# Series-type ohmmeter

by ranju
Tags: ohmmeter, seriestype
PF Gold
P: 3,748
 Quote by ranju but we are knowing the value of I.. since its full scale deflection and we are knowing the full scale deflection current i.e., 0.5mA so we should put it over here..!!
No we don't know the current I . 0.5 ma through the meter movement is just part of it.
We do know that 0.5 is the current through the meter movement.
We do not know yet how much more current flows through R2 in parallel with the meter movement.

The current through R1 and Runknown is the sum of those two currents.
That's I. It's the SUM of current through the meter movement AND the current through R2.

That's why you must divide the two equations to get that unknown I out of them.
After you've solved for R1 and R2//Rmeter, you can go back and solve for current I with any value of Runknown.

I think you'll find full scale current I is one ma, half of it through R2 and the other half through your meter movement. Halfscale current will be 0.5 ma, again half through R2 and half through your meter.

Try your algebra and see if it takes you there.

old jim
P: 92
there's something wrong in your previous expression R1=14.0*R2//Rm...
 Quote by jim hardy Think of it as a resistive voltage divider. Voltage divides in proportion to the resistances as you know from your basic circuit class.
because according to voltage divider formula.. it'll be R1 = 14 *(R1+R2//Rm)..& not tht you'hv written..!!
VR1=V(R2//Rm) * (R1/R1+R2//Rm)...
PF Gold
P: 3,748
 Quote by ranju there's something wrong in your previous expression R1=14.0*R2//Rm... because according to voltage divider formula.. it'll be R1 = 14 *(R1+R2//Rm)..& not tht you'hv written..!! VR1=V(R2//Rm) * (R1/R1+R2//Rm)...
??

Are you speaking of the earlier thought experiment?

 Let us assume midscale is 5 ohms. Also that Vbattery is 1.5 volts. And that we have a meter movement that takes 1 milliamp for full deflection and has Rmeter of 100ohm, so drops 100 millivolts. That is not an unusual movement to find in cheap meters. Better ones are more like 50 microamps 50 millivolts. In your mind, short meter leads together. Vbattery divides across R1 and R2//Rmeter. We know that R2//Rmeter sees 0.1 volt so R1 sees 1.40 volts meaning R1 = 14.0 X R2//Rmeter or R2//Rmeter = R1/14
What could be wrong with that ?
Same current flows through both resistances
one drops 0.1 volts, the other drops 1.4 volts

1.4 = I X R1
0.1 = I X R2//Rm

divide top equation by bottom equation to cancel unknown I
and you get
14 = R1/(R2//Rm)
R1 = 14*(R2//Rm)

Show how you got your expression:
 R1 = 14 *(R1+R2//Rm)
which resolves to
R1 = 14*R1 + 14*R2///Rm
-13R1 = 14*R2//Rm
and a negative number for either resistance seems an unlikely result.
 P: 92 ohh..I am so sorry... !! I told you I was so confused regarding it that I was behaving like a moron..I was nor getting even basic things... I apologise for it..you were right..!! Now..finallyy its all clear ..I'have solved it out... R1 is coming to be 2975 & R2=50ohm... I got the points which were'nt clear ... thanks a lot for your guidance..
PF Gold
P: 3,748
 Now..finallyy its all clear
I knew it'd click soon.
Do not apologize that's unnecessary . I remember very well my own stumbling up the learning curve. As you saw i still make dumb mistakes...
You have been most polite and i'm happy to 'see the light come on.'
That you persevered i find heartwarming.
In learning we need to push ideas back and forth until our thinking leads us to math that works. Arm waving and exaggeration are useful tools.

You'll be surprised how many engineers have never studied how an ohm-meter works.
If you understand how your test equipment works you won't get fooled by it.
And you're less apt to blow it up in a lab course.

good luck in your studies. I think that with your combined curiosity and perseverance you will excel.

old jim

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