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I have a 9VDC with two resistors in series. R1 = 3900 ohms, R2 = 100 ohms.both 5% tollerance
Upon testing resistance of each
R1 = 3840 Ohms
R2 = 100 Ohms
Both are still within +/- 5% range
Upon testing my power source is in fact 9.81 volts
Total Resistance is 3930 ohms ( range : 3800 - 4200 )
according to formula current should be 2.5mA, upon test it is in fact 2.5mA.
Voltage Drop ( formula)
R1 = 9.6 Volts
R2 - 0.25 volts
Total 9.85 // is this acceptable since it differs only slightly??
Voltage Drop ( Test)
R1 - 9.38 volts = 9.4 Volts
R2 - 0.24 volts = 0.2 Volts
Total . . . . . . . 9.6 Volts
Are these decrepencies ok and a normal thing?
Battery reads 9.81 Volts, the formula produces 9.85 volts, and actual testing proves it is 9.6 Volts.

Total resistance is 3930 but individual resistances adds to 3940. Should i not worry about this as long as it is still within the tollerance range?

2007-03-09 15:43:21 · 5 answers · asked by Anonymous in Science & Mathematics Physics

5 answers

There are a couple of other things that you haven't accounted for.
A.
How much 'stray' resistance is in the wiring in your circuit. The average solder joint is around .05 to as high as .5 ohm in resistance. If you built your circuit on one of those 'breadboards' with the little spring clips to hold component leads, those spring clips can look like another ohm or two. And with 2.5 mA in the circuit, each ohm accounts for 2.5 mV (.0025 V) Those can start to add up in a hurry.

B.
How accurate is your meter? When was the last time it was calibrated? And how linear is it? The average 'student' VOM or DVM is, typically, only about ± 3% or so on accuracy (and frequently not the same inaccuracy on the voltage, current, and resistance ranges) and many of them are as much as 3 or 4 percent non-linear.

All in all, if your reading are consistantly within 1% in a student lab environment, you're doing very well indeed.

But it's also good that you notice such things and ask about them. Ignorance and confusion really *are* the first steps on the road to knowledge and understanding.

HTH ☺

Doug

Comment for LeAnn: If I had a nickel for every time I've literally 'blown' the lid off of a TO-3 package and 'buried' it in the overhead acustical tile, or vaporized a trace on a PCB, I could have retired a *long* time ago. If my Engineers and Techs don't manage to blow something up in the Lab at least once a week or so, it means they're setting on their butts and not doing anything ☺

2007-03-09 16:26:45 · answer #1 · answered by doug_donaghue 7 · 0 0

Your actual test results on this circuit are remarkably close to your computed results.

The very minor differences are most likely due to internal battery resistance and a slight voltage drop at the battery when the load is applied.

In real circuit design, the case is often true that if you establish on paper that you need a 1/4 W resister, for instance, I highly recommend a 1/2 W be used in the actual circuit - the best laid plans on paper often go up in smoke when the actual circuit is subjected to a turn on/off surge or voltage variations for a number of reasons.

It's no fun when you let the smoke out of a component or two on the initial test.

2007-03-10 00:24:14 · answer #2 · answered by LeAnne 7 · 0 0

For your purposes, your answers are just fine. There is no need to be concerned so long as the circuit is not a critical one. The values of your components are with in tolerances, so there is no need for concern. How accurate you meter is, is also of little concern, for the circuit that you were testing. If this were a circuit with transistors, and the meter was an old analog type with a 20,000 ohms per volt rating for the voltage circuits, then I would be very concerned. As it is, you have done excellent work.

2007-03-10 04:19:45 · answer #3 · answered by Anonymous · 0 0

you forgot the internal resistance of the Battery .

2007-03-10 04:00:33 · answer #4 · answered by Kiamehr 3 · 0 0

Dont worry, be happy

2007-03-09 23:46:44 · answer #5 · answered by cattbarf 7 · 0 0

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