Find the current in the 5Ω resistor from the given circuit using Mesh analysis.

Steps to Solve

1. Identify Mesh Currents Assume mesh currents for the two loops in the circuit:

• Let $I_1$ be the current in the left mesh (25V source).
• Let $I_2$ be the current in the right mesh (45V source).

2. Apply Kirchhoff's Voltage Law (KVL) to the First Loop For the left mesh: $$ -25V + 5Ω(I_1) + 6Ω(I_1 - I_2) = 0 $$ Rearranging gives: $$ 11I_1 - 6I_2 = 25 \quad (1) $$

3. Apply KVL to the Second Loop For the right mesh: $$ -45V + 6Ω(I_2 - I_1) + 4Ω(I_2) = 0 $$ Rearranging gives: $$ -6I_1 + 10I_2 = 45 \quad (2) $$

4. Solve the System of Equations You now have two equations:

5. $11I_1 - 6I_2 = 25$

6. $-6I_1 + 10I_2 = 45$

To solve, isolate $I_2$ in either equation and substitute: From (1): $$ I_2 = \frac{11I_1 - 25}{6} \quad (3) $$

Substituting into (2): $$ -6I_1 + 10\left(\frac{11I_1 - 25}{6}\right) = 45 $$

Now clear the fraction by multiplying through by 6: $$ -36I_1 + 110I_1 - 250 = 270 $$

Rearranging gives: $$ 74I_1 = 520 $$ Thus, $$ I_1 = \frac{520}{74} \approx 7.03A $$

5. Substitute Back to Find $I_2$ Using equation (3): $$ I_2 = \frac{11(7.03) - 25}{6} \approx 3.76A $$

6. Find the Current through the 5Ω Resistor The current through the 5Ω resistor is simply $I_1$, which we found to be approximately: $$ I = I_1 \approx 7.03A $$