Solve for I1 and I2 using the equations in the image.
Understand the Problem
The question involves circuit analysis using Kirchhoff's Voltage Law (KVL) to determine loop currents I1 and I2. You've established equations based on KVL around different loops and you're solving for the current values.
Answer
$I_1 \approx 0.15 \, \text{A}$ $I_2 \approx -0.53 \, \text{A}$
Answer for screen readers
$I_1 \approx 0.15 , \text{A}$ $I_2 \approx -0.53 , \text{A}$
Steps to Solve
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Write down the first equation. From KVL in the first loop: $6 - 8I_1 - 7(I_1 - I_2) = 0$ Simplifying gives: $6 - 8I_1 - 7I_1 + 7I_2 = 0$ $15I_1 - 7I_2 = 6$
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Write down the second equation. The equation from the second loop is: $7(I_1 - I_2) + 9I_2 = 0$ Simplifying yields: $7I_1 - 7I_2 + 9I_2 = 0$ $7I_1 + 2I_2 = 0$
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Solve for $I_1$ in the second equation $7I_1 = -2I_2$ $I_1 = -\frac{2}{7}I_2$
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Substitute $I_1$ into the first equation $15(-\frac{2}{7}I_2) - 7I_2 = 6$ $-\frac{30}{7}I_2 - 7I_2 = 6$ $-\frac{30}{7}I_2 - \frac{49}{7}I_2 = 6$ $-\frac{79}{7}I_2 = 6$
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Solve for $I_2$ $I_2 = -\frac{42}{79} \approx -0.53$
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Solve for $I_1$ using the value of $I_2$ $I_1 = -\frac{2}{7}I_2 = -\frac{2}{7}(-\frac{42}{79}) = \frac{84}{553} = \frac{12}{79} \approx 0.15$
$I_1 \approx 0.15 , \text{A}$ $I_2 \approx -0.53 , \text{A}$
More Information
The currents $I_1$ and $I_2$ are loop currents, circulating in the loops defined in the circuit. $I_2$ is negative, which means the current is flowing in the opposite direction to what was initially assumed.
Tips
A common mistake is making sign errors when applying KVL around the loops. Another common mistake is to make errors when solving the system of equations.
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