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I am doing self-study on circuit analysis and thought a good way to check my results would be through MultiSim and see if I am solving the circuit problems correctly. The circuit below is shown:

Circuit Diagram

Where I received ix = 50mA, Vg = 2.5V, V1 = 1.5V, and i1 = -10mA.

(1) My first question would be if these values seem accurate.

(2) My second question is how I would transcribe this circuit into Multisim/Spice to check the values computationally. The independent voltage source, Vg, is something that is unknown and is the only power source available. I decided to add an independent current source in place of Vg, since ix is given, and added another independent voltage source for Vx, since it is also given:

Spice Diagram

I'm not sure how to properly build my circuits in Multisim to test my values (I'm hoping to be able to practise lots of problems and test my solutions this way) since it looks like the center wire is completely removed from the circuit.

This was one of the simplest diagrams I could find, so any tips or advice would be greatly appreciated in case I need to go a different route. Thank you!

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4 Answers 4

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One thing to recognize is that, probably somewhat unusually, this problem is overconstrained. You don't need vx= 3.5V to solve it, and if vx is anything other than 3.5V the problem cannot be solved. ix = 50mA and the resistor values are sufficient information.

Normally to check you'd just shove in the known values and the unknown would be calculated. Your replacement of the unknown voltage with a current source is clever, simulating what you'd do on the bench (wire the resistors up and turn up the voltage until the current is 50mA).

If we do that and ignore the 3.5V the other numbers will drop out from the simulation, including the 3.5V which should appear where it is expected (hint: you made an error somewhere).


Here is that done in Circuitlab. All the numbers in the problem are shown or can be calculated in your head (power)

schematic

simulate this circuit – Schematic created using CircuitLab

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Your computed values of I1, V1, Vg seem inaccurate. Please use KVL for all loops & KCL for all nodes to check by yourself whether the answer is correct.

For simulation, you can do a DC sweep of Vg from 0 to say 20V and see at what value of Vg you get the ix and vx mentioned in the question. At the correct Vg, you can then check i1, v1 also.

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\$v_x\$ is proportional to \$i_x\$, so you only need to "set" one of those, and the other one must follow. That's probably why they both share the same subscript \$x\$: they are linearly related.

This is the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

\$v_x\$ cannot be anything but 3.5V, as shown.

Now you can add voltage and current probes to measure the voltages and currents as needed.

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Your approach isn't good. The simulation should be only a verification after the calculations (needless, according to Bob Pease :-) ), then you should know the basic theory.
In this case Ohm law is enough.

1) Your values are, intuitively, not accurate.
You you think, first of all, that \$ v_x = 3.5V\$ and the only source is \$v_g\$ you can undestand that the voltage drop across the 175Ω resistor should be less than the source, because there's also the voltage drop across 50Ω and 200Ω resistors. What you need to know, to approach this type of problems, before KVL or KCL, is just: - Ohm law; - parallel and series resitors (the foundations of the Kirchhoff laws) - voltage divider - current divider

You need, basically, the first few pages of Horowitz. (I suggest you to start with a good book, like the Horowitz-Hill, Art of Electronic.
Otherwisee there's a lot of free resources, like this or this)

2. To simulate your circuit (and this is where you need the basic knowledge I mentioned), you have to do some thinking about the circuit.
The only source is the \$v_g\$ but you know only the current delivered from it. Then you can replace it with a current generator as you, correctly, did.
But the \$ v_x\$ voltage drop is an effect of the current delivered by \$v_g\$, not a real voltage source then you don't need to add 3.5V voltage source.

You can see what I did in the circuit below.

schematic

simulate this circuit – Schematic created using CircuitLab

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