Calculating Resistances in Parallel and Series: A Tricky Problem Solved

[NotaPhysicsMan]

oK here it is!

Three identical resistors are connected in parallel. The equivalent resistance increases by 700 ohms when one resistor is removed and connected in series with the remaining two, which are still in parallel. Find the resistance of each resistor.

Ok my thoughts:

Ok in situation one, we have:

1/Req=1/R1+1/R2+1/R3.

In situation two we have:

1/Req +700 ohm=(R1+R2)+1/R3, R1+R2=Reff.

So 1/Req +700 ohm=1/Reff+1/R3.

Ok, now I'm stuck...Any tips welcome!

 

[Justin Lazear]

Let the resistance of the resistors be R.

For the first situation, we have R_{eq1} = (\frac{1}{R} + \frac{1}{R} + \frac{1}{R})^{-1}.

For the second situation, we have R_{eq2} = R + (\frac{1}{R} + \frac{1}{R})^{-1}.

And we know that R_{eq1} + 700\Omega = R_{eq2}. Can you take it from here?

--J

 

[NotaPhysicsMan]

Ok let's try:

So I'm assuing I can sub Req2 in for the bottom equation.

Req1+700ohm=R+ (2/R)^-1

From Req1=(3/R)^-1.

Ok so (3/R)^-1 +700 ohm=R +(2/R)^-1

Solve for R.

(-1/R)^-1+700=R
-R+700=R
2R=700
R=350?

 

[Justin Lazear]

You made an error solving for R. Try it again.

You're good once you get R = 600 ohms.

--J

 

[NotaPhysicsMan]

Ah HA! I got it now. Ok so R/3+700ohm=R+R/2.
Ok now 700=R+R/2-R/3
Then 700=7R/6
And R=600!

Thanks a bunch :)