Question 3.5: For the circuit in Fig. 3.18, find the branch currents I1, ......
For the circuit in Fig. 3.18, find the branch currents I_{1}, I_{2} , and I_{3} using mesh analysis.
we first obtain the mesh currents using KVL. For mesh 1,
-15 + 5i_{1} + 10(i_{1} – i_{2}) + 10 = 0
or
3i_{1} – 2i_{2} = 1 (3.5.1)
For mesh 2,
6i_{2} + 4i_{2} + 10(i_{2} – i_{1}) – 10 = 0
or
i_{1} = 2i_{2} – 1 (3.5.2)
■ METHOD 1 Using the substitution method, we substitute Eq. (3.5.2) into Eq. (3.5.1), and write
6i_{2} – 3 – 2i_{2} = 1 ⇒ i_{2} = 1 AFrom Eq. (3.5.2), Thus i_{1} = 2i_{2} – 1 = 2- 1 = 1 A . Thus ,
I_{1} = i_{1} = 1 A , I_{2} = i_{2} = 1 A , I_{3} = i_{1} – i_{2} = 0■ METHOD 2 To use Cramer’s rule, we cast Eqs. (3.5.1) and (3.5.2) in matrix form as
\begin{bmatrix} 3 & -2\\ -1 & 2\end{bmatrix} \begin{bmatrix} i_{1} \\ i_{2} \end{bmatrix} = \begin {bmatrix} 1 \\ 1 \end {bmatrix}We obtain the determinants
Δ= \begin {vmatrix} 3 & -2 \\ -1 & 2 \end {vmatrix}= 6 – 2 = 4
Δ_{1} = \begin {vmatrix} 1 & -2 \\ 1 & 2 \end {vmatrix}= 2 + 2 = 4, Δ_{2} = \begin {vmatrix} 3 & 1 \\ -1 & 1 \end {vmatrix}= 3 + 1 = 4,
Thus,
i_{1} = \frac{Δ_{1}}{Δ} = 1 A , i_{2} = \frac{Δ_{2}}{Δ} = 1 Aas before .