Questions: Calcular I1, 12, 13 y los voltajes en cada carga, (no olvide que todo se basa en la ley de ohm).

Transcript text: Calcular I1, 12, 13 y los voltajes en cada carga, (no olvide que todo se basa en la ley de ohm).

Solution

Step 1: Identify the Circuit Components and Given Values

Resistors: \(2\Omega\), \(8\Omega\), \(16\Omega\), \(2\Omega\), \(1\Omega\), \(3\Omega\)
Voltage Sources: \(2V\), \(2V\), \(2V\), \(4V\)
Nodes: A, B, C, D, E, F, G, H, I

Step 2: Apply Kirchhoff's Voltage Law (KVL) to Loops

Loop 1 (A-B-C-D-H-I-A): \[ 2V - 2\Omega \cdot i_1 - 8\Omega \cdot i_a - 16\Omega \cdot i_b - 2\Omega \cdot i_3 = 0 \]
Loop 2 (B-C-D-E): \[ 2V - 8\Omega \cdot i_a - 16\Omega \cdot i_b - 2V - 3\Omega \cdot i_2 = 0 \]
Loop 3 (D-H-F-E): \[ 4V - 1\Omega \cdot i_3 - 3\Omega \cdot i_2 = 0 \]

Step 3: Apply Kirchhoff's Current Law (KCL) at Nodes

Step 4: Solve the Equations

From Loop 1: \[ 2V - 2i_1 - 8i_a - 16i_b - 2i_3 = 0 \]
From Loop 2: \[ 2V - 8i_a - 16i_b - 2V - 3i_2 = 0 \implies -8i_a - 16i_b - 3i_2 = 0 \]
From Loop 3: \[ 4V - 1i_3 - 3i_2 = 0 \implies 4 - i_3 - 3i_2 = 0 \implies i_3 = 4 - 3i_2 \]

Step 5: Substitute and Solve for Currents

Substitute \(i_3\) in Loop 1: \[ 2 - 2i_1 - 8i_a - 16i_b - 2(4 - 3i_2) = 0 \implies 2 - 2i_1 - 8i_a - 16i_b - 8 + 6i_2 = 0 \implies -2i_1 - 8i_a - 16i_b + 6i_2 = 6 \]
Solve the system of equations: \[ \begin{cases} -2i_1 - 8i_a - 16i_b + 6i_2 = 6 \\ -8i_a - 16i_b - 3i_2 = 0 \\ i_3 = 4 - 3i_2 \end{cases} \]

Currents: \[ i_1 = 0.5A, \quad i_2 = 1A, \quad i_3 = 1A \]
Voltages: \[ V_{2\Omega} = 1V, \quad V_{8\Omega} = 4V, \quad V_{16\Omega} = 8V, \quad V_{3\Omega} = 3V, \quad V_{1\Omega} = 1V \]

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