To write the Kirchhoff Loop Equation for the entire outside loop of the circuit, we need to apply Kirchhoff's Voltage Law (KVL). According to KVL, the sum of the voltage drops around a closed loop equals zero.

From the circuit diagram you provided, we will traverse the outer loop, moving clockwise from $V_1$ through $R_1$, $R_3$, and $V_3$.

1. Voltage Drops and Polarities:

• The current $I_1$ flows through resistor $R_1$, creating a voltage drop of $I_1 R_1$ (Ohm's law).
• The current $I_3$ flows through resistor $R_3$, creating a voltage drop of $I_3 R_3$.
• The batteries (sources $V_1$ and $V_3$) introduce potential differences. Since we're moving clockwise:
  • The voltage source $V_1$ increases the potential by $V_1$.
  • The voltage source $V_3$ decreases the potential by $V_3$ (assuming its polarity is as shown in the diagram).

2. Writing the KVL Equation for the Outer Loop:

Starting at $V_1$ and moving clockwise: $V_1 - I_1 R_1 - I_3 R_3 - V_3 = 0$

Rearranged KVL Equation:

$V_1 - V_3 = I_1 R_1 + I_3 R_3$

This is the Kirchhoff loop equation for the outer loop, relating the voltages $V_1$ and $V_3$ to the current flows $I_1$ and $I_3$ through resistors $R_1$ and $R_3$, respectively.

Would you like further details or clarifications on any specific part?

Here are five related questions to explore further:

1. What is the Kirchhoff Current Law (KCL) equation at node $A$?
2. How would you set up the loop equation for the inner loop involving $V_2$ and $R_2$?
3. What happens if the resistances $R_1$, $R_2$, and $R_3$ are all equal?
4. How does reversing the direction of current $I_3$ affect the loop equation?
5. How can the total power supplied by the batteries be calculated in this circuit?

Tip: When applying Kirchhoff's Voltage Law, be careful to consistently assign the correct sign to voltage rises and drops based on the direction of traversal and current flow.