By the passive reference convention, the current enters the positive terminal of a passive element such as a resistor. The voltage source polarity is given by the symbol.
So, KVL clockwise around loop 1 is:
\$U_{AB} - E_5 - I_2 \cdot R_2 - E_1 - I_1 \cdot R_1 = 0\$
Or,
\$U_{AB} = E_5 + I_2 \cdot R_2 + E_1 + I_1 \cdot R_1\$
Consider how you would measure the voltage across R2. How do you choose which polarity to measure? You have a choice of where to place the red and black leads so which one is correct? Obviously, both are. If the red lead is on placed on the more positive lead of the resistor, the measured voltage will be positive. If the red lead is placed on the more negative lead, the measured voltage will be negative. Choosing the reference polarity of the voltage variable (the variable you solve for in your equations) for R2 works in exactly the same way.
You are free to choose the reference polarity and, after you solve the equations, if you find that the voltage across R2 is negative, you know that the end you chose to be positive is actually the more negative end. The point is, just as with the voltmeter leads, the reference polarity is arbitrary. However, it is always a good practice to use the passive reference convention so, as in your example, if you want the voltage across R2 in terms of I2, you should should choose your reference polarity such that I2 enters the positive terminal.
Label Vc and Vl in your diagram... current always flows from higher potential to lower potential (voltage). So depending on the direction your draw your current flowing, the voltage always drops in the direction of that current flow.
Simply put, in your first picture there is only one current loop. The current flowing through all the elements must be equal. I'll call the upper left corner node Vc and the upper right corner node Vl. Then namely:
(Vc - 0) / Zc = (Vl - Vc) / R = (0 - Vl) / Zl
Here Zc and Zl are the complex impedance of the capacitor and inductor respectively.
Alternatively, if we assume a current i0 is flowing through the circuit clockwise, and remember that voltage always drops across an element when current flows through it, then:
Vc = 0 - i0 * Zc
Vl = Vc - i0 * R
0 = Vl - i0 * Zl
Therefore:
0 = Vc - i0 * R - i0 * Zl = - i0 * Zc - i0 * R - i0 * Zl
Changing the sign:
i0 * Zc + i0 * R + i0 * Zl = 0
So yes, all elements should have the + sign.
Best Answer
Kirchhoff's current law (KCL): the sum of the currents in a node is zero.
Say you have 5 wires coming together in a node as shown, and \$I_1\$, \$I_3\$ and \$I_4\$ supply current to the node. This current has to go somewhere, and will go from the node by \$I_2\$ and \$I_5\$:
such that
(The minus signs for \$I_2\$ and \$I_5\$ are due to the reversed arrows for those currents.)
A more general form of KCL says that the current entering a closed boundary is equal to the current leaving it:
Kirchhoff's voltage law (KVL): the sum of the voltages in a closed circuit is zero. If you have a circuit consisting of a battery and a resistor as a load then the voltage over the resistor is \$\mathrm{-V_{bat}}\$ (the minus sign means that if you go clockwise through your circuit you go from \$-\$ to \$+\$ for the battery, but from \$+\$ to \$-\$ for the resistor).
Total voltage: \$\mathrm{+V_{bat} - V_{bat} = 0}\$.
This goes for every closed loop path you can find in a design, no matter how complicated and how many branches there are.