The sum of the decreases of the electric potential along the path will be 2.004V.
From the complete information, the total current(I) is given as:
= 2/(4 + 3 + 5)
= 2/12
= 0.167A
V3 = I × 3
= 0.167 × 3
= 0.501V
V4 = I × 4
= 0.167 × 4
= 0.668V
V5 = I × 5
= 0.167 × 5
= 0.835V
Therefore, the sum of the decreases of the electric potential along the path will be:
= 0.501V + 0.668V + 0.835V
= 2.004V.
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