To solve for $I_1$ and $I_2$, we need to set up the mesh current equations for each loop based on the given circuit. Let's label the loops and apply Kirchhoff's Voltage Law (KVL) to each loop.

Given Data

• $V_1 = 10 \, \text{V}$
• $R_1 = 20 \, \Omega$
• $R_2 = 27 \, \Omega$
• $R_3 = 15 \, \Omega$
• $R_4 = 5 \, \Omega$
• $V_2 = 0 \, \text{V}$ (assuming it's a placeholder for zero voltage drop or a reference)

Loop Equations

1. Loop 1 (with $I_1$ as the loop current):

   • KVL for Loop 1: $V_1 - I_1 R_1 - (I_1 - I_2) R_2 = 0$
   • Substituting the values: $10 - 20I_1 - 27(I_1 - I_2) = 0$ Simplify this equation: $10 - 20I_1 - 27I_1 + 27I_2 = 0$ $-47I_1 + 27I_2 = -10$ Rearrange to get the first equation: $47I_1 - 27I_2 = 10$

2. Loop 2 (with $I_2$ as the loop current):

   • KVL for Loop 2: $-(I_2 - I_1) R_2 - I_2 R_3 - I_2 R_4 = 0$
   • Substituting the values: $-27(I_2 - I_1) - 15I_2 - 5I_2 = 0$ Simplify this equation: $-27I_2 + 27I_1 - 15I_2 - 5I_2 = 0$ $27I_1 - 47I_2 = 0$

System of Equations

Now we have the following two equations:

1. $47I_1 - 27I_2 = 10$
2. $27I_1 - 47I_2 = 0$

We can solve this system to find $I_1$ and $I_2$.

Solving the Equations

Let's solve these equations:

1. From the second equation, solve for $I_1$: $I_1 = \frac{47}{27} I_2$
2. Substitute $I_1 = \frac{47}{27} I_2$ into the first equation: $47 \left(\frac{47}{27} I_2\right) - 27I_2 = 10$ Simplifying further will provide the values of $I_1$ and $I_2$.

I'll proceed with the calculations to obtain the final values.The solution to the system of equations gives us:

• $I_1 = \frac{47}{148} \approx 0.3176 \, \text{A}$
• $I_2 = \frac{27}{148} \approx 0.1824 \, \text{A}$

So, the mesh currents are approximately:

• $I_1 = 0.3176 \, \text{A}$
• $I_2 = 0.1824 \, \text{A}$

Would you like further details on the calculations, or do you have any additional questions?

Here are five related questions for further exploration:

1. How would the values of $I_1$ and $I_2$ change if $V_2$ were non-zero?
2. What impact would altering $R_1$ or $R_3$ have on the current distribution?
3. Can we find the power dissipated in each resistor based on these current values?
4. How does the mesh current analysis differ when using a different orientation or direction for $I_1$ and $I_2$?
5. Could we apply nodal analysis to solve for $I_1$ and $I_2$ instead of mesh analysis?

Tip: Always double-check the signs when applying Kirchhoff’s laws, as they can significantly affect the results.