Question

We name currents I_1, I_2, and I_3 as shown. From Kirchhoff's current rule, I_3 = I_1 + I_2. Applying Kirchhoff's voltage rule to the loop containing I_2 and I_3. 12.0 V - (4.00)I_3 - (6.00)I_2 - 4.00 V = 0 8.00 = (4.00)I_3 + (6.00)I_2 Applying Kirchhoff's voltage rule to the loop containing I_1 and I_2. -(6.00)I_2 - 4.00 V + (8.00)I_1 = 0 (8.00)_I_1 = 4.00 + (6.00)I_2. Solving the above linear system, we proceed to the pair of simultaneous equations: {8 - 4I_1 + 4I_2 + 6I_2 8I_1 = 4 + 6I_2 or {8 = 4I_1 + 10I_2 I_2 = 1.331I_1 - 0.667 and to the single equation 8 = 4I_1 + 13.3 I_1 - 6.67 I_1 = 14.7 V/17.3 Ohm = 0.846 A. Then I_2 = 1.33 (0.846 A) - 0.667 and I_3 = I_1 + I_2 give I_1 = 846 mA, I_2 = 462 mA, I_3 = 1.31 A All currents are in the directions indicated by the arrows in the circuit diagram.

Answer 1

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