To find the value of $I_3$ in this circuit, we can use Kirchhoff's Voltage Law (KVL). The circuit shows two voltage sources, $V_1 = 5 \, V$ and $V_2 = 7 \, V$, along with three resistors $R_1 = 1 \, k\Omega$, $R_2 = 2.2 \, k\Omega$, and $R_3 = 3.3 \, k\Omega$. Here’s the approach:

1. Apply KVL in the loop that includes the entire circuit. The loop has two sources of voltage and three resistors. The current $I_3$ is the current flowing through $R_3$.

2. Define the loop and assign current directions:

   • Assume the current flows from $V_1$ through $R_1$, $R_3$, $R_2$, and then to $V_2$.

3. Write the KVL equation around the loop: $V_1 - I \cdot R_1 - I_3 \cdot R_3 - I \cdot R_2 - V_2 = 0$

4. Since $I$ and $I_3$ can be treated as the same current due to the single loop, the equation simplifies to: $5 - (I \cdot 1 \, k\Omega) - (I \cdot 3.3 \, k\Omega) - (I \cdot 2.2 \, k\Omega) - 7 = 0$

5. Simplify the equation: $5 - 7 = I \cdot (1 + 3.3 + 2.2) \, k\Omega$ $-2 = I \cdot 6.5 \, k\Omega$ $I = \frac{-2}{6.5 \, k\Omega}$

6. Calculate $I$: $I = -0.3077 \, \text{mA}$

Thus, the value of $I_3$, which is the same as $I$, is approximately $-0.3077 \, \text{mA}$.

Would you like a detailed breakdown of each step?

Here are related questions for further exploration:

1. How does Kirchhoff's Voltage Law help in solving complex circuits?
2. What is the significance of the negative sign in the calculated current $I_3$?
3. How would the solution change if there was an additional resistor in parallel?
4. How can Ohm's law be applied to verify the voltage drops across each resistor?
5. What are the implications of changing the values of $V_1$ or $V_2$ on $I_3$?

Tip: In circuit analysis, always double-check resistor values and the polarity of voltage sources before setting up KVL equations to avoid errors.