Part C: Solving Resistance in a Complex Circuit

[edgarpokemon]

Homework Statement

Part C of the following problem
http://www.chegg.com/homework-help/questions-and-answers/problem-1-find-equivalent-resistance-rab-circuits-figp36-q8156743

2. Homework Equations

The Attempt at a Solution

Hi, to solve this problem, first i tried to ignore the two resistors in the middle sincr they are not even connected. I am asked to find the total resistance of the circuit. I got an answer of 64 when ignoring the two middle resistors, but that is not the answer. So i assume those two resistors have an effect on the answer. But i don't understand how. Help!

 

[anorlunda]

Hint: The problem wants you to find the total resistance between nodes a and b, which are the ends of those two resistors in the middle.

 

[edgarpokemon]

Hint: The problem wants you to find the total resistance between nodes a and b, which are the ends of those two resistors in the middle.

I end up with a circuit in series with three resistors, a 15ohm, and the 3ohm and the 2ohm. So is it 20? I read somewhere that when a resistor is not connected to the circuit, it can be ignored, so why is this not the case?

 

[phinds]

I end up with a circuit in series with three resistors, a 15ohm, and the 3ohm and the 2ohm. So is it 20? I read somewhere that when a resistor is not connected to the circuit, it can be ignored, so why is this not the case?

If it were not connected to the circuit you COULD ignore it. Did you not understand anorlunda's comment? If you start at a and b, how can you believe that those resistors can be ignored? How could any current flow to/from a and b without going through those resistors?

Also, your 15ohms is wrong. Go carefully, step by step, in combining the other resistors. If you continue to get 15 ohms, show your work (step by step) here.

 

[NascentOxygen]

to solve this problem, first i tried to ignore the two resistors in the middle sincr they are not even connected.

The problem involves putting electrons into the network at point A, letting them make their way around the circuit and having them emerge from point B. Your task is to determine what effective resistance they would encounter in going around the resistor arrangement that is in place between A and B.

 

[CWatters]

Show your working.

 

[FactChecker]

For any homework, you need to show your work. To show your work, you can post the diagram with each resister labeled (number them) and show step-by-step which ones are combined to make an equivalent resister. For instance, if you add the resistance of the two in the upper right corner and they are R(1) and R(2), you can say that R(1,2) = R(1) + R(2) = 30Ω + 24Ω = 54Ω
That is one way to do it. In any case, you need to make it clear what you are doing.

 

[FactChecker]

Also, your 15ohms is wrong. Go carefully, step by step, in combining the other resistors. If you continue to get 15 ohms, show your work (step by step) here.

I also got 15 ohms before the 2 and 3 were added. So I got the same final answer of 20 ohms as he did.

 

[phinds]

I also got 15 ohms before the 2 and 3 were added. So I got the same final answer of 20 ohms as he did.

Hm. Checked it twice. Must be making the same error. I got 11 point something.

 

[gneill]

Please refrain from further postings until the OP has shown his work in sufficient detail to critique it.

 

[cnh1995]

I also got 15 ohms before the 2 and 3 were added. So I got the same final answer of 20 ohms as he did.

Me too. I believe the OP has it right.

 

[FactChecker]

Just for fun, I wrote a Perl program to calculate these answers. I only checked it on this example, but I was pleased with how easily it seems to have worked out.

# define the resister network. '+' for series.  '|' for parallel
#   use parentheses to the show nesting order of network connections
#   Spaces, tabs and newlines do not matter.  Use them as desired for clarity.
$network = '
(
   (
      (30+50)|20
   )
   +14
)
|
(
   (
      (30+24)|27
   )
   +12
)
+3 + 2
';

# =====================  BEGIN CODE  ===================
sub series {
    return $1 + $2;
}
sub parallel {
    return 1/(1/$1 + 1/$2);
}

# ============= BEGIN LOOP TO PROCESS NETWORK ==========
$changed = 'y';
while($changed){
    $orig = $network;
    $changed='';
    $network =~ s/\s//g;  # remove any spaces to make everything easier to parse
    $network =~ s/\(([\d.]+)\)/$1/g; # remove unneeded parentheses
 
    #  process one pass of two in series
    $network =~ s/([\d.]+)\+([\d.]+)/&series($1,$2)/eg;
    $network =~ s/\(([\d.]+)\)/$1/g; # remove unneeded parentheses
 
    #  process one pass of two in parallel
    $network =~ s/([\d.]+)\|([\d.]+)/&parallel($1,$2)/eg;
    $network =~ s/\(([\d.]+)\)/$1/g; # remove unneeded parentheses
 
    #  if a change was made, there may be more remaining to do.
    if( $orig ne $network ){$changed='y'}
#    print "network=$network\n"; # can add printouts to verify calculation steps
}
# ==============  PRINT FINAL ANSWER  ==================
print "$network\n";

# ==========  PAUSE BEFORE CLOSING PROGRAM  ============
$ans=<STDIN>;

 

[phinds]

Found my mistake, now also agree w/ 15 + 2 + 3