Let's solve the problem step by step using the **branch current method** (which is essentially applying Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL)). --- ## (a) **Setting Up the Equations** Let: - \( I_1 \) = current through \( R_1 \) (left branch) - \( I_2 \) = current through \( V_5 \) (middle branch) - \( I_3 \) = current through \( R_2 \) (right branch) ### **Assigning Directions** - All currents are assigned as shown in the figure (to the right or downwards). ### **1. Kirchhoff’s Current Law (KCL) at the Top Node** At the top node: \[ I_1 + I_2 + I_3 = 0 \] However, since the currents are flowing out of the node, you may also write: \[ I_1 = I_2 + I_3 \] But for systematic analysis, let's just use KVL for independent loops. --- ### **2. Kirchhoff’s Voltage Law (KVL) for Loops** #### **Loop 1 (Left Loop: \( V_1 \to R_1 \to V_2 \to V_4 \to V_1 \))** \[ V_1 - I_1 R_1 - V_2 = V_4 \] or rearranged: \[ V_1 - I_1 R_1 - V_2 - V_4 = 0 \] #### **Loop 2 (Right Loop: \( V_3 \to R_2 \to V_5 \to V_4 \to V_3 \))** \[ V_3 - I_3 R_3 - V_5 = V_4 \] or rearranged: \[ V_3 - I_3 R_3 - V_5 - V_4 = 0 \] #### **Loop 3 (Outer Loop: \( V_1 \to R_1 \to V_2 \to R_2 \to V_3 \to V_1 \))** Here, the current in \( R_1 \) is \( I_1 \), in \( R_2 \) is \( I_3 \), the direction is important. Let's walk counter-clockwise: \[ V_1 - I_1 R_1 - V_2 - I_2 R_2 - V_3 = 0 \] But \( I_2 \) passes through \( V_5 \), not a resistor. The only resistors are \( R_1 \) and \( R_2 \). Alternatively, **use the KCL at the top node**: \[ I_1 = I_2 + I_3 \] --- ### **Minimum Unique Equations** So our three equations are: 1. \( I_1 = I_2 + I_3 \) (KCL) 2. \( V_1 - I_1 R_1 - V_2 - V_4 = 0 \) (KVL Left Loop) 3. \( V_3 - I_3 R_3 - V_5 - V_4 = 0 \) (KVL Right Loop) --- ## (b) **Solving for \( I_1, I_2, I_3 \) with Given Values** Given: \[ \begin{align*} V_1 &= 12\, \text{V} \\ V_2 &= 8\, \text{V} \\ V_3 &= 4\, \text{V} \\ V_4 &= 4\, \text{V} \\ V_5 &= 6\, \text{V} \\ R_1 &= 6\, \Omega \\ R_2 &= 3\, \Omega \\ R_3 &= 2\, \Omega \\ \end{align*} \] Let's write the equations using these values: ### **Equation 1 (KCL)** \[ I_1 = I_2 + I_3 \tag{1} \] ### **Equation 2 (KVL Left Loop)** \[ 12 - 6I_1 - 8 - 4 = 0 \implies 0 = 6I_1 \] But this simplifies to \( 0 = 6I_1 \rightarrow I_1 = 0 \). This suggests a need to re-express the loop equations to include all elements along each current's path. #### Let's re-express the three mesh currents: #### **Mesh Currents \( I_1, I_2, I_3 \):** - \( I_1 \) through \( V_1, R_1, V_2 \) - \( I_2 \) through \( V_5 \) - \( I_3 \) through \( R_2, V_3 \) But we need to account for all voltage sources and resistors each current passes through. #### **Let's assign mesh currents for the loops:** - Mesh 1 (left loop, current \( I_1 \)): \( V_1 \to R_1 \to V_2 \to \) bottom node \( \to V_4 \to \) back to \( V_1 \) - Mesh 2 (right loop, current \( I_3 \)): \( V_3 \to R_3 \to \) bottom node \( \to V_4 \to \) back to \( V_3 \) - The center branch: \( V_5 \) (vertical branch), current \( I_2 \) #### **Redraw the loops and write equations for each mesh:** ### **Left Mesh (clockwise):** \[ V_1 - I_1 R_1 - V_2 - V_4 = 0 \] \[ 12 - 6I_1 - 8 - 4 = 0 \] \[ 12 - 6I_1 - 12 = 0 \] \[ -6I_1 = 0 \implies I_1 = 0 \] ### **Right Mesh (clockwise):** \[ V_3 - I_3 R_3 - V_5 - V_4 = 0 \] \[ 4 - 2I_3 - 6 - 4 = 0 \] \[ 4 - 2I_3 - 10 = 0 \] \[ -2I_3 = 6 \implies I_3 = -3 \] ### **Current in the center branch (\( I_2 \)):** From KCL: \[ I_1 = I_2 + I_3 \] \[ 0 = I_2 + (-3) \implies I_2 = 3 \] --- ## **Final Answers** \[ \boxed{ \begin{align*} I_1 &= 0 \, \text{A} \\ I_2 &= 3 \, \text{A} \\ I_3 &= -3 \, \text{A} \end{align*} } \] - \( I_1 \) flows through \( R_1 \) (left branch): **0 A** - \( I_2 \) flows through \( V_5 \) (center branch): **3 A** (downwards) - \( I_3 \) flows through \( R_3 \) (right branch): **-3 A** (the negative sign means the actual direction is opposite to the arrow shown in the diagram) --- ### **Summary Table** | Branch | Current (A) | Direction (relative to figure) | |---------------|-------------|-------------------------------| | \( R_1 \) | 0 | rightward | | \( V_5 \) | 3 | downwards | | \( R_3 \) | -3 | rightward (actually leftward) | --- **If you need clarification on mesh current setup or further explanation, let me know!**