bigu01
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How to Determine Current Flow in a Circuit with a Shorted Voltage Source?
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If there is no voltage source, then there is no current.
Just redraw the circuit diagram with a short in the place of the voltage source and work the resistor combinations normally.
http://en.wikipedia.org/wiki/Thévenin's_theorem#Calculating_the_Th.C3.A9venin_equivalent
... don't mix up the two versions of step 2.
Just redraw the circuit diagram with a short in the place of the voltage source and work the resistor combinations normally.
http://en.wikipedia.org/wiki/Thévenin's_theorem#Calculating_the_Th.C3.A9venin_equivalent
... don't mix up the two versions of step 2.
bigu01
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Doing as you suggested i am getting Rt(thevenin equivalent resistance) as 4.2k ohms while book solution is 4.0k and i haven't done any calculation error.at least I think I haven't.
- 17,871
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I get 4.0k - please show your working and describe your reasoning.
i.e. which resistors are left and what is the circuit?
i.e. which resistors are left and what is the circuit?
bigu01
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Is the eq res. of 1M and 2k parallel to 3k? If yes can you please tell me why?because calculating it as parallel i got the result right
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The 1M, 2k and 3k are, indeed, parallel.
But they are also disconnected from the rest of the circuit by the short.
Even if you did include it, you would not get exactly 4.0k. Be careful about rounding off too soon.
Actually physically draw the circuit (without the load) with the short in place of the 20V source.
If you put an ohmmeter between a and b, what would it read?
But they are also disconnected from the rest of the circuit by the short.
Even if you did include it, you would not get exactly 4.0k. Be careful about rounding off too soon.
Actually physically draw the circuit (without the load) with the short in place of the 20V source.
If you put an ohmmeter between a and b, what would it read?
bigu01
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Well if i calculate only the right side of the circuit I'm getting 4.4 considering the two 6k's are in series then in parallel with 3k then in series with 2k.
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One of the 6k resistors is in parallel with a short circuit though.
bigu01
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So i should ignore every resistance affiliated with the short circuit,draw the circuit with only three resistances(6k in parallel with 3k then in series with 2k) and get 4.0k
- 17,871
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You technically use all the resistances - but you have to include the effect of the short circuit.
The effect of the short is to give a large chunk of the circuit an equivalent resistance of zero.
That leaves only the three by the terminals having any effect on the resistance seen at the terminals.
Sometimes it helps to see which resistors have any effect if you imagine you replace the load with a small current source and trace where the current would go. All the resistors the current pass through are the ones that you use.
The effect of the short is to give a large chunk of the circuit an equivalent resistance of zero.
That leaves only the three by the terminals having any effect on the resistance seen at the terminals.
Sometimes it helps to see which resistors have any effect if you imagine you replace the load with a small current source and trace where the current would go. All the resistors the current pass through are the ones that you use.
bigu01
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https://www.physicsforums.com/showthread.php?t=727410 Can you please check this thread, its just a simple question that has got me stopped in moving forward.I have a midterm tomorrow and I need to finish the final chapter.