# Finding the current it i(0-)

Discussion in 'Homework Help' started by StuckStudent123, Oct 13, 2016.

1. ### StuckStudent123 Thread Starter New Member

Sep 23, 2016
10
0
Hi,
I'm trying to find iot on the diagram just before the switch is opened. I know that the answer is 556mA from a simulation but I can't find out why. Initially I thought it would be as simple as 4v/12ohms but then I saw that this is the incorrect answer. I also tried to calculate the voltage between the 4V, 6ohms, and 2H but I'm not sure how using nodal analysis because the 4V doesn't have a component between it and the node.
Would anyone be able to help?

Thanks

2. ### ci139 Member

Jul 11, 2016
343
39
your 4Ω is eighter the series resistance for 2H ? coreless inductor ? oh, boy
or the circuit is idealized - the inductor has only an inductance no resistance and the R4 is standalone . . . resistor

if the currents are balanced the nameless 2H has no voltage drop (because i.const · 0Ω = 0V as if it was not there)

3. ### WBahn Moderator

Mar 31, 2012
20,225
5,749
You are making one of the classic mistakes in that you see a voltage source and you see a resistance and so you just threw them at Ohm's Law. This generally indicates that you are approaching things from the viewpoint of just grabbing equations that have the right letters in them instead of understanding what those equations mean. Ohm's Law, for instance, relates a resistance to the voltage across THAT resistance and the current flowing through THAT resistance. Does the 4 V source appear ACROSS the 12 Ω resistor? If not, then you can't just throw those numbers at Ohm's Law and expect any meaningful result.

Have you tried mesh analysis?

Have you considered using a Thevenin equivalent?

Do you know either of those techniques?

If so, the Thevenin equivalent circuit can be written down by inspection, treating the 12 Ω as the load. The open circuit voltage is particularly easy to calculate in your head. The equivalent resistance is only just a bit more difficult to do mentally.

Redraw the circuit as it effectively appears just before the switch is opened? What do you know about the behavior of the inductor in steady state?

You need to show your work in order for us to help you along. Otherwise we have little to go on as far as what you are thinking that is right and what is wrong in your approach.

4. ### DGElder Member

Apr 3, 2016
350
87
Why should that matter? The current through the 4V 12ohm branch is the same whether the voltage source is above the resistor or below the resistor. The voltage at that node (with respect to the bottom node at the switch) will be equal to the i(-0)*R voltage drop + the voltage of the voltage source, i.e. V = 12*Io - 4V. If you want to do a nodal analysis you need to apply KCL and define the node voltage in terms of all three branch currents. So you would have 3 equations in 3 unknown currents to solve.

Or you could do a mesh analysis instead, which would require only 2 equations in two unknown mesh currents. Or you could simplify with Norton equivalents and/or superposition - depending on what techniques you have learned so far and what is expected of you for this problem.

Last edited: Oct 13, 2016
5. ### ci139 Member

Jul 11, 2016
343
39
Last edited: Oct 13, 2016
6. ### MrAl Distinguished Member

Jun 17, 2014
3,733
790

Hello there,

From your description it sounds like you just dont know how to handle a floating voltage source, or just a voltage source that happens to be part of a circuit where in the past you normally encountered a resistor or other passive. It also sounds like you knew enough to short out the inductor for this analysis so i assume you did that already.

You were on the right track actually. You tried 4/12 but you dont get the right answer because the voltage at the top of the 12 ohm resistor is NOT 4 volts.

So what is that voltage then if it is not 4 volts?
Well, you know that just to the left of that source you have another node and that node will have a voltage of it's own, which we can call v1 for convenience. Now that we "know" the voltage at that node is v1 and we see the 4v source in series with that, we can calculate the symbolic voltage at the top of the 12 ohm resistor (say we call it v2) because when a voltage is in series with another voltage the two voltages add.
With this in mind, we now know that the voltage at the top of the 12 ohm resistor is:
v2=v1+4

So what is the current now? Since the voltage is v2 the current is v2/12, or (v1+4)/12.

You see how this works now? Now you can apply what you know about Nodal analysis to finish writing the equation for v1, and once you have that actual numerical voltage, you can add 4 and then calculate the current through the 12 ohm resistor.

Try that and see how you make out.

That method is fairly intuitive, but the more conventional way is to consider the 4v source a 'super node'. A super node is created by shorting out the voltage source and writing the equations as usual for nodal analysis, then adding one more equation that describes the voltage that actually is between those two nodes. This second equation would be something like v2-v1=4, but that is combined with the equation you find for v1 after converting the 4v source into a super node (by shorting it out). So you end up with two equations, and then solve for v1 and v2, then you can calculate any current you need.
One small note about the super node:
When you write the equation with the super node you have to continue to use the symbolic values v1 and v2, even though they are shorted together. That means that your statement for the current though the inductor is v1/4 but the current through the 12 ohm resistor is v2/12, even though v1 is temporarily shorted to v2.

We could do a couple more examples after this if you like. Once you do two or three of these kinds of circuit it becomes old hat

Last edited: Oct 13, 2016