Switching from parallel to series and keeping restistance the same.

[lilcho]

Homework Statement

Two resistors of X and Y are connected in parallel with one another and in series with a 200 Ω resistor and a battery $$\epsilon$$=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.

Homework Equations

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.

 

[n.karthick]

I can give u a hint.
Because the current is same
(i) when Y is connected and
(ii) Y disconnected and 50 ohm connected in circuit
it means that total resistance in the circuit is same for both the cases.
So you can equal the total resistance in 2 cases and solve for Y.

 

[ehild]

The current is the same through resistor X in both cases. It does not mean that the whole current drawn by the battery is the same.

ehild

 

[gneill]

Attached is a circuit diagram which may help. The blue components are the ones that change between configurations. Note that in order for the current through RX to remain the same, the voltage across RX must also remain the same -- you might find it convenient to treat the circuit(s) as a voltage divider.

 

[n.karthick]

Oh I didn't notice that it is the current through X which is same, sorry. Equating voltage across X in both cases and solving that should give Y.