Nice work on providing an example and a thorough question. You're obviously quite confused, but by talking it out in detail it's easy to see where you've gone wrong.
First thing, your colleague is not helping at all. Best to expel that advice.
Next, study this maxim:
- current flows through a circuit - the current in equals the current out
- voltage changes across components - the voltage goes from high to low around a circuit
So when you look at you battery connected with a resistor, 50mA flows out of the battery, 50mA flows through the resistor and 50mA flows back into the other terminal of the battery. Voltage starts high (5V) at the positive terminal, drops across the resistor and is low (0V) by the time it reaches the negative terminal. Physics makes it so.
Energy on the other hand, leaves the battery and is consumed by the resistor. The battery will not charge in this scenario, because the current direction is out of the positive terminal. The fact that it goes back into the negative terminal is a given (due to #1 above) and doesn't change anything.
A battery is a voltage source, so the current in the circuit is dictated by the resistance in that circuit. So if you have a 5V battery connected to a 50 Ohm load, there will be 100mA flowing through it - there's no argument there, no short circuit. Physics makes it so.
A short circuit only occurs when you put zero resistance across a voltage source.
Now you wonder, in a short circuit, how can the voltage go from 5V to 0V if there's no resistance in the circuit? The answer is that there's always something, even in a "short circuit". Say 0.01 Ohm. So you drop 5V across 0.01 Ohm and produce an enormous current, sparks, fire and all the fun stuff, and we call it a short circuit.
I get the following four equations:
$$\begin{align*}
0\:\text{V}+24\:\text{V}-5\:\Omega\cdot\left(I_1-I_3\right)-V_{4\text{A}}-3\:\Omega\cdot 4\:\text{A}&=0\:\text{A}
\\\\
0\:\text{V}+3\:\Omega\cdot 4\:\text{A}+V_{4\text{A}}-10\:\Omega\cdot\left(I_2-I_3\right)-20\:\Omega\cdot I_2&=0\:\text{A}
\\\\
0\:\text{V}-5\:\Omega\cdot\left(I_3-I_1\right)-5\:\Omega\cdot I_3-10\:\Omega\cdot\left(I_3-I_2\right)&=0\:\text{A}
\\\\
I_1-I_2&=4\:\text{A}
\end{align*}$$
where \$V_{4\text{A}}\$ is the voltage across the current source.
I used the + sign on the top node and - on the bottom node of the current source. As shown below:
No need for any super- anything.
These solve out (using sympy):
var('i1 i2 i3 v4a')
eq1 = Eq( 0 + 24 - 5*(i1-i3) - v4a - 3*4, 0 )
eq2 = Eq( 0 + 3*4 + v4a - 10*(i2-i3) - 20*i2, 0 )
eq3 = Eq( 0 - 5*(i3-i1) - 5*i3 - 10*(i3-i2), 0 )
eq4 = Eq( i1 - i2, 4 )
solve( [ eq1, eq2, eq3, eq4 ], [ i1, i2, i3, v4a ] )
{i1: 24/5, i2: 4/5, i3: 8/5, v4a: -4}
Check your work against mine to see if you agree. Then check all this against the stated solution.
Toss the textbook into the trash can. And with it, the 8th edition of "Electronic Principles" by Malvino & Bates. And probably dozens of other terrible books on the topic of electronics. I've no idea how they get so bad, but they do. Even after 8 editions! The world is flush with garbage. I'm guessing it has to do, in part, with grad students. But I'm not sure.
Best Answer
No, the authors have not declared that the current \$i_8\$ is flowing downward. They have defined some current named \$i_8\$ that is flowing downward. Once the circuit is solved it may be found that the value of \$i_8\$ is negative. In that case, we know that there is actually a positive current flowing up.
We must label some assumed direction for all currents and some assumed polarity for voltages so we can write meaningful and consistent equations for the circuit. If the resulting values come out negative it just means that our assumptions were not correct, but it is a perfectly valid way to start a problem.