To solve this question using the upper loop in the circuit, we apply Kirchhoff's Voltage Law (KVL), which states that the sum of voltages around any closed loop in a circuit must equal zero.

Given Data:

• $V_1 = 5 \, \text{V}$
• $V_2 = 25 \, \text{V}$
• $R_1 = 4 \, \Omega$
• $R_2 = 3 \, \Omega$
• $R_3 = 6 \, \Omega$

In the upper loop, the relevant elements are:

1. Voltage source $V_1$
2. Resistor $R_1$ (with current $I_1$)
3. Resistor $R_2$ (with current $I_2$)
4. Voltage source $V_2$

KVL Equation for the Upper Loop:

The voltage drops across the resistors are:

• $V_{R_1} = I_1 R_1 = I_1 \times 4 \, \Omega$
• $V_{R_2} = I_2 R_2 = I_2 \times 3 \, \Omega$

The sum of voltages for the upper loop:

• Starting at $V_1$, going through $R_1$, then $R_2$, and finally $V_2$, we get: $V_1 - (I_1 R_1) - (I_2 R_2) - V_2 = 0$ Substituting the values: $5 - 4I_1 - 3I_2 - 25 = 0$ Simplifying: $-4I_1 - 3I_2 = 20$ $4I_1 + 3I_2 = -20$

This matches none of the given options, as none correspond to $4I_1 + 3I_2 = -20$. Therefore, the correct choice is "none of these".

Would you like further details or clarifications?

Here are 5 related questions to explore:

1. How do you apply Kirchhoff’s Current Law (KCL) in circuits like this?
2. What are the steps to solve a circuit with more than two loops?
3. How do you solve for unknown currents using simultaneous equations in a circuit?
4. How do resistor values affect the current distribution in parallel circuits?
5. How would the equation change if the polarity of $V_2$ was reversed?

Tip: Always double-check the polarity of voltage sources when applying Kirchhoff’s Voltage Law (KVL).