To explain: I'm sure this is an easy question. But I've tried so many times, I have 5 pages of attempts. I just can't work out how to attack it. I've even been given the answer by the tutor (200 ohms) and tried working backwards from the answer and I still can't get it.
Any help greatly appreciated. Please, I need this for my own sanity.
Well I could scan the sheets but most of it just dead ends.
I tried coming at it two ways
Current in parallel section = current in 48Ω resistor
Desired current in 30Ω resistor= V/R1+R2+R3 = 2/ (30+48+72)= 1/75
Current in 48Ω resistor = 2/48 = 1/24
1/75 - 1/24 = 17/600 = current in Rx
R= V/I = 2/(17/600) = 70.59Ω which was wrong
and then I tried a different approach and didn't know what to do with it:
150 = 48 + 30*Rx/30+Rx
When I tried working backwards I worked out the equivalent single resistance of the parallel section 30*200/(30+200) and got 2.61Ω the current should therefore be V/R= 2/(2.61+48) = 0.04 which didn't really help me as I could not achieve this figure.
In the series arrangement, the total R is 30 + 72 + 48. The total current is therefore given by I = E/R. That current is the desired current through the 30 ohm resistor when paralleled by Rx.
If you know the current through a resistor, you can use Ohm's law to determine the voltage across it. Since the drop across the 30 ohm resistor is the same for Rx in parallel, you can see what it has to be for the 48 ohm resistor in series. That gives you the total current and therefore the equivalent resistance of Rx in parallel with the 30 ohms.
I disagree with your tutor. I get 20 ohms for Rx instead of 200.
I looked at your calculations and the only thing that seems wrong is your calculation for current through the 48 ohm resistor. You might want to give it another look - particularly at the voltage used.
Ah yeah, volt drop . Well I got your 20Ω. Seems quite likely it could be a typo on answer sheet.
Thanks a lot for the help, I know it's an easy question but I've just been banging my head against the wall because of it. Especially trying to get 200Ω.