1. **State the problem:** We have a resistor network with resistors $R_1=4\Omega$, $R_2=2\Omega$, $R_3=5\Omega$, $R_4=2\Omega$, voltage sources $E_1=27V$, $E_2=5V$, and current sources $I_1=8A$, $I_2=5A$. We want to find the loop currents $i_1, i_2, i_3, i_4$ with $i_1$ given as 8A. 2. **Known:** - $i_1 = I_1 = 8A$ (given) - We need to find $i_2, i_3, i_4$. 3. **Set up equations using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL):** - Loop 1 current $i_1$ is known. - For loops 2, 3, and 4, write KVL equations considering voltage drops across resistors and voltage sources. 4. **Write the system of linear equations:** Using mesh analysis, the equations for loops 2, 3, and 4 are: $$\begin{cases} R_2(i_2 - i_4) + R_4(i_2 - i_3) + E_2 = 0 \\ R_3(i_3 - i_2) + R_4(i_3 - i_2) - E_2 = 0 \\ R_1(i_4 - i_1) + R_2(i_4 - i_2) - E_1 = 0 \end{cases}$$ Substitute values: $$\begin{cases} 2(i_2 - i_4) + 2(i_2 - i_3) + 5 = 0 \\ 5(i_3 - i_2) + 2(i_3 - i_2) - 5 = 0 \\ 4(i_4 - 8) + 2(i_4 - i_2) - 27 = 0 \end{cases}$$ 5. **Simplify each equation:** Equation 1: $$2i_2 - 2i_4 + 2i_2 - 2i_3 + 5 = 0 \Rightarrow 4i_2 - 2i_3 - 2i_4 = -5$$ Equation 2: $$5i_3 - 5i_2 + 2i_3 - 2i_2 - 5 = 0 \Rightarrow 7i_3 - 7i_2 = 5$$ Equation 3: $$4i_4 - 32 + 2i_4 - 2i_2 - 27 = 0 \Rightarrow 6i_4 - 2i_2 = 59$$ 6. **Rewrite system:** $$\begin{cases} 4i_2 - 2i_3 - 2i_4 = -5 \\ -7i_2 + 7i_3 = 5 \\ -2i_2 + 6i_4 = 59 \end{cases}$$ 7. **Solve for $i_2, i_3, i_4$:** From second equation: $$7i_3 = 7i_2 + 5 \Rightarrow i_3 = i_2 + \frac{5}{7}$$ Substitute $i_3$ into first equation: $$4i_2 - 2\left(i_2 + \frac{5}{7}\right) - 2i_4 = -5$$ Simplify: $$4i_2 - 2i_2 - \frac{10}{7} - 2i_4 = -5 \Rightarrow 2i_2 - 2i_4 = -5 + \frac{10}{7} = -\frac{25}{7}$$ Divide both sides by 2: $$\cancel{2}i_2 - \cancel{2}i_4 = -\frac{25}{7} \div 2 = -\frac{25}{14}$$ So: $$i_2 - i_4 = -\frac{25}{14}$$ From third equation: $$-2i_2 + 6i_4 = 59$$ Divide by 2: $$-i_2 + 3i_4 = \frac{59}{2}$$ Rewrite as: $$-i_2 + 3i_4 = 29.5$$ 8. **Solve the system:** From $i_2 - i_4 = -\frac{25}{14}$, we get: $$i_2 = i_4 - \frac{25}{14}$$ Substitute into $-i_2 + 3i_4 = 29.5$: $$-\left(i_4 - \frac{25}{14}\right) + 3i_4 = 29.5$$ Simplify: $$-i_4 + \frac{25}{14} + 3i_4 = 29.5 \Rightarrow 2i_4 + \frac{25}{14} = 29.5$$ Subtract $\frac{25}{14}$: $$2i_4 = 29.5 - \frac{25}{14} = \frac{59}{2} - \frac{25}{14} = \frac{413}{14}$$ Divide by 2: $$i_4 = \frac{413}{28} \approx 14.75 A$$ 9. **Find $i_2$:** $$i_2 = 14.75 - \frac{25}{14} = 14.75 - 1.79 = 12.96 A$$ 10. **Find $i_3$:** $$i_3 = i_2 + \frac{5}{7} = 12.96 + 0.714 = 13.67 A$$ **Final answers:** $$i_1 = 8 A$$ $$i_2 \approx 12.96 A$$ $$i_3 \approx 13.67 A$$ $$i_4 \approx 14.75 A$$