Question:

Please convert the below content to rendered equations without any changes and errors. We are told that mesh analysis leads to the following equations: [ a i_1 - 20 i_2 = b ] [ -20 i_1 + 35 i_2 = -15 ] and we must **find the value of (b)**. --- # 1. Identify Mesh Currents From the circuit there are three meshes: * (i_1): bottom-left loop * (i_2): bottom-right loop * (i_3): top loop All mesh currents are **counter-clockwise**. --- # 2. Important Shared Elements ### Between mesh (i_1) and (i_2) Shared element: [ 20\Omega ] So the voltage drop is: [ 20(i_1-i_2) ] This produces the **−20 coupling term** in the equation. --- ### Between mesh (i_1) and (i_3) Shared resistor: [ R_v ] But the problem already simplifies the equations to remove (i_3). --- # 3. Apply KVL to Mesh (i_1) Traverse mesh (i_1) counter-clockwise. Elements encountered: 1. (R_v) 2. (20\Omega) 3. (10V) source 4. (5V) source --- ## Voltage drops Across (R_v): [ R_v(i_1-i_3) ] Across (20\Omega): [ 20(i_1-i_2) ] --- ## Voltage sources Left source: [ +10V ] Center source: Going **top → bottom** [ -5V ] --- ### KVL [ R_v(i_1-i_3)+20(i_1-i_2)-10+5=0 ] Simplify source terms [ -10+5=-5 ] Thus [ R_v(i_1-i_3)+20(i_1-i_2)=5 ] --- # 4. Eliminate (i_3) Look at the **top mesh**. The top loop contains a **2A current source**. Therefore the mesh current is fixed: [ i_3 = 2A ] --- Substitute into mesh-1 equation. [ R_v(i_1-2)+20(i_1-i_2)=5 ] Expand: [ R_v i_1 -2R_v +20i_1 -20i_2 =5 ] Group terms: [ (R_v+20)i_1 -20i_2 =5+2R_v ] --- # 5. Compare With Given Equation Given form: [ a i_1 -20 i_2 = b ] Thus [ a = R_v+20 ] [ b = 5 + 2R_v ] --- # 6. Given in Problem The question states [ R_v=[Rv]\Omega ] which means the value used in the exam gives the final constant. From the provided equation set [ -20i_1 +35i_2=-15 ] This confirms that [ R_v = 15\Omega ] (because (R_v+20 = 35)). --- # 7. Compute (b) [ b = 5 + 2R_v ] Substitute [ R_v=15 ] [ b = 5 + 2(15) ] [ b = 5 + 30 ] [ b = 35 ] --- # ✅ Final Answer [ \boxed{b = 35} ] --- If you'd like, I can also show a **much faster 30-second trick professors expect for this type of exam problem** (avoids writing the full mesh equations).