2 = 4A
Am I good?
No. I've attached an image which shows the circuit with phasor quantites and complex impedances filled in. This is just a collection of all the information you've found up to now.

Each node in the circuit is labled with a number in a circle.
* Node 0 - This is the reference node (ground), this node is taken to be 0V and the voltages of all other nodes are stated with respect to this node. Selection of this node is arbitrary, although it is typically best to pick the node connected to the most elements and/or most voltage sources.
* Node 3 - This is the node between the resistor and inductor, however because we are treating the two components as a single complex impedance, we can ignore this node.
* Node 2 - This node is at the top terminal of the voltage source (+). Since the bottom of the voltage source is at 0 (ground) the top must be at 8<(∏/4) volts
* Node 1 - This is the unknown node we must solve for, V1

V1 is not 8<(∏/4) volts because, in going from the voltage source to V1, there is an additional voltage drop across the resistor (Z1).

When writing the node equation, you are basically stating KCL for the node using ohm's law.
For example, KCL on node 1 would be: I = I1 + I2 + I3

Now, replace each current with its Ohm's law equivalent. I = V/Z

For example, doing the left-most branch, containing current I:

I = (V2 - V1)/Z1

Note that the voltage across the element is used when setting these up (i.e. the difference between the node voltages at each end.) V2 is the node voltage at node 2, but as mentioned above, this is already known and is just 8<(∏/4). V1 is our unknown voltage. And, Z1 is the impedance for that branch (2Ω)

So I = [8<(∏/4) - V1]/2

Putting this into the original KCL equation:

I = I1 + I2 + I3 becomes [8<(∏/4) - V1]/2 = I1 + I2 + I3

You can replace I1, I2 and I3 by the same process. In the end you will have a single equation with only one variable, V1. Then it is just algebra to get V1 isolated.

From there you can use ohm's law to get all circuit values directly.
 

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Zazoo

Joined Jul 27, 2011
114
Setup looks good, but in going from line 3 to line 4 you made a mistake in adding V1 from the LHS to 2V1 from the RHS:
 

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Thread Starter

lemon

Joined Jan 28, 2010
125
oh wow. I think I just had a brain surge!
So the reason for doing this node analysis and starting at the V1 node in the position between as many elements as possible is because we now have the voltage which will be the same across all the parallel components and we can then deal with the final element on its own no problem.
But Vr and Vl are in series and I need to do the calculation of their combined impedance Z4 to find the current and then I can find their individual values. But that calculation looks impossible and I don't know what j is. Is j -1?
I have attached images of work done so far.
 

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t_n_k

Joined Mar 6, 2009
5,455
oh wow. I think I just had a brain surge!
So the reason for doing this node analysis and starting at the V1 node in the position between as many elements as possible is because we now have the voltage which will be the same across all the parallel components and we can then deal with the final element on its own no problem.
That's a good observation.

But that calculation looks impossible and I don't know what j is. Is j -1?
On the matter of the complex operator 'j' .

By definition j=√(-1) ..... hence its complex nature.

To progress in AC circuit analysis you will need a reasonable grounding in complex number operations. Presumably you've not previously studied complex numbers.
 

Thread Starter

lemon

Joined Jan 28, 2010
125
oh - I had a brain fart - Yes, of course j=√(-1). I have studied complex numbers previously but only at a basic level. I need to finish this for some coursework though and my module is graded entirely on this work so I gotta go for it.
I just need to get this into polar or rectangular form though so I can use Ohm's law to find I.
 
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