# Missing Resistor Value!

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Hello everyone. I am an electronics student, and I am having trouble with o ne the problems in my text. The problem is a simple circuit with a 6V suppl y voltage, and two resistors in series. One resistor (R1) is 500kOhms and t he other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series cir cuit, but I tried using the current I calculated just using the info I have ...6V and a 500k Ohm, but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volt s. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

Thanks, Michael

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Would there there not be ( 6v - 1v ) across 500K so I = 5v/500k ?

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"Michael Covington"

Hello everyone. I am an electronics student, and I am having trouble with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm, but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

** This is math problem involving a simple equation.

The current flow is given by 6/(R1+R2) and also by 1/R2

Do you need any more ?

.... Phil

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The problem is a simple circuit with a 6V supply voltage, and two resistors in series.

One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value,

but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit,

but I tried using the current I calculated just using the info I have...6V and a 500k Ohm,

but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts.

I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

So what do you know?

6 volts supply voltage 1 volt across R2 How much voltage across R1? ______Volts So now you know the Voltage and the resistance of R1, What is the current?

______amps, or mamps, or micro amps.

So now you know the Voltage and the current of R2, What is the resistance. _____ohms.

Mikek

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Not much, else ACA would NEVER have gotten into place!?

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The current does NOT stay the same! That wouldn't make sense.

The current will be I = 6/(R1+R2), because R1+R2 is the resistance that the battery sees.

So you can do a couple of simultaneous equations, one the thing above and another one for R2. But that's a nuisance, and doesn't train your intuition.

Imagine a long skinny strip of resistive material, out in the open. Let's make it 6 inches long for fun.

Apply 6 volts on one end, and ground the other.

Now get a voltmeter, ground the - probe, and slide the + probe along the strip. Obviously you'll have 0 volts at one end and 6 volts at the other end.

How far along the strip will you see 3 volts? And where will you see 1 volt?

You could actually do this. A piece of wet paper would pretty much work.

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John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    ```
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Sympel:

1) Vtotal = I*R1 + I*R2 Substituting with known values: 2) 6V = I*500K + 1V Do a bit of math: 3) 5V = I*500K Solve for one unknown: 4) I= .01mA or I=100uA Write new equation from (1) and (2): 5) 1V = I*R2 Substituting with known values: 6) 1V = 0.01mA * R2 Solve for one unknown: 7) R2=100K fin
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? Affordable Care Act ? ? American Correctional Association ? ? American Canoe Association ? ? American Camp Association ? ? American Chiropractic Association ?

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Not much, else ACA would NEVER have gotten into place!? ? Affordable Care Act ? ? American Correctional Association ? ? American Canoe Association ? ? American Camp Association ? ? American Chiropractic Association ?

I dunno what anyone else thinks, but I deem that kinda not so helpful.

Simplify.

-

FUCK all that. And f*ck Kirschoff as well. He's right but that shit is for when you are designing some other shit. forget it for now.

Ummmmm

What voltage are you starting with ? yoiu got a voltafge at the top and you expect a portion of that voltage out of like a resistive divider, no ? if you have 100 bvolts at the top, and say one volt at the botom, if the top r esistor is 500K, the bottom one, or whatever the combined load to ground at that point of the circuit would be 5K.

But you got 0.6 volt.

Well, when everything is resistive, is was so nice. Unicorns danced all ove r my front lawn and shit, you know. Good old days.

So that would be 4K or whatever, but it probably ain't. It is probebly the latent drop of a semiconductor junction. And you hav to be aware that it ca n fool you when the equipment is off becus everything and its brother in la w has all kinds of suppresion diodes built into the substrate, so when powe r goes out to one thing, everything it is connected to might read a "short" with an ohmmeter or DVM.

Pain in thre ass, it really is. Thsat's why I try to do things powered, and try to have a scope with tunnel diode triggering. That way I can see it ha ppen. I do not care if it is 11.4 volts or 11.7 volts, I care if it is 3.4 volts or 52.9 volts. In most cases, my reading simply does not havbe to be all that exact.

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Be careful with the terminology. The voltage appears *across* R2, not "at" it.

