I got stuck on this problem on last year's exam, and apparently I still can't solve it The text reads: 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?
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!
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.