Two resistors of X and Y are in parallel with one another and in series with a 200 Ω resistor and a battery whose emf is 1.5 volt and whose internal resistance is 0.5 Ω. The resistance of X is 100 Ω. When Y is disconnected from X, an additional resistance of 50 Ohms must be inserted in the circuit in order to keep the current through X is unchanged. Find the resistance of Y. Compute the value of Y again for the same case considering the internal resistance of the battery negligible.
R1 = 200 Ω + 0.5 Ω + (XY/X+Y)
R2 = 200 Ω + 0.5 Ω + 50 Ω + X
R1 = R2
The Attempt at a Solution
In R1, X and Y are parallel, but in R2, Y is removed, so it is just in series (I think?). If the 50 Ω was just added in place of Y so they are still parallel, then Y would be 50 Ω and it doesn't seem right.
Since the resistance should still be the same, I set R1 and R2 equal to each other to get:
200.5 + 100Y/(100+Y) = 350.5
I then solved for Y to get Y = 100
Am I solving this correctly? I thought I would need the 1.5V in an equation and compared to the rest of the problems I am doing, this seems much easier. Any help is appreciated. Thanks.