Using thevenin s theorem through direct formula clearly step by step please

Okay, let's re-evaluate the problem using Thevenin's theorem with a direct, step-by-step formula application. We are referring to the circuit diagram from the previous question (Q1b). Goal: Determine the current, p.d., and power across the 10 load resistance using Thevenin's theorem. Step 1: Calculate the Thevenin Resistance (R_Th). To find R_Th, we first remove the load resistor (the 10 resistor between terminals A and B). Then, we turn off all independent voltage sources (replace them with short circuits) and independent current sources (replace them with open circuits). Replace the 6V voltage source with a short circuit. This connects node C directly to node D. Replace the 9V voltage source with a short circuit. This connects node F directly to node D. Now, nodes C, D, and F are all electrically connected to the same point (let's call it ground for simplicity). Looking into terminals A and B: The 4 resistor is connected between terminal A and node C (which is now ground). The 6 resistor is connected between terminal B and node D (which is now ground). The 5 resistor is connected between node C and node F. Since C and F are both connected to D (ground), the 5 resistor is connected between two points that are at the same potential. This means the 5 resistor is shorted out and does not contribute to R_Th between A and B. The effective path between A and B is through the 4 resistor to ground (C/D/F) and then from ground (C/D/F) through the 6 resistor to B. Therefore, the 4 and 6 resistors are in series when viewed from terminals A and B. R_Th = R_4 + R_6 R_Th = 4 \, + 6 \, R_Th = 10 \, Step 2: Calculate the Thevenin Voltage (V_Th). To find V_Th, we determine the open-circuit voltage across the terminals where the load resistor was removed (terminals A and B). This is V_AB. With the 10 load removed, no current flows through the 4 resistor (connected to A) or the 6 resistor (connected to B) because the circuit is open at A-B. Let's set node D as the reference node (V_D = 0V). The 6V source is connected between C and D, with C being positive. So, V_C = 6V. The 9V source is connected between F and D, with F being positive. So, V_F = 9V. The current flowing through the 5 resistor (between C and F) is I_CF: I_CF = (V_C - V_F)/(5 \, ) = (6V - 9V)/(5 \, ) = (-3V)/(5 \, ) = -0.6A This means 0.6A flows from F to C. Since no current flows through the 4 resistor, there is no voltage drop across it. Thus, the voltage at terminal A is the same as the voltage at node C: V_A = V_C = 6V Since no current flows through the 6 resistor, there is no voltage drop across it. Thus, the voltage at terminal B is the same as the voltage at node D: V_B = V_D = 0V The Thevenin voltage V_Th is the potential difference between A and B: V_Th = V_A - V_B = 6V - 0V V_Th = 6V Step 3: Connect the Thevenin equivalent circuit to the load and calculate the current (I_L). The Thevenin equivalent circuit consists of V_Th in series with R_Th. Now, reconnect the 10 load resistor (R_L) to this equivalent circuit. The total resistance in the series circuit is R_total = R_Th + R_L. R_total = 10 \, + 10 \, = 20 \, The current flowing through the load resistor is given by Ohm's Law: I_L = V_ThR_total I_L = (6V)/(20 \, ) I_L = 0.3A The current across the 10 resistance is 0.3 A. Step 4: Calculate the potential difference (p.d.) across the load (V_L). Using Ohm's Law for the load resistor: V_L = I_L × R_L V_L = 0.3A × 10 \, V_L = 3V The p.d across the 10 resistance is 3 V. Step 5: Calculate the power consumed by the load (P_L). Using the formula for power: P_L = I_L^2 × R_L P_L = (0.3A)^2 × 10 \, P_L = 0.09 A^2 × 10 \, P_L = 0.9W The power consumed by the 10 resistance is 0.9 W. 3 done, 2 left today. You're making progress.