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A handy little chart for anyone new to resistors  or has a memory as bad
as
mind:)
http://www.lalena.com/audio/electronics/color/
Scroll to the bottom of the page, the last image can be printed out cleanly
and covers all the E12 and E24 resistors as a quick reference beside the
work table.
Later,
Dave S. 0 Attachment
Hey Dave
That helps but I like this link.
http://webhome.idirect.com/~jadams/electronics/resist_calc.htm
My PC is next to my bench so I use it a lot.
Gord
 David Simmons <devs66@...> wrote:> A handy little chart for anyone new to resistors  or has a memory
__________________________________________________
> as bad
> as
> mind:)
>
> http://www.lalena.com/audio/electronics/color/
>
> Scroll to the bottom of the page, the last image can be printed out
> cleanly
> and covers all the E12 and E24 resistors as a quick reference beside
> the
> work table.
>
> Later,
> Dave S.
>
>
>
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Hi everyone,
Sorry about the nonrobotics question, but I had a question today on a
physics test (it was bonus), and I have a slight disagreement with the
teacher about the answer. Attached is a tiny gif that is the diagram for
this question. The question is  what is the current at each of the +
terminals of the cells? Yes, those are 10V cells, and no, I don't know where
you can buy those :)
I would greatly appreciate if one of the many eengineers on the list could
answer that question.
Thanks,
Ori 0 Attachment
This network can be solved intuitively using ohms law.Since the voltage drops across both resistors are the same, you can see right away that the current through the 1 ohm resistor must be double the current of the 2 ohm resistor .With equal 10V supplies (just one supply could be used) , the two resistors are effectively in parallel with a resistance of 1.5 ohms and with the 4 ohm resistor in series, the total resistance is 5.5 ohm.Using ohms law I=E/R, the total current is 10V/5.5ohm=1.818...A .The current through the 2 ohm resistor is 1/3 of the total current or 0.606A and the 1 ohm takes 2/3 of the total current or 1.212A.wilf Original Message From: OriSent: Thursday, January 16, 2003 6:37 PMSubject: [beam] ResistorsHi everyone,
Sorry about the nonrobotics question, but I had a question today on a
physics test (it was bonus), and I have a slight disagreement with the
teacher about the answer. Attached is a tiny gif that is the diagram for
this question. The question is  what is the current at each of the +
terminals of the cells? Yes, those are 10V cells, and no, I don't know where
you can buy those :)
I would greatly appreciate if one of the many eengineers on the list could
answer that question.
Thanks,
Ori
To unsubscribe from this group, send an email to:
beamunsubscribe@egroups.com
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I'll take a stab at that... please correct me if I am wrong.
I have redrawn the diagram with symbols to make the math a little
easier....
Know that Kirchoff said all voltages in a path = 0 and all currents at a
node = 0. So, build a few equations....
0=E1Er1Er3
0=E2Er2Er3  solve this one for Er3 and substitute into the first
because it appears in both equations
Er3=E2Er2
and simplify....
E1=Er1+E2Er2
E1E2=Er1Er2
expand to find the currents:
E1E2=i1r1 i2r2
1010=i1(1)  i2(2)
0=i12*i2
i1=2*i2 SO... Current 1 = 2x Current 2.. now you only need to find
one of the currents and you will also have the other. :)
Assuming that the cells are ideal and have no internal resistance, you
can just put the 2 branch resistors in parallel and do the math.
For the Current at A, it is I=E/R so
i1=E1/totalR
i1=10/ 1+ (2*4/2+4) (resistor law)
i1= 10/2.333 = 4.292 Amps
and we know from before that i1=2*i2, so i2=i1/2 = 4.292/2 = 2.146 Amps.
Therefore the Current at point A is 4.292 Amps and at point B it is
2.146 Amps.
TOM
On Thu, 20030116 at 19:37, Ori wrote:
> Hi everyone,
>
> Sorry about the nonrobotics question, but I had a question today on a
> physics test (it was bonus), and I have a slight disagreement with the
> teacher about the answer. Attached is a tiny gif that is the diagram for
> this question. The question is  what is the current at each of the +
> terminals of the cells? Yes, those are 10V cells, and no, I don't know where
> you can buy those :)
>
> I would greatly appreciate if one of the many eengineers on the list could
> answer that question.
>
> Thanks,
>
> Ori
>
> To unsubscribe from this group, send an email to:
> beamunsubscribe@egroups.com
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
> 0 Attachment
Hi Ori;
Ori wrote:
> Sorry about the nonrobotics question, but I had a
Lets see:
> question today on a physics test (it was bonus), and
> I have a slight disagreement with the teacher about
> the answer. Attached is a tiny gif that is the diagram
> for this question. The question is  what is the
> current at each of the + terminals of the cells?
