Short Circuit, and open circuit question

In summary, the difference between an open circuit and a short circuit is that an open circuit has no current flowing through it, while a short circuit has a very large current flowing through it. In circuit analysis, an open circuit can be ignored for analysis as it will have no effect on the circuit. A short circuit, on the other hand, must be considered as it can significantly affect the circuit. In terms of finding the Thevenin equivalent, an open circuit can be ignored while a short circuit must be taken into account. The voltage across an open circuit is equal to the voltage at the source, while the voltage across a short circuit is close to 0. Transient behavior of energy-storage components must also be taken into consideration when analyzing
  • #1
th3plan
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Homework Statement


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 let's 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 let's take a look at some problems i have right here, and i can't do...

http://i454.photobucket.com/albums/qq261/integralx2/2.jpg

Now let's 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, I am confused on what to do , since Its open circuit , i ignore the 8 olhm resitor correct ?

Now another problem i have is this one ...:

http://i454.photobucket.com/albums/qq261/integralx2/1.jpg

Now if swithc is t<0 then the isc is just 40/5k which is .008A, now what is the Voltage open circuit, i don't understand how to calculate it, cause when i take Vth/Isc i don't 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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  • #2
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).
 
  • #3


I can provide a response to your question about the difference between an open circuit and a short circuit. An open circuit is a circuit where there is a break or interruption in the path of the current, causing there to be no flow of electricity. This can occur when a switch is open or when a wire is disconnected from a circuit. In an open circuit, the current is zero, so any resistor in series with it will have no current flowing through it, and therefore can be ignored for analysis using Ohm's law (V=IR).

On the other hand, a short circuit is a circuit where there is a very low resistance path for the current to flow through, causing a large flow of electricity. This can occur when a wire is connected directly from the positive to the negative terminal of a battery. In a short circuit, the current will take the path of least resistance, so any resistor in series with it will have a very small current flowing through it, and can also be ignored for analysis.

It is important to note that a short circuit can have voltage across it, as long as there is a difference in potential between the two points where the short circuit is connected. However, the voltage will be very small due to the low resistance of the short circuit.

In order to find the Thevenin equivalent at the 4H inductor in the first problem, you are correct in using the Isc (short circuit current) to find the equivalent resistance. For the voltage open circuit, you are correct in ignoring the 8 ohm resistor, as there is no current flowing through it in an open circuit. However, you can still use Ohm's law to calculate the voltage open circuit by multiplying the equivalent resistance by the Isc.

In the second problem, you are also correct in using the Isc to find the equivalent resistance. However, for the voltage open circuit, you need to use the total resistance (5k + 2k = 7k) instead of just the 2k resistor, as the open circuit voltage will be across the entire circuit. So the voltage open circuit would be Vth = Isc * Rth = 0.008A * 7k = 56V.

I hope this helps clarify the difference between open circuits and short circuits, and how to approach problems involving them. If you continue to have trouble, I suggest consulting with your teacher or a tutor for further assistance.
 

1. What is a short circuit?

A short circuit is a fault in an electrical circuit that occurs when there is an unintended connection between two points in the circuit, causing a current to flow along an unintended path. This can result in overheating and damage to the circuit and connected devices.

2. How does a short circuit happen?

A short circuit can happen due to a variety of reasons such as damaged or faulty wiring, incorrect installation, or malfunctioning components. It can also occur when two conductors come into direct contact with each other, bypassing the intended load in the circuit.

3. What are the consequences of a short circuit?

A short circuit can cause damage to the circuit and connected devices, including overheating, melting, and even fire. It can also result in power outages and disruption of electrical systems.

4. What is an open circuit?

An open circuit is a fault in an electrical circuit that occurs when there is a break or discontinuity in the path of the current, preventing it from flowing. This can be caused by a broken wire, a switch in the off position, or a disconnected component.

5. How is an open circuit different from a short circuit?

An open circuit and a short circuit are two different types of faults in an electrical circuit. While a short circuit creates an unintended path for the current to flow, an open circuit interrupts the flow of current completely. This means that in a short circuit, there is still current flowing, while in an open circuit, there is no current flowing at all.

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