![]() ![]() 2 The rest of the entries set the type of sweep (here logarithmic) and the range of values over which to sweep. So the result of all this is that the component property value Id1 of the current source s property I will be swept through a range of values as determined by our parameter sweep function named SW1. ![]() If you look at the current source driving our diode you will see that it just happens to be labeled Idrive. 2Ħ The first two items to take note of are the Simulation entry (here DC1, corresponding to the name of the simulation box) and the Sweep Parameter entry, here entered as Id1. If you really want to be thorough you could then also build the circuit and measure the result. 1 I could tell you the value my simulation gave, but why should I spoil your fun. To set all the resistances to the correct value I would have had to open the Properties Editor window twelve times here is how it looks. 1 Those with good attention to detail will be complaining about now that I haven t really solved the problem, as the question mentioned one ohm resistors while I have used fifty ohms. If I use a current of one amp, the output voltage will be equal to the resistance in ohms. As I am forcing a constant current through the cube from one corner to another, Ohm s Law tells me that the voltage between those corners will give me the resistance. With my cube of resistors accurately drawn, I only have to hit the simulation button and the tabulated results will show me the voltage at the corner node. And, I might add, it is ALWAYS worth checking that we have done it right simulate the wrong circuit and it will tell you lies. The Rule is: if we can correlate the junctions of our components with those of the real circuit, we are accurately representing the physical circuit. Sometimes we have to bend and squeeze things to get it into a format that our simulator will accept, which leaves us wondering whether we are working with an accurate representation. 1ģ Which all might seem trivial, but is a good reminder right at the beginning that we are creating a virtual representation of a physical circuit. Which I m sure is topologically the same as a cube. Then I wired two sets of four into squares, then connected the remaining four between the corners of the squares. ![]() Figure 1: resistor cube schematic All I did was select resistance in the left hand component window and paste them down, rotating as necessary, until I had twelve on the schematic. Here is my attempt to make a cube in Qucs anyone is welcome to try and improve it. What is the resistance between opposite corners of the cube? The intention may have been to teach soldering, as more than one student solved it by making just such a cube! These days we can do that without touching the soldering iron we simulate the circuit. 2 DC Static Circuits A favourite question in electronics courses used to be: You have twelve one ohm resistors you connect them together so that each resistor lies along the edge of a cube. ![]()
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