
#1
Oct2808, 08:39 PM

P: 92

1. The problem statement, all variables and given/known data
What is the difference between an open circuit and a short circuit? Open circuit has no current, so does that mean any resistor in series with it, has no current ,so it can be ignored for analysis( v=ir so 0 current means 0 V) for finding lets say the Thevenin equivalent.? Now for Short circuit, do we ignore a resistor in series with the short, because current will take path of least resistance and ignore that resistor? Also can short circuit have voltage through it ? Now lets take a look at some problems i have right here, and i cant do.... Now lets say i want Thevenin equivalent at the 4H inductor, so i do the Isc(t<0 which makes it closed switch to charge the inductor), i would use mesh analysis and find the current through the 8 ohlm resistor, which will give me my Isc, this correct ? NOw for the Voltage open circuit, im confused on what to do , since Its open circuit , i ignore the 8 olhm resitor correct ? Now another problem i have is this one ...: Now if swithc is t<0 then the isc is just 40/5k which is .008A, now what is the Voltage open circuit, i dont understand how to calculate it, cause when i take Vth/Isc i dont get teh 2k olhms resistance i should have in the back of the book, what am i doign wrong ? Any help would be great , thanks 



#2
Oct2908, 05:42 AM

Sci Advisor
P: 1,724

Well, a glance at Wikipedia's article on short circuits gives you the answer to the first question regarding the difference between open and short circuits:
http://en.wikipedia.org/wiki/Short_circuit As for the questions at hand... Well, I would suggest that these questions are more about the transient behaviour of energystorage components (inductors and capacitors) rather than circuit analysis techniques. When you're approaching questions like these, you have to determine: 1) what the situation is right before "the change" (in steadystate) 2) what happens immediately afterwards, and general shortterm behaviour (transient response), and 3) what happens a long time after "the change" (once the circuit returns to steadystate) So, since you're just working on step 1)... For your first circuit, I'm not sure what you mean by "charging up the inductor" (an inductor in steadystate DC acts as a coiled piece of wire), but yes, i_sc would be the same as the 8 ohm resistor. In the open circuit case, you know that no current will flow through the 8 ohm resistor and hence, the voltage at the top of the 8 ohm resistor is at the same potential as the bottom of the battery (0 volts). In that sense, you can ignore the 8 ohm resistor (by pretending it's a wire). Now, what's the potential at the top of the 4 ohm resistor? With those two numbers, you can figure out the potential between these two points. For the second circuit (the text is cut off, so I assume that the switch is open before t=0, and that the two resistors on the left side of the switch are connected togetherotherwise, you need to specify which terminal the switch connects to at t=0), you don't actually give your method of figuring out V_oc, but I'll assume you did a voltage divider. Unless the switch actually moves between one terminal and the other (though it doesn't look that way). With the above assumptions (and I hope somebody else will confirm), the answer in the back of the textbook is incorrect (by both this and the inspection method). 


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