# Series DC problem

Discussion in 'Homework Help' started by TwoPlusTwo, Mar 2, 2012.

1. ### TwoPlusTwo Thread Starter Member

Oct 14, 2010
51
0
I got stuck on this problem on last year's exam, and apparently I still can't solve it

a. How big is Rx if Va = - 40V
b. Find I.

The only way I can make it work is if Rx is not a resistor but a 5V battery.

I get I = 30/4 = 7.5A because the "drop" across the top resistor is 30. But then the next drop has to be 45, and the two add up to more than 70. And I can't have a drop in the opposite direction across Rx, because there is no such thing as a negative resistance.

What am I missing here?

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2. ### BillB3857 AAC Fanatic!

Feb 28, 2009
2,493
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The analysis you gave is exactly what I would have done. Did you get an answer from the instructor?

3. ### TwoPlusTwo Thread Starter Member

Oct 14, 2010
51
0
That's a relief. I'll ask my teacher on Monday and see what he has to say for himself..

4. ### BillB3857 AAC Fanatic!

Feb 28, 2009
2,493
389
Let us know. Never too old to learn something new!

5. ### Paul Kerry Active Member

Jan 9, 2012
36
4
Just had a play with this and the voltage at Va is -42V with Rx at 0Ω and
the current flowing is 7A. I think Mr. Instructor may have made a mistake with the question!

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,923
1,381
In your circuit we can get -40V at point Va if we replace Rx with negative resistance.

Rx = - 2/3Ω ≈ -0.666Ω

In theory and in real live also we can build a device that has a negative resistance.
Here you have a diagram. I use a VCVS to convert positive resistance Rx to the negative one.

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