Question

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

Answer 1

Applying KVL in loop 1, we get:
3i + 6i₁ = 4.5 ...(1)
Applying KVL in loop 2, we get:
$$\begin{gathered} \left(i-i_1\right)10+3-6i_1=0 \\ 10i-16i_1=-3...\left(2\right) \end{gathered}$$
Multiplying equation (1) by 10 and (2) by 3 and then, subtracting (2) from (1), we get:
$$\begin{gathered} -108i_1=-54 \\ \Rightarrow i_1=\frac{54}{108}=\frac{1}{2}=0.5 \end{gathered}$$
Substituting the value of i₁ in (1), we get:
$$\begin{gathered} 3i+6\times \frac{1}{2}-4.5=0 \\ 3i-1.5=0 \\ \Rightarrow i=\frac{1.5}{3}=0.5 \end{gathered}$$
So, current flowing through the 10 Ω resistor = i - i₁ = 0.5 - 0.5 = 0 A