Circuit Problem electrical engineering

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  • Thread starter izelkay
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  • #1
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Homework Statement



Here's the question:
http://puu.sh/171ls [Broken]


The Attempt at a Solution



I have no idea where to begin. It says R(AB) = R(L) = 301 ohms. What does that mean? Does R(AB) mean the resistance from A to B? Is that the total resistance? The total resistance is equal to 301 ohms which is also equal to R(L)? I don't know. Any help would be appreciated.
 
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Answers and Replies

  • #2
gneill
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2,858

Homework Statement



Here's the question:
http://puu.sh/171ls [Broken]


The Attempt at a Solution



I have no idea where to begin. It says R(AB) = R(L) = 301 ohms. What does that mean? Does R(AB) mean the resistance from A to B? Is that the total resistance? The total resistance is equal to 301 ohms which is also equal to R(L)? I don't know. Any help would be appreciated.
Yes, RAB is the equivalent resistance of the entire network as "seen" from the terminals AB.
 
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  • #3
115
3
Ok, I think I got it. R should equal roughly 174 ohms.
 
  • #4
gneill
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20,909
2,858
Looks good :smile:
 
  • #5
Hi! Note that the circuit can be drawn as a resistance R in series with the parallel combination of: (a) a resistance of R and (b) a series combination of the resistances R and R(L).

Then, the circuit can be drawn as a resistance of R in series with the parallel combination of R and (R + R(L)).

First, determine the equivalent resistance of the parallel portion of the circuit, R and R + R(L).

[tex]\frac{1}{R(Parallel)} = \frac{1}{R + R(L)} + \frac{1}{R}[/tex]
[tex]\Rightarrow R(Parallel) = \frac{1}{\frac{1}{R + R(L)} + \frac{1}{R}}[/tex]

Multiply the numerator and denominator of the right side of the equation by [itex]\frac{R(R + R(L))}{R(R + R(L))}[/itex] to get the following:

[tex]R(Parallel) = \frac{R(R + R(L))}{R + R + R(L)} = \frac{R^2 + R(L)}{2R + R(L)}[/tex]

You are then told that R(AB) must equal R(L). This becomes the following:

[tex]R + \frac{R^2 + R(L)}{2R + R(L)} = R(L)[/tex]

Then, multiply both sides of the equation by (2R + R(L)) to get the following:

[tex]R(2R + R(L)) + R^2 + R(L) = (2R + R(L))R(L) = 2RR(L) + (R(L))^2[/tex]
[tex]\Rightarrow 2R^2 + RR(L) + R^2 + R(L) = 2RR(L) + (R(L))^2[/tex]
[tex]\Rightarrow 3R^2 - RR(L) + R(L) - (R(L))^2 = 0[/tex]
[tex]\Rightarrow 3R^2 - 301R + (301) - (301)^2 = 0[/tex]
[tex]\Rightarrow 3R^2 -301R + 301 - 90601 = 0[/tex]
[tex]\Rightarrow 3R^2 - 301R - 90300 = 0[/tex]

Applying the quadratic equation (I'll leave the details up to you) yields the following:

[tex]R = 230.7675926 ohms ≈ 231 ohms[/tex]

You will want to double check my result; but I think that I did everything okay.
 
  • #6
115
3
^If I plug 231 in for R, Rab comes out to be about 390 ohms.

174 for R gives me 301 ohms for Rab though.
 

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