Solve in English

Let's define two clockwise mesh currents:

• $i_1$ for the left loop.

• $i_2$ for the right loop.

• KVL for Mesh 1 (Left Loop): Starting from the bottom left corner and moving clockwise, we sum the voltage changes:

  $$+10\text{V} - (3\Omega)i_1 - (5\Omega)(i_1 - i_2) = 0$$ $$10 - 3i_1 - 5i_1 + 5i_2 = 0$$ $$10 - 8i_1 + 5i_2 = 0$$ $$8i_1 - 5i_2 = 10 \quad \text{(Equation 1)}$$

• KVL for Mesh 2 (Right Loop): Starting from the bottom right corner and moving clockwise, we sum the voltage changes:

  $$-(9\Omega)i_2 - 15\text{V} - (5\Omega)(i_2 - i_1) = 0$$ $$-9i_2 - 15 - 5i_2 + 5i_1 = 0$$ $$5i_1 - 14i_2 = 15 \quad \text{(Equation 2)}$$

We have the following system of two linear equations:

1. $8i_1 - 5i_2 = 10$
2. $5i_1 - 14i_2 = 15$

To solve for $i_1$ and $i_2$, we can use the elimination method. Multiply Equation (1) by 14 and Equation (2) by 5 to eliminate $i_2$:

Multiply Equation (1) by 14:

$$14(8i_1 - 5i_2) = 14(10)$$ $$112i_1 - 70i_2 = 140 \quad \text{(Equation 3)}$$

Multiply Equation (2) by 5:

$$5(5i_1 - 14i_2) = 5(15)$$ $$25i_1 - 70i_2 = 75 \quad \text{(Equation 4)}$$

Subtract Equation (4) from Equation (3):

$$(112i_1 - 70i_2) - (25i_1 - 70i_2) = 140 - 75$$ $$112i_1 - 25i_1 = 65$$ $$87i_1 = 65$$ $$i_1 = \frac{65}{87} \text{ A}$$

Now substitute the value of $i_1$ back into Equation (1) to find $i_2$:

$$8\left(\frac{65}{87}\right) - 5i_2 = 10$$ $$\frac{520}{87} - 5i_2 = 10$$ $$-5i_2 = 10 - \frac{520}{87}$$ $$-5i_2 = \frac{870 - 520}{87}$$ $$-5i_2 = \frac{350}{87}$$ $$i_2 = \frac{350}{87 \times (-5)}$$ $$i_2 = \frac{-70}{87} \text{ A}$$

The current $I_1$ is indicated in the diagram as flowing downwards through the $9\Omega$ resistor. Our mesh current $i_2$ is defined as flowing clockwise through the right loop, which means it flows downwards through the $9\Omega$ resistor.
Therefore, $I_1 = i_2$.

$$I_1 = \frac{-70}{87} \text{ A}$$

As a decimal, $I_1 \approx -0.804597... \text{ A}$.

Rounding to two decimal places:

$$I_1 = \boxed{-0.80 \text{ A}}$$