Question

Find the current through the 10 Ω resistor shown in the figure (32-E14).
Figure 32-E14

Answer 1

Answer:

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Explanation:

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Answer 2

Explanation:

In loop 1, when KVL is applied, we obtain:

6i1 + 3i = 4.5    …(1)

In loop 2, when KVL is applied, we obtain:  

I - i110 + 3 - 6i1 = 010i - 16i1 = -3    …(2)

When equation (1) is multiplied by 10 and equation is (2) multiplied by 3 and after subtracting equation (2) from equation (1), we obtain:

-108i1 = -54 ⇒i1 =0.5

When the the value of i1 is substituted in (1), we obtain:

3i + - 4.5 + 6 × 12 = 03i - 1.5 = 0

⇒ I = 1.53 =0.5

Thus, via the 10 Ω resistor, the flowing current = i – i1 = 0 A (0.5 – 0.5).