Imagine the circuit is already set up and running with the correct value for R2 (whatever it may be). You have a 6v supply and 1v is being lost across R2, this leaves 5v across R1.

The value of R1 is 500kOhms, so the current through it (by Ohms Law) must be 5/500000 = 10 microamps (1 x 10^-5 amps)

To develop 1 volt across R2 with a 10 microamp current through it, Ohms Law says it would have to be 1/ (1 x 10^-5) Ohms = 100kOhms

As a quick check, you now have a potential divider of a 500k resistor in series with a 100k resistor, the 500k drops 5v, the 100k drops 1v and the total voltage is 6v. This looks correct

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1.  For the circuit you describe: (View using a fixed-pitch font)

.     E1 ```
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one the problems in my text. The problem is a simple circuit with a 6V sup ply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series c ircuit, but I tried using the current I calculated just using the info I ha ve...6V and a 500k Ohm, but still cannot figure out how to get the resistan ce of R2!! I tried Kirchoff's loop law and simply have a difference of 5 vo lts. I would really appreciate it if someone could give me a hand with this , so I will know what i am doing in the future.

V=IR so for a given i, V is proportional to R. So in a chain where you ha ve 5v across 500k, you have 1v across 100k.

NT

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h one the problems in my text. The problem is a simple circuit with a 6V su pply voltage, and two resistors in series. One resistor (R1) is 500kOhms an d the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I h ave...6V and a 500k Ohm, but still cannot figure out how to get the resista nce of R2!! I tried Kirchoff's loop law and simply have a difference of 5 v olts. I would really appreciate it if someone could give me a hand with thi s, so I will know what i am doing in the future.

Maybe in your inebriated state...a series circuit by definition means "the same" current passes through all the components. If this "same" current has magnitude I, then I=V1/R1=V2/R2 from which R2/R1=V2/V1 or the resist ance ratios are equal to the corresponding voltage drop ratios. Since the v oltage drop ratio of V2/V1=1V/5V= 1/5 then that's the resistor ratio, m aking R2= 1/5 R1= 100K.

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It looked to me like the OP calculated the current with just 500K in the circuit, and then added R2 and assumed that the current wouldn't change. Which is why his math didn't work.

Haven't had any alcohol in days. Two days, in fact. Half a beer on Sunday.

Another simple way to do it:

Assume the result, 1 volt across R2. That leaves 5 volts across R1. NOW calculate the current. Then compute R2.

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John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    ```
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m

with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a seri es circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm, but still cannot figure out how to get the resi stance of R2!! I tried Kirchoff's loop law and simply have a difference of

5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

he same" current passes through all the components.

Which

.

It's an educational problem, the student is supposed to walk away from it w ith something more than just a number. The problem leads the student into d iscovering voltage drops are in the same ratio as resistor values in a seri es circuit.

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but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

Thanks for spoon feeding the future techs that won't be worth anything at job interview.

I suppose however if ever asked to do a simple little math quiz he could always pop open his smart phone and ring your bell here!

Personally I think you got more out of it trying to prove to the audience that you could do it. Those types usually are lacking and know it.

Jamie

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You are being given a story problem, and are expected to turn it into a problem in pure math, and then to solve that.

You don't have sufficient information to calculate just the current, because while it is the same in R1 and R2, it depends on the value of R2. You don't have sufficient information to calculate just the value of R2 -- but you would if you knew the current.

Try writing out the equation for the voltage at (I think you mean across) R2 as a function of the current (which you do not know -- use a symbol). Then write out the equation for the current as a function of the supply voltage and the two resistances. Then see if you have enough information to correctly solve the problem.

I _highly suggest_ that you retain your units everywhere -- ohms, amps, volts, etc. If you end up finding that the value of R2 is in volts and the value of the current is in ohms squared, then you know you've messed up and you can go try again (dimensional analysis has saved my butt so many times I couldn't even begin to count the number of times -- but the count would be in butt-saves).

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Tim Wescott
Wescott Design Services ```
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That really helped, and makes sense to me now!! Thank you very much.

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Well, now you'll be able to claim you know how to design a voltage
divider, so for the next job interview you go to maybe how to scoop ```
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...and those tunnel diodes go all the way to China.

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