1. **State the problem:** a) Derive the equations to find Y-connected resistor parameters ($R_a$, $R_b$, $R_c$) from a given Δ-connected resistor network ($R_1$, $R_2$, $R_3$). b) For the given network in Fig.03, determine equivalent resistance $R_T$, voltages $V_1$ and $V_4$, and currents $I_3$ and $I_5$. c) Use mesh analysis on Fig.04 to find loop currents $I_1$ and $I_2$. 2. **Deriving Y from Δ (Part a):** The Δ network resistors are $R_1$, $R_2$, $R_3$ connected in a triangle. The equivalent Y resistors $R_a$, $R_b$, $R_c$ are connected between nodes such that: $$R_a = \frac{R_2 R_3}{R_1 + R_2 + R_3}$$ $$R_b = \frac{R_1 R_3}{R_1 + R_2 + R_3}$$ $$R_c = \frac{R_1 R_2}{R_1 + R_2 + R_3}$$ These formulas ensure the same resistances between any two terminals in both Y and Δ configurations. 3. **Calculating $R_T$ for the network (Part b):** Given: $R_1 = 16\ \Omega$, $R_2 = 8\ \Omega$, $R_3 = 4\ \Omega$, $R_4 = 32\ \Omega$, $R_5 = 16\ \Omega$, and source voltage $E = 32 V$. First, calculate the equivalent resistance of $R_4$ and $R_5$ in parallel: $$R_{45} = \frac{32 \times 16}{32 + 16} = \frac{512}{48} = 10.6667\ \Omega$$ Then, $R_3 = 4\ \Omega$ is in series with $R_{45}$: $$R_{345} = R_3 + R_{45} = 4 + 10.6667 = 14.6667\ \Omega$$ Now, $R_2 = 8\ \Omega$ is in series with $R_{345}$: $$R_{2345} = R_2 + R_{345} = 8 + 14.6667 = 22.6667\ \Omega$$ Finally, $R_1 = 16\ \Omega$ is in parallel with $R_{2345}$: $$R_T = \frac{16 \times 22.6667}{16 + 22.6667} = \frac{362.667}{38.6667} = 9.38\ \Omega$$ 4. **Determine currents $I_1$, $I_5$ and voltages $V_1$, $V_4$ (Part b continues):** Total current from the source: $$I_t = \frac{E}{R_T} = \frac{32}{9.38} = 3.41\ A$$ Voltage $V_1$ across $R_1$ (since $R_1$ and $R_{2345}$ are in parallel, voltage across both is $V_1$): $$V_1 = I_t \times R_{2345} = 3.41 \times 22.6667 = 77.25\ V$$ But this is impossible since source is 32 V; error is assuming voltage across $R_1$ equals total voltage — rather, source voltage splits between parallel branches. Correct approach: Voltage across $R_1$ and $R_{2345}$ is source voltage $E = 32 V$. Therefore: Current through $R_1$ ($I_1$): $$I_1 = \frac{V}{R_1} = \frac{32}{16} = 2\ A$$ Current through $R_{2345}$: $$I_{2345} = \frac{32}{22.6667} = 1.41\ A$$ Current $I_5$ flows through $R_5$ as part of $I_{2345}$ branch. Current divides at $R_4$ and $R_5$ parallel branch: $$I_5 = I_{2345} \times \frac{R_4}{R_4 + R_5} = 1.41 \times \frac{32}{32 + 16} = 1.41 \times \frac{32}{48} = 0.94\ A$$ Voltage $V_4$ across $R_4$ and $R_5$: $$V_4 = I_5 \times R_5 = 0.94 \times 16 = 15.08 V$$ 5. **Mesh analysis for loop currents $I_1$ and $I_2$ (Part c):** Define two loops: left loop with $R_1$ and $R_2$, right loop with $R_3$, $R_4$, and $R_5$. Loop 1 equation (clockwise): $$16 I_1 + 8 (I_1 - I_2) = 32$$ Simplify: $$16 I_1 + 8 I_1 - 8 I_2 = 32$$ $$24 I_1 - 8 I_2 = 32$$ Loop 2 equation: $$4 I_2 + 32 I_2 + 16 I_2 - 8 (I_1 - I_2) = 0$$ Simplify: $$52 I_2 - 8 I_1 + 8 I_2 = 0$$ $$60 I_2 - 8 I_1 = 0$$ Rearrange: $$-8 I_1 + 60 I_2 = 0$$ 6. **Solving equations:** From second: $$8 I_1 = 60 I_2 \implies I_1 = \frac{60}{8} I_2 = 7.5 I_2$$ Substitute into first: $$24 (7.5 I_2) - 8 I_2 = 32$$ $$180 I_2 - 8 I_2 = 32$$ $$172 I_2 = 32 \implies I_2 = \frac{32}{172} = 0.186 A$$ Then: $$I_1 = 7.5 \times 0.186 = 1.395 A$$ **Final answers:** - Y resistors from Δ: $R_a=\frac{R_2 R_3}{R_1 + R_2 + R_3}$, $R_b=\frac{R_1 R_3}{R_1 + R_2 + R_3}$, $R_c=\frac{R_1 R_2}{R_1 + R_2 + R_3}$ - $R_T = 9.38\ \Omega$ - $I_t = 3.41 A$ - $I_1 = 2 A$ - $I_5 = 0.94 A$ - $V_1 = 32 V$ - $V_4 = 15.08 V$ - Loop currents by mesh: $I_1 = 1.395 A$, $I_2 = 0.186 A$