# Effective Resistance of a circuit

#### logearav

Joined Aug 19, 2011
243
Revered members,
I need to find the effective resistance of this circuit which i have given in attachment.
Now let me explain how i carried out.
I proceeded from right to left
1) R s = 12 + 6 = 18Ω
2) Rp = R1 R2/R1 + R2 which is (18 x 6)/24 = 4.5Ω
3) R = R1 + R2 = 18 + 4.5 = 22.5Ω
I have given the above steps in my second attachment
Now, for the same problem when i proceed from left to right
1) I took 18Ω and 6Ω as parallel i got 4.5 Ω and then this 4.5 is in series with
12 Ω and 6Ω so 4.5 + 12 + 6 = 22.5Ω
I have given the above steps in my third attachment
Now my question is why should not i take 18 Ω and 6Ω as series connection and proceed from left to right, because when i take those 18 Ω and 6Ω as described in my last attachment, i get wrong answer. Where do i go wrong?

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#### logearav

Joined Aug 19, 2011
243
Revered members,
Is this not a series circuit? But when i proceed from left to right taking these resistances as series, i get wrong answer. Why?

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#### Mysteryname

Joined Oct 19, 2010
4
From the looks of it where you made the final 2 resistors as an equivalent circuit (22.5ohm) is on the right track.

remember V=IR.

you have a 30volt supply and 22.5ohm resistance. with both of those you have the current. Using the current (I) you can find the voltage drop across the 18ohm resistor.

#### Georacer

Joined Nov 25, 2009
5,182
@logearav

Pay more attention to the resistor configurations. Your first attempt is correct. However, in your second one, you started by merging the 18Ω and 6Ω resistors as parallel. This is wrong, as they don't have the same endpoints.

In your first attempt, merge the last two remaining resistors (18Ω and 6Ω) and find the current flowing through their combination. As they are in series, that same current will flow through each one.

Now that you know the current running through the 18Ω resistor, you can find the voltage drop across it.

Is that clear?

#### logearav

Joined Aug 19, 2011
243
you started by merging the 18Ω and 6Ω resistors as parallel. This is wrong, as they don't have the same endpoints.

Thank you both for your replies.
Georacer,
I agree with you that they are not parallel. Then it should be series. When i proceed like this, taking these resistors as series, i don't get the correct answer 22.5 ohm.

#### Georacer

Joined Nov 25, 2009
5,182
Two resistors don't necessarily have to be either parallel or in series. They can just be in a non-reducible configuration.

Only series and parallel resistors can be reduced. If they are not, you must find other pairs that are.

#### logearav

Joined Aug 19, 2011
243
Two resistors don't necessarily have to be either parallel or in series. They can just be in a non-reducible configuration.
This is news to me. Thanks Georacer, for giving new insight into the domain of resistors. All along, i thought resistors would be either in series or parallel. Thanks again.

#### JoeJester

Joined Apr 26, 2005
4,390

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#### logearav

Joined Aug 19, 2011
243
Thanks for the reply JoeJester. Can you tell me how you drew the circuit, I mean the software, because, everytime i have to scan my work and post that, if i need help wrt to diagrams and circuits. It would be helpful for me in future.

#### #12

Joined Nov 30, 2010
18,224
Jester did what I was thinking. Make consecutive drawings as you reduce the resistors to their equivalents keeps it from getting confusing.

#### JoeJester

Joined Apr 26, 2005
4,390
That particular drawing was made with Visio ... a CAD software. I was debating whether or not to do it in my simulation software (TINA). I haven't used Visio in a while so I chose that since I recently upgraded to Visio 2007.

You need to take your time and draw each section as you do it. That will maintain your line of thought and you won't get so confused in the process. I numbered the drawings to keep them straight as well as annotating what each equvalent resistor combo was. Can you see where you went astray with the second and third drawings of yours? That is the most important part of this thread ... you seeing where you went astray. A little general housekeeping with your drawings would have paid dividends.

#### logearav

Joined Aug 19, 2011
243
Thanks JoeJester. In fact, i solved the problem as given in your image. But, my teacher said it was wrong and drawn the image as given below and assumed to be parallel circuit and got 4.5 ohm as resistance and then he added this with remaining 12 and 6 ohm resistors to get 22.5 ohm as effective resistance. I got doubt, but i avoided discussing with him for the fear of trashing from him. Thats why i raised here, which is of very immense help to gain knowledge through learned members

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#### JoeJester

Joined Apr 26, 2005
4,390
I'm having a difficult time believing that the instructor explained that was the way to work the problem. They were truely fortunate that R3 and R4 totaled the same as R1, which lead to the correct solution. How you explained what they said does not make sense except in that narrow case ... where R3 + R4 = R1.

Change the values of R3 and R4 to 3 and 4 ohms respectively. You will not be able to duplicate the correct answer, solving the problem in the manner described by your instructor as stated above.

Solve the problem both ways, the way you described above and the way I described in the my first graphic. My answers are included.

p.s. before you ask, the graphic was made in paint.net ... by pasting the excel table plus the schematic in my simulation software.

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#### logearav

Joined Aug 19, 2011
243
Thanks a lot again, Mr. Joe Jester, for your lovely explanation. Fortunate I am to be a member of this group which has excellent members like you, who take pains to help learners like me. Now i got a clear idea about this problem. Thank you so much, sir.

#### JoeJester

Joined Apr 26, 2005
4,390
Is there a TA (Teachers Assistant) teaching you or is it the actual professor listed on the course syllabus?

#### logearav

Joined Aug 19, 2011
243
TA is teaching me, sir.

#### JoeJester

Joined Apr 26, 2005
4,390
Thanks for answering that, it has some illuminating effects on what you stated earlier.

Joined Dec 26, 2010
2,148
Thanks JoeJester. In fact, i solved the problem as given in your image. But, my teacher said it was wrong and drawn the image as given below and assumed to be parallel circuit and got 4.5 ohm as resistance and then he added this with remaining 12 and 6 ohm resistors to get 22.5 ohm as effective resistance. I got doubt, but i avoided discussing with him for the fear of trashing from him. Thats why i raised here, which is of very immense help to gain knowledge through learned members
Surely you do not mean THRASHING, as in corporal punishment, or perhaps more accurately physical assault or abuse, as a possible sanction for asking for clarification of some point of theory from your teacher?

I am really quite shocked. While some people think that the abandonment of beating in some regions may have contributed to difficulties with school discipline, the idea of its use over a theoretical disagreement seems absurd. I suppose this may reflect cultural differences (East vs. West?) but it seems like something that should be long dead.

That the teacher should insist on the primacy of his opinion is not completely unreasonable, as after all he should generally know better than his students. This kind of bullying though seems to speak of insecurity on his part.

#### Georacer

Joined Nov 25, 2009
5,182
Or it could be that some of us aren't native English speakers and might misuse some words.

#### JoeJester

Joined Apr 26, 2005
4,390
I took it as a verbal thrashing ... belittleing et al; after allowing for all the language diversity we have here.