Engineering Circuit Problem electrical engineering

[izelkay]

Homework Statement

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

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.

 

[gneill]

Homework Statement

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

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, R_AB is the equivalent resistance of the entire network as "seen" from the terminals AB.

 

[izelkay]

Ok, I think I got it. R should equal roughly 174 ohms.

 

[gneill]

Looks good

 

[NoPhysicsGenius]

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).

\frac{1}{R(Parallel)} = \frac{1}{R + R(L)} + \frac{1}{R}
\Rightarrow R(Parallel) = \frac{1}{\frac{1}{R + R(L)} + \frac{1}{R}}

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

R(Parallel) = \frac{R(R + R(L))}{R + R + R(L)} = \frac{R^2 + R(L)}{2R + R(L)}

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

R + \frac{R^2 + R(L)}{2R + R(L)} = R(L)

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

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

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

R = 230.7675926 ohms ≈ 231 ohms

You will want to double check my result; but I think that I did everything okay.

 

[izelkay]

^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.