Two resistors of X and Y are connected in parallel with one another and in series with a 200 Ω resistor and a battery [tex]\epsilon[/tex]=1.5 V and internal resistance r=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 unchanged. Find the resistance of Y. Compute the value of Y again for the same case considering the internal resistance of the battery negligible.
Resistance Total (parallel) = ((1/r1)+(1/r2))^(-1)
Resistance Total (series) = (r1+r2)
The Attempt at a Solution
I have drawn the diagram and calculated:
Rt (parallel) = 200 + ((1/100)+(1/y))^(-1)
Rt (series) = 200 + 50 + 100 = 350
I am confused when removing resistor y and adding 50 ohms of resistance.