4.11 use the mesh-current method to find the mesh current $i_a$ in the circuit shown.

Explanation:

Step1: Identify mesh currents and constraints

Define mesh currents $i_a$ (bottom-left), $i_b$ (top), and $i_c$ (bottom-right). From the current sources: $$i_b = -10\text{ A}$$ $$i_c = \frac{2v_\phi}{5}$$

Step2: Express the dependent variable

The voltage $v_\phi$ is across the $5\ \Omega$ resistor, shared by $i_a$ and $i_c$: $$v_\phi = 5(i_a - i_c)$$

Step3: Substitute $v_\phi$ into the $i_c$ equation

Substitute the expression for $v_\phi$ into the constraint for $i_c$: $$i_c = \frac{2}{5}[5(i_a - i_c)] = 2i_a - 2i_c \implies 3i_c = 2i_a \implies i_c = \frac{2}{3}i_a$$

Step4: Apply KVL to mesh $a$

Sum the voltages around the bottom-left loop: $$-75 + 2(i_a - i_b) + 5(i_a - i_c) = 0$$

Step5: Substitute known values and solve

Substitute $i_b = -10$ and $i_c = \frac{2}{3}i_a$ into the KVL equation: $$-75 + 2(i_a + 10) + 5(i_a - \frac{2}{3}i_a) = 0$$ $$-75 + 2i_a + 20 + 5(\frac{1}{3}i_a) = 0$$ $$-55 + 2i_a + \frac{5}{3}i_a = 0$$ $$\frac{11}{3}i_a = 55$$ $$i_a = \frac{55 \times 3}{11} = 15\text{ A}$$

Answer:

$i_a = 15\text{ A}$