# "Find the missing resistance" type question. Wrecking my brain.

Discussion in 'Homework Help' started by Ecilious, May 31, 2010.

May 31, 2010
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2. ### Ecilious Thread Starter New Member

May 31, 2010
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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.

Apr 5, 2008
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Hello,

Can you show us one of your efforts on this question?
In that way we can see where you got stuck and will be able to help you.

Greetings,
Bertus

4. ### Ecilious Thread Starter New Member

May 31, 2010
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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.

I'm flapping in the wind really.

5. ### beenthere Retired Moderator

Apr 20, 2004
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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.

Apr 5, 2008
18,691
3,639
Hello,

Here are the circuits for the two cases:

Bertus

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7. ### Ecilious Thread Starter New Member

May 31, 2010
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Ok, I've gone through this again. Pretty much what I had before:

desired I in 30Ω resistor = 1/75

V in 30Ω resistor = 1/75 * 5 = 0.4V

Current in circuit = 2/48 = 1/24

R= V/I= 0.4/(1/24-1/75)= 14.12Ω

a far shot from the correct answer 200Ω

I still can't see where I'm going wrong and I can see it should be very easy. Very frustrating.

Thanks for your help any way guys.

8. ### GuitarLightning New Member

Jan 29, 2008
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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.

9. ### Ecilious Thread Starter New Member

May 31, 2010
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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.