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Short Circuit, and open circuit question

by th3plan
Tags: circuit
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th3plan
#1
Oct28-08, 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
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MATLABdude
#2
Oct29-08, 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 energy-storage 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 steady-state)
2) what happens immediately afterwards, and general short-term behaviour (transient response), and
3) what happens a long time after "the change" (once the circuit returns to steady-state)

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 steady-state 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 together--otherwise, 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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