Three identical resistors: parallel to mixed parallel-series

[vaxopy]

Please help me with resistors in parallel/series - with many variables :)

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

Im totally confused how to do this..
I know in parallel, voltage remains the same, wheras in series, current stays the same

please kick in the right direction on how to solve this (with many unknowns :) )
Thanks!

 

[danesh]

Do you really mean 700 W as in power?
Because then you would have to many unknowns, like the current for instead.
If you mean $$700 \ohm$$, as in resistance, then it is not to hard to calculate the answer.

 

[thepatientmental]

No problem, let's break this down step by step. First, let's label the resistors as R1, R2, and R3. We know that when they are connected in parallel, the equivalent resistance (Req) is given by:

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

We also know that when one resistor is removed and connected in series with the remaining two, the equivalent resistance increases by 700 W. This means that the new equivalent resistance (Req') is given by:

Req' = Req + 700

Now, let's substitute in the equation for Req into the equation for Req':

Req' = (1/R1 + 1/R2 + 1/R3) + 700

Simplifying this equation, we get:

1/Req' = 1/R1 + 1/R2 + 1/R3 + 1/700

Now, we can see that the only difference between this equation and the one for Req is the addition of 1/700. This means that the resistance of the removed resistor (let's call it Rx) is equal to 700 W. So now we have two equations:

1/Req = 1/R1 + 1/R2 + 1/R3
1/700 = 1/Rx

We can solve for Rx by taking the reciprocal of both sides:

700 = Rx

So we now know that the removed resistor has a resistance of 700 W. But we still have three unknown resistances (R1, R2, and R3). To solve for these, we need one more equation. Luckily, we have one from the given information: the equivalent resistance increases by 700 W when one resistor is removed. This means that the equivalent resistance of the remaining two resistors in parallel is equal to the resistance of the removed resistor:

Req = Rx

Substituting in the values we know, we get:

1/R1 + 1/R2 = 1/700

Now we have two equations with two unknowns, so we can solve for R1 and R2. Once we have those values, we can use the equation for Req to solve for R3 as well. I hope this helps guide you in the right direction! Remember to always use the equations for parallel and series connections to solve problems like this. Good luck!