To find the value of $i_3$ in the given circuit, we can analyze the circuit using Kirchhoff’s Voltage Law (KVL). Here’s how to solve it:

Circuit Information

• $V_1 = 5 \, \text{V}$
• $V_2 = 7 \, \text{V}$
• $R_1 = 1 \, \text{k}\Omega$
• $R_2 = 2.2 \, \text{k}\Omega$
• $R_3 = 3.3 \, \text{k}\Omega$

The current $i_3$ flows through $R_3$, and we assume current directions based on the polarity of the voltage sources.

Steps to Solution

1. Assign Loop Currents:

   • Let $i_1$ be the current flowing from $V_1$ through $R_1$.
   • Let $i_2$ be the current flowing from $V_2$ through $R_2$.
   • By Kirchhoff's Current Law (KCL), at the node where $R_1$, $R_2$, and $R_3$ meet, we have: $i_3 = i_1 - i_2$

2. Apply Kirchhoff’s Voltage Law (KVL):

   • For the left loop containing $V_1$, $R_1$, and $R_3$: $V_1 = i_1 R_1 + i_3 R_3$
   • For the right loop containing $V_2$, $R_2$, and $R_3$: $V_2 = i_2 R_2 + i_3 R_3$

3. Substitute Values:

   • Left loop: $5 = 1000 \cdot i_1 + 3300 \cdot i_3$
   • Right loop: $7 = 2200 \cdot i_2 + 3300 \cdot i_3$

4. Solve the Equations:

   • We now have two equations with three unknowns: $5 = 1000 \cdot i_1 + 3300 \cdot i_3$ $7 = 2200 \cdot i_2 + 3300 \cdot i_3$
   • Using $i_3 = i_1 - i_2$, we can substitute $i_3$ in terms of $i_1$ and $i_2$ to solve the system of equations.

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