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I was reading Hayt Kemmerly Engineering Circuit Analysis Book,(I tried others, but this is the most comprehensible to me.), And I came across this circuit. I understand the first two, but I don't understand how, in the 3rd circuit (c), there is negative voltage through the resistor $R_3$ , while the current through it from $+$ to $-$ is positive $7A$. I don't understand how resistors can supply voltage. My guess is this is only a mathematical model, not real.

Edit: The Answers are shuffled in this book.

In a passive device, negative absolute resistance cannot exist. However, negative differential resistance, where an increase in voltage leads to a decrease in current or vice versa, is observed in a number of rather common systems, such as neon signage and fluorescent lighting, as well as some more esoteric ones like tunnel diodes. Below is a figure showing an I-V curve for a generic electrical discharge; notice the region between points D and G where the voltage decreases as the current increases. This is the region in which both fluorescent lighting and neon signage normally operate.

(image source)

Negative absolute resistance can exist over limited ranges by using active elements. There's an op amp circuit commonly called a negative impedance converter that simulates a negative resistance, capacitance, or inductance by using an op amp and feedback:

^{simulate this circuit – Schematic created using CircuitLab}

The two circuits above are equivalent, provided the op amp does not saturate.

My apologies to everyone, the original solution was wrong, I had the direction of the currents through R2 and R3 reversed. Solution now edited.

If we Measure all voltages relative to the common junction of the 2 ohm resistor, R2, the 8A current source and the 3 ohm resistor then:

1. Summing the currents at the node at the top right gives the 37V
   voltage source supplying a current of 7A. (15-8)


2. At the - end of the 37V source the 2 ohm resistor has 6A flowing through it therefore the current through R2 is 1A to make up the 7A.


3. The top end of the 2 ohm resistor is at -12V (2 ohms X 6A). and hence R2 = 12 ohms (12V / 1A).


4. The node at the top right is at 45V as 15A flows through the 3 ohm resistor.


5. The other end of R3 (the + end of the voltage source) is at 37-12 = +25V (the voltage across R2 and the 2 ohm resistor -12V + the 37V source)


6. The voltage across R3, Vs = 20V (45-25).


7. -7A is flowing through R3 and hence R3 = -20/7 ohms, or approximately -2.86 ohms.
   

The more I look at it, the more I think the "-" in front of the 4 and 20 in the answers is just a dash (hyphen) not a minus sign.

To get the current and voltage values that were shown in the book.

You indeed need a negative resistance in the circuit.
The R3 needs to have negative resistance $R_3 = - frac20V7A = -2.857Omega$

Because for the positive resistance we get this result:

As you can see the result is not even a close to the book solution.

But if we use a "real" negative resistance (negative impedance converter) instead.

The simulation result will match the book solution:

This won't answer the question, but will show how you find that R3 must have negative resistance.

Here's the circuit diagram, with a couple of annotations:

First, from Ohm's law, we know that the voltage across the 2 ohm resistor is 12 V, and the voltage across the 3 ohm resistor is 45 V.

If you take KVL around the loop indicated with the orange arrow, you get

$$ -45 V + (-12 V) + 37 V - v_x = 0$$

This gives you $v_x = -20 V$.

Defining $I_x$ as the current through R3 (flowing left to right according to the passive sign convention), and using KCL at node "A" you get

$$ I_x + 8 A - 15 A = 0$$

From which, $I_x =7 A$.

You now have

$$R_3 = fracv_xI_x = frac-20 V7 A = -2.86 rm Omega$$

It does not matter if you reverse the direction of $v_x$. If you do that (and also reverse the direction of $I_x$ to maintain the passive current convention) you'll just get $v_x=+20 V$ and $I_x=-7 A$.

Ohmic negative resistance doesnt exist.However there is negative resistance.Zener diodes at their breakdown voltages have negative resistance since they create current by quantum tunneling.