[johnconnor92]

Algebra mistake? The answer is11.8ohm for (a).

Thank you!

[Stonebridge]

The question is much simpler than you imagine.
For a)
If the current in the circuit is given as 0.4A use the formula provided
0.4 = 0.2V³ to find V the pd across the rod
Subtract this from 6V to find the pd across the resistor. (The battery has negligible internal resistance)
You now know the pd across, and current through the resistor.
V=IR will find its resistance.

[johnconnor92]

[QUOTE=Stonebridge;38449280]The question is <b>much </b>simpler than you imagine.<br />/QUOTE]

(FACEPALM) What is wrong with me wtf. I corrected the algebra mistake and managed to solve the first part but I took such a big turn to get to the answer.

I can't seem to solve the second part! Why?

[Stonebridge]

(Original post by johnconnor92)

(Original post by Stonebridge)
The question is <b>much </b>simpler than you imagine.<br />/QUOTE]

(FACEPALM) What is wrong with me wtf. I corrected the algebra mistake and managed to solve the first part but I took such a big turn to get to the answer.

I can't seem to solve the second part! Why?

Because you are not thinking it through logically.
You seem to over complicate matters.
Simple circuit problems like this are not usually mega maths problems. They just involve applying the standard circuit rules.
Part b
If the power in the rod is twice that in R then VI for the tube is 2 VI for the resistor. As the current is the same in both it means the pd across the rod is twice that across the resistor. As the pds add to give 6 volts that means the pd across the rod is 4V and the resistor is 2V
Knowing the pd across the rod gives the current in it using the formula given in the question. This current in the resistor, together with the known pd across it, gives its resistance.