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Finding Current and Voltage In A Schematic

10/15/2011 11:33 PM

http://img263.imageshack.us/img263/822/77751684.jpg I need to find current (labeled I) and voltage (labeled V).

I got I = 7.426 mA and V = 7.426V but I think it's wrong.

I know the formulas and all, but I just need to know how to combine the resistors (which ones are in parallel and which ones are in series?) I know that R3 and R4 are in series so I combine them and get 2k equivalent resistance. After that point, I don't know how to combine the rest of the resistors. Can someone help me out?

NOTE: I'm not asking anyone to do all the math and tell me the numbers!! I just want to know how to combine the rest of the resistors after R3 and R4 have been combined

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#1

Re: Finding current and voltage in a schematic

10/16/2011 1:32 AM

First off, I worked this out using mesh analysis and Kirchoff's current law. So I ended up with three equations and three variables. The variables are the currents in each of the three meshes. I came up with I = 0.585 mA and V = 0.172 volts. It has been a long time since I worked a problem of this sort so I could have made an error. If you don't already know mesh and nodal analyses, I suggest that you learn these methods as they can make life easier in complex circuits.

Converting the series/parallel resistors and arriving at Thevinan equivalent:

Yes, R3 + R4 = 2k. Then R5 (3k) parallel to 2k = 3k x 2k / (3k + 2k) = 1.2k.

R2 (4k) is in series with 1.2k = 5.2k

R6(5k) parallel to 5.2k = 5k x 5.2k / (5k + 5.2k) = 2.549 k.

So I = 5 / (6k + 2.549K) = 0.585 mA. This confirms my answer using mesh analysis. Now just reassemble the circuit, mesh by mesh to get V. But believe me, mesh analysis is easier for solving this circuit!

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#2
In reply to #1

Re: Finding current and voltage in a schematic

10/16/2011 3:58 AM

Agreed. GA.

[It can all be done by repetitions of series resistance Ra and Rb = Ra + Rb, and parallel resistance = RaRb/(Ra + Rb).]

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#5
In reply to #1

Re: Finding current and voltage in a schematic

10/17/2011 8:13 AM

"It has been a long time since I worked a problem of this sort so I could have made an error."

I did it the long way, just for the exercise, and got 172 mV as well so either we've somehow made different mistakes and gotten the same wrong answer or you've got it right.

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#3

Re: Finding Current and Voltage In A Schematic

10/16/2011 10:28 PM

He must be desperate. Seems to have posted in many fora.

http://www.crazyengineers.com/forum/electrical-electronics-engineering/48331-find-current-voltage-using-schematic.html#post187600

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#4
In reply to #3

Re: Finding Current and Voltage In A Schematic

10/16/2011 11:16 PM

Homework due on Monday.

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#6

Re: Finding Current and Voltage In A Schematic

10/17/2011 9:24 AM

This is just to explain which resistances are to be taken in series and which in parallel. Those resistances which are connected together only at one end and carry same current (not equal current) should be taken in series and the resistances which are connected together at their both the ends (they will have same voltage across their two ends) should be taken in parallel. Hope this clarifies your doubt.

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#7

Re: Finding Current and Voltage In A Schematic

10/18/2011 10:47 AM

Regards.

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#8

Re: Finding Current and Voltage In A Schematic

10/20/2011 2:14 PM

I am not quite sure whatthe question or should I say answer should be.

Basicaly If you have a resistor in the track that effects current.

If you have a resistor from the track V+ to ground V- then that effects voltage.

If you have a row of resistors the current or voltage will be reduced between each resistor untill you com to the final load which gives the resultant total.

Standard ohms law covers it all but the actual result will depend on the the actual value of the resistors and should be measured if accuracy of current or voltage is required,eg I am repairing a servo design that has been causing problems and we found that the comparator was sending 2 slightly differant voltage signals (ANALOGE) and the manufacturer used 10% tollerance resistors when they should have used 1%

once the resistors were changed to equal values no more problems.

HOPE THIS HELPS

Rachael

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