> Yes, those are 10V cells, and no, I don't know where
> you can buy those :)
> I would greatly appreciate if one of the many
> eengineers on the list could answer that question.
You can rearrange the circuit so a single 10 V battery
feeds the 2 resisters in parallel.
1 / ( 1/1 + 1/2 ) = .667 ohms.
This makes the total 4.667 ohms.
This total resistance has 10 volts across it so the
total current is.
10 / 4.667 = 2.1428571 = 15/7ths
1/3rd of the current goes through the 2 ohm resister:
2.1428571 / 3 = .7142857 = 5/7ths
Thus the rest of the current goes through the 1 ohm resister:
2.1428571  .7142857 = 1.4285714 = 10/7ths
> Thanks, Ori
Duane

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Using ohms law I=E/R, the total current is 10V/5.5ohm=1.818...A .
The current through the 2 ohm resistor is 1/3 of the total current or 0.606A
and the 1 ohm takes 2/3 of the total current or 1.212A.
wilf
Yes, I knew I had done it right! Thanks Wilf! Also, did you see the message
I sent you off the list regarding heads?
Thanks again,
Ori 0 Attachment
OK Ori... are you more confused than ever?  3 different answers from 3
different sources..
On Thu, 20030116 at 20:59, Duane C. Johnson wrote:
> Hi Ori;
>
> Ori wrote:
>
> > Sorry about the nonrobotics question, but I had a
> > question today on a physics test (it was bonus), and
> > I have a slight disagreement with the teacher about
> > the answer. Attached is a tiny gif that is the diagram
> > for this question. The question is  what is the
> > current at each of the + terminals of the cells?
> > Yes, those are 10V cells, and no, I don't know where
> > you can buy those :)
>
> > I would greatly appreciate if one of the many
> > eengineers on the list could answer that question.
>
> Lets see:
> You can rearrange the circuit so a single 10 V battery
> feeds the 2 resisters in parallel.
> 1 / ( 1/1 + 1/2 ) = .667 ohms.
>
> This makes the total 4.667 ohms.
> This total resistance has 10 volts across it so the
> total current is.
> 10 / 4.667 = 2.1428571 = 15/7ths
>
> 1/3rd of the current goes through the 2 ohm resister:
> 2.1428571 / 3 = .7142857 = 5/7ths
>
> Thus the rest of the current goes through the 1 ohm resister:
> 2.1428571  .7142857 = 1.4285714 = 10/7ths
>
> > Thanks, Ori
>
> Duane 0 Attachment
Sorry Ori, I made a mistake.1 ohm in parallel with 2 ohms is 0.666 ohms (must be less than 1 ohm) so the total is 4.666 ohms and the total current is 2.143 A, split 0.714A and 1.428A respectively. Original Message From: OriSent: Thursday, January 16, 2003 8:09 PMSubject: Re: [beam] ResistorsUsing ohms law I=E/R, the total current is 10V/5.5ohm=1.818...A .
The current through the 2 ohm resistor is 1/3 of the total current or 0.606A
and the 1 ohm takes 2/3 of the total current or 1.212A.
wilf
Yes, I knew I had done it right! Thanks Wilf! Also, did you see the message
I sent you off the list regarding heads?
Thanks again,
Ori
To unsubscribe from this group, send an email to:
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Hi Ori;
Ori wrote:>
But that would be wrong.
> Using ohms law I=E/R, the total current is 10V/5.5ohm=1.818...A .
The parallel combination of 1 ohm and 2 ohm is .667 ohm.
And a total of 4.667 ohms.
The total current is 10 / 4.667 = 2.1428 amps.
> The current through the 2 ohm resistor is 1/3 of the total current or 0.606A
Duane
> and the 1 ohm takes 2/3 of the total current or 1.212A.
>
> wilf
>
> Yes, I knew I had done it right! Thanks Wilf! Also, did you see the message
> I sent you off the list regarding heads?
>
> Thanks again, Ori

Home of the $35 LED solar tracker.
