Let’s solve the mesh equation for **mesh 2** step by step: ### **Step 1: Identify mesh 2 and the elements in its loop** Mesh 2 (right loop) contains: - \(15\,\Omega\) (top right) - \(5\,\Omega\) (bottom right) - \(10\,\Omega\) (middle, shared with mesh 1) - Voltage source \(40\,V\) (right side) Mesh currents: - \(i_2\) (mesh 2, clockwise) - \(i_1\) (mesh 1, left mesh, also clockwise) ### **Step 2: Assign polarities and write KVL for mesh 2** Start at the bottom and go clockwise: #### **Bottom resistor (5Ω):** Voltage drop: \(5i_2\) #### **Right voltage source (40V):** Going from - to + (bottom to top in loop direction): **Voltage gain**: \(+40\) #### **Top resistor (15Ω):** Voltage drop: \(15i_2\) #### **Middle resistor (10Ω, shared):** - Current through this resistor is: \(i_2 - i_1\) (since \(i_2\) goes right, \(i_1\) goes left). - Voltage drop: \(10(i_2 - i_1)\) ### **Step 3: Write the KVL equation** Sum of voltage drops should be zero: \[ 10(i_2 - i_1) + 15i_2 + 40 + 5i_2 = 0 \] ### **Step 4: Rearrange the equation** Combine like terms: \[ 10(i_2 - i_1) + 15i_2 + 5i_2 + 40 = 0 \\ 10i_2 - 10i_1 + 15i_2 + 5i_2 + 40 = 0 \\ (10i_2 + 15i_2 + 5i_2) - 10i_1 + 40 = 0 \\ 30i_2 - 10i_1 + 40 = 0 \] But **the equation as written in the options** is not simplified, so let's check the form: \[ 10(i_2 - i_1) + 15i_2 + 40 + 5i_2 = 0 \] ### **Step 5: Match to options** Option **D** is: \[ 10(i_2 - i_1) + 15i_2 + 40 + 5i_2 = 0 \] #### **This matches exactly!** --- ## **Final Answer:** \[ \boxed{D} \] **10(i₂ - i₁) + 15i₂ + 40 + 5i₂ = 0**