Let's solve the first **two steps** with explanations: --- ## **Step 1: Kirchhoff’s Junction Rule** **Apply the Junction Rule at point B:** Let: - \( I_1 \) be the current through \( P \) - \( I_2 \) be the current through \( Q \) - \( I_3 \) be the current through \( R \) - \( I_4 \) be the current through \( S \) - \( I_G \) be the current through the galvanometer (between B and D) **At point B:** \[ I_1 = I_2 + I_G \] **At point D:** \[ I_3 + I_G = I_4 \] **Balanced condition:** When the galvanometer shows zero current, \( I_G = \). So, the relationships become: - \( I_1 = I_2 \) - \( I_3 = I_4 \) **Explanation:** At the balanced condition (no current through the galvanometer), the current entering node B equals the current leaving through Q, and the current entering node D equals that leaving through S. This is Kirchhoff's Junction Rule (conservation of charge). --- ## **Step 2: Kirchhoff’s Loop Rule** **Apply the Loop Rule to two loops:** - **Loop 1: A-B-D-A** (Left loop) - **Loop 2: B-C-D-B** (Right loop) Let's write the voltage equations (clockwise): ### **Loop A-B-D-A:** \[ \text{Voltage drop across } P + \text{Voltage drop across } R = \] \[ I_1 P - I_3 R = \] But since \( I_1 = I_3 \) (from Step 1, at balance), this simplifies to: \[ I_1 P = I_1 R \implies P = R \] But let's keep it general for now. ### **Loop B-C-D-B:** \[ \text{Voltage drop across } Q + \text{Voltage drop across } S = \] \[ I_2 Q - I_4 S = \] Again, since \( I_2 = I_4 \) (from Step 1, at balance), this simplifies to: \[ I_2 Q = I_2 S \implies Q = S \] But generally, we write: \[ I_2 Q = I_4 S \] **Explanation:** Kirchhoff’s Loop Rule states that the sum of potential differences (voltage) around any closed loop is zero. At balance, the voltage drop from A to B equals that from A to D, and similarly, from B to C equals from D to C. --- ## **Summary of Steps So Far:** - **Step 1:** At balanced condition, \( I_1 = I_2 \) and \( I_3 = I_4 \). - **Step 2:** The voltage drops across \( P \) and \( R \) are equal for their respective branch currents, as are the drops across \( Q \) and \( S \). --- ## **Final Answer (for Steps 1 & 2):** - **Step 1:** At balance, \( I_1 = I_2 \), \( I_3 = I_4 \) (by Junction Rule). - **Step 2:** For the loops, \( I_1 P = I_3 R \) and \( I_2 Q = I_4 S \) (by Loop Rule). At balance, these simplify further as the currents in each branch are equal. --- Let me know if you'd like to proceed with Steps 3 and 4!