http://www.redrok.com/electron.htm#led3
CUL8ER \ \ \ \ \ \\ \ \ Receiver
Powered by\ \ \ \ \ \\ \ \ [*]
Thermonuclear \ \Solar\Energy\from the Sun \ /////
Energy(the Sun) \ \ \ \ \\ \ / / /\/ / /
\ \ \ \ \ /\ / \/ / / / 
WA0VBE \ \ \ \ / /\ \/ / / \/ /
Ziggy \ \ \/ / / \ \/ \/ /\ 
\ / \ \/ / /\ \\ / \ / / 
"Red Rock Energy" === ===\ / \ / \ === \ / ===
Duane C. Johnson, Designer=== === \ \ === / 
1825 Florence St Mirrors,Heliostats,Controls & Mounts
White Bear Lake, Minnesota \ \ / 
USA 551103364 \ \ 
(651)6355O65 work \ \ / 
(651)4264766 home use Courier New Font \ \ 
(413)556659O Fax copyright \ / 
(651)5832O62 Red Rock Energy Site (C)980907 ===\ 
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redrok2@... (Hotmail address) \ 
duane.johnson@... (Unisys address) \ 
http://www.redrok.com/index.htm (My New Web site) \
These are my opinions, and not that of Unisys Corp. === 0 Attachment
i just finished superpostion in school and using it my results wereat the 1 ohm resistor 1.43 A and 1.43 Vat the 2 ohm resistor 710mA and 1.42 Vat teh 4 ohm resistor 716mA and 2.86 Vmike Original Message From: Tom MairsSent: Thursday, January 16, 2003 10:56 PMSubject: Re: [beam] ResistorsI'll take a stab at that... please correct me if I am wrong.
I have redrawn the diagram with symbols to make the math a little
easier....
Know that Kirchoff said all voltages in a path = 0 and all currents at a
node = 0. So, build a few equations....
0=E1Er1Er3
0=E2Er2Er3  solve this one for Er3 and substitute into the first
because it appears in both equations
Er3=E2Er2
and simplify....
E1=Er1+E2Er2
E1E2=Er1Er2
expand to find the currents:
E1E2=i1r1 i2r2
1010=i1(1)  i2(2)
0=i12*i2
i1=2*i2 SO... Current 1 = 2x Current 2.. now you only need to find
one of the currents and you will also have the other. :)
Assuming that the cells are ideal and have no internal resistance, you
can just put the 2 branch resistors in parallel and do the math.
For the Current at A, it is I=E/R so
i1=E1/totalR
i1=10/ 1+ (2*4/2+4) (resistor law)
i1= 10/2.333 = 4.292 Amps
and we know from before that i1=2*i2, so i2=i1/2 = 4.292/2 = 2.146 Amps.
Therefore the Current at point A is 4.292 Amps and at point B it is
2.146 Amps.
TOM
On Thu, 20030116 at 19:37, Ori wrote:
> Hi everyone,
>
> Sorry about the nonrobotics question, but I had a question today on a
> physics test (it was bonus), and I have a slight disagreement with the
> teacher about the answer. Attached is a tiny gif that is the diagram for
> this question. The question is  what is the current at each of the +
> terminals of the cells? Yes, those are 10V cells, and no, I don't know where
> you can buy those :)
>
> I would greatly appreciate if one of the many eengineers on the list could
> answer that question.
>
> Thanks,
>
> Ori
>
> To unsubscribe from this group, send an email to:
> beamunsubscribe@egroups.com
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
To unsubscribe from this group, send an email to:
beamunsubscribe@egroups.com
Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service. 0 Attachment
Don't you have a corelation problem here....?
If you check your numbers by putting them back in the circuit, it does
not seem to work.
if there is .606A across the 2 ohm resistor, that leaves 8.788 Volts
across the 4 ohm resister, which DOES agree with the other side.
HOWEVER, 8.788V across the 4ohm resister is 2.197 Amps. But you only
have 1.212+0.606 = 1.818Amps Calculated.
Wazzup ??
TOM
On Thu, 20030116 at 20:47, Wilf Rigter wrote:
> This network can be solved intuitively using ohms law.
>
> Since the voltage drops across both resistors are the same, you can
> see right away that the current through the 1 ohm resistor must be
> double the current of the 2 ohm resistor .
>
> With equal 10V supplies (just one supply could be used) , the two
> resistors are effectively in parallel with a resistance of 1.5 ohms
> and with the 4 ohm resistor in series, the total resistance is 5.5
> ohm.
>
> Using ohms law I=E/R, the total current is 10V/5.5ohm=1.818...A .
>
> The current through the 2 ohm resistor is 1/3 of the total current
> or 0.606A and the 1 ohm takes 2/3 of the total current or 1.212A.
>
> wilf
>
>
>  Original Message 
> From: Ori
> To: beam@yahoogroups.com
> Sent: Thursday, January 16, 2003 6:37 PM
> Subject: [beam] Resistors
>
> Hi everyone,
>
> Sorry about the nonrobotics question, but I had a question
> today on a
> physics test (it was bonus), and I have a slight disagreement
> with the
> teacher about the answer. Attached is a tiny gif that is the
> diagram for
> this question. The question is  what is the current at each
> of the +
> terminals of the cells? Yes, those are 10V cells, and no, I
> don't know where
> you can buy those :)
>
> I would greatly appreciate if one of the many eengineers on
> the list could
> answer that question.
>
> Thanks,
>
> Ori
>
>
> To unsubscribe from this group, send an email to:
> beamunsubscribe@egroups.com
>
>
>
> Your use of Yahoo! Groups is subject to the Yahoo! Terms of
> Service.
