from mesh (1) 2I_1 - I_2 - 4I_2 + 3I_1 = 10 - 5 5I_1 - 5I_2 = 5 I_1 - I_2 = 1 (A

Question

from mesh (1) 2I_1 - I_2 - 4I_2 + 3I_1 = 10 - 5 5I_1 - 5I_2 = 5 I_1 - I_2 = 1 (A) from Mesh (2) I_2 + 10I_3 + 4I_2 = 5 5 I_2 + 10 I_3 = 5 from (A) I_3 = I_1 + I_2 5I_2 + 10 (I_1 + I_2) = 5 10I_1 + 15 I_2 = 5 2I_1 + 3I_2 = 1 (B) from (B) & (B) 3 times I_1 - I_2 = 1 1 times 2_I + 3 I_2 = 1 copper branch 5I_1 =4 I_1 = 0.8A middle branch I_2 = I_1 - 1 = 0.8 - 1 = 0.2 A lower branch I_3 = I_1 + I_2 = 0.8 - 0.2 = 0.6 A V_av =10 + 2(0.8) - 0.2 - 5 = 6.4 V In the circuit shown in the figure (Figure 1), find the magnitude of current in the upper branch. Find the magnitude of current in the middle branch. Find the magnitude of current in the lower branch. What is the potential difference V_ab of point a relative to point b?

Explanation / Answer

Here ,

part D) for the voltage across the points ab

Vab = R3 * current in the lower branch

Vab = 10 * 0.60

Vab = 6 V

the current in the potential ab is 6 V

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