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.oh wait wait!! I think we covered that right. V1 is 8V.
So:
I1 = E1/R1 = 8/2 = 4A
Am I good?
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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