In the figure, if I = 80 mA, determine the resistance R.
simulate this circuit – Schematic created using CircuitLab
Answer is 600Ohm.
My Steps :
$$
I2=-0.08$$
$$20+R\left(I1-I2\right)+100I1=0\tag1$$
$$200I2+40+R\left(I2-I1\right)=0\tag2$$
So
$$20+R\left(I1+0.08\right)+100I1=0\tag1$$
$$24-R\left(I1+0.08\right)=0\tag2$$
Then
$$
R\left(I1+0.08\right)=24\tag{a}
$$
Now I sub (a) to (1) and get
$$
20+24+100I1=0\\
100I1=-44$$
$$I1=-0.44\tag{b}
$$
Sub (b) back to (a),
$$
R\left(-0.44+0.08\right)=24\\
R=-66.67 \Omega\\
$$
How to obtain the answer 600 Ohm?
Best Answer
The answer is definitely wrong. Here's a quick way to tell. If you have R=infinite, then you have 40V/300ohms = 0.133 amps if you ignore current provided by the 20V supply. That's the absolute minimum current that will flow through I. Any lowering of R from infinite will only increase the amount of current flowing through I. That means that their initial statement of I=80mA is impossible.
The only exception to that is if we do allow negative resistances as you have calculated. A negative resistance would be kind of like a voltage/current source. You're likely correct in your negative resistance calculation.