>
> Yahoo! Groups Sponsor
> ADVERTISEMENT
> HGTV Dream Home Giveaway
>
> To unsubscribe from this group, send an email to:
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>
>
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Whoops! I was looking at Duane's numbers when I agreed with Wilf... :)
Anyways, I got the same numbers as Duane, not the same as Wilf. Which one is
right?
Now I need a Tylenol :)
Ori
 Original Message 
From: "Duane C. Johnson" <redrok@...>
To: <beam@yahoogroups.com>
Sent: Thursday, January 16, 2003 11:14 PM
Subject: Re: [beam] Resistors
> Hi Ori;
>
> Ori wrote:
> >
> > Using ohms law I=E/R, the total current is 10V/5.5ohm=1.818...A .
>
> But that would be wrong.
> The parallel combination of 1 ohm and 2 ohm is .667 ohm.
> And a total of 4.667 ohms.
> The total current is 10 / 4.667 = 2.1428 amps.
>
> > The current through the 2 ohm resistor is 1/3 of the total current or
0.606A
> > and the 1 ohm takes 2/3 of the total current or 1.212A.
> >
> > wilf
> >
> > Yes, I knew I had done it right! Thanks Wilf! Also, did you see the
message
> > I sent you off the list regarding heads?
> >
> > Thanks again, Ori
>
> Duane
>
> 
> Home of the $35 LED solar tracker.
> http://www.redrok.com/electron.htm#led3
> CUL8ER \ \ \ \ \ \\ \ \ Receiver
> Powered by\ \ \ \ \ \\ \ \ [*]
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> redrok2@... (Hotmail address) \ 
> duane.johnson@... (Unisys address) \ 
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>
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> 0 Attachment
>Which one is right?Duane or if you like my second post. Original Message From: OriSent: Thursday, January 16, 2003 8:41 PMSubject: Re: [beam] ResistorsWhoops! I was looking at Duane's numbers when I agreed with Wilf... :)
Anyways, I got the same numbers as Duane, not the same as Wilf.
Now I need a Tylenol :)
Ori
 Original Message 
From: "Duane C. Johnson" <redrok@...>
To: <beam@yahoogroups.com>
Sent: Thursday, January 16, 2003 11:14 PM
Subject: Re: [beam] Resistors
> Hi Ori;
>
> Ori wrote:
> >
> > Using ohms law I=E/R, the total current is 10V/5.5ohm=1.818...A .
>
> But that would be wrong.
> The parallel combination of 1 ohm and 2 ohm is .667 ohm.
> And a total of 4.667 ohms.
> The total current is 10 / 4.667 = 2.1428 amps.
>
> > The current through the 2 ohm resistor is 1/3 of the total current or
0.606A
> > and the 1 ohm takes 2/3 of the total current or 1.212A.
> >
> > wilf
> >
> > Yes, I knew I had done it right! Thanks Wilf! Also, did you see the
message
> > I sent you off the list regarding heads?
> >
> > Thanks again, Ori
>
> Duane
>
> 
> Home of the $35 LED solar tracker.
> http://www.redrok.com/electron.htm#led3
> CUL8ER \ \ \ \ \ \\ \ \ Receiver
> Powered by\ \ \ \ \ \\ \ \ [*]
> Thermonuclear \ \Solar\Energy\from the Sun \ /////
> Energy(the Sun) \ \ \ \ \\ \ / / /\/ / /
> \ \ \ \ \ /\ / \/ / / / 
> WA0VBE \ \ \ \ / /\ \/ / / \/ /
> Ziggy \ \ \/ / / \ \/ \/ /\ 
> \ / \ \/ / /\ \\ / \ / / 
> "Red Rock Energy" === ===\ / \ / \ === \ / ===
> Duane C. Johnson, Designer=== === \ \ === / 
> 1825 Florence St Mirrors,Heliostats,Controls & Mounts
> White Bear Lake, Minnesota \ \ / 
> USA 551103364 \ \ 
> (651)6355O65 work \ \ / 
> (651)4264766 home use Courier New Font \ \ 
> (413)556659O Fax copyright \ / 
> (651)5832O62 Red Rock Energy Site (C)980907 ===\ 
> redrok@... (my primary email: address) \ 
> redrok2@... (Hotmail address) \ 
> duane.johnson@... (Unisys address) \ 
> http://www.redrok.com/index.htm (My New Web site) \
> These are my opinions, and not that of Unisys Corp. ===
>
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Yikes this question brings back memories, by the end of the first semester
we had to do circuits with a dozen resistors and three different voltage
sources...
Just for fun, the current given by the left 10V supply is 1.429A and the
right supply is 0.714A.
Bonus question? we only ever had one class that gave bonus questions!
GrantM
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