Calculate the current across the 8 ohm resistor

Step 1: Label nodes and define Kirchhoff's Current Law (KCL) equations. Let the bottom wire be the ground (0 \, V). Let V_A be the voltage at the node after the 2 \, resistor (R_6) and before the 8 \, (R_5) and 12 \, (R_4) resistors. Let V_B be the voltage at the node after the 8 \, resistor (R_5) and before the 6 \, resistor (R_1). Let V_C be the voltage at the node after the 12 \, resistor (R_4) and before the 4 \, resistor (R_3). Let V_D be the voltage at the node after the 6 \, resistor (R_1) and before the 3 \, resistor (R_2). The voltage source is 28 \, V. KCL at Node A: (28 \, V - V_A)/(2 \, ) = (V_A - V_B)/(8 \, ) + (V_A - V_C)/(12 \, ) Multiply by 24 (LCM of 2, 8, 12): 12(28 - V_A) = 3(V_A - V_B) + 2(V_A - V_C) 336 - 12V_A = 3V_A - 3V_B + 2V_A - 2V_C 336 = 17V_A - 3V_B - 2V_C (Equation 1) KCL at Node B: (V_A - V_B)/(8 \, ) = (V_B - V_D)/(6 \, ) Multiply by 24: 3(V_A - V_B) = 4(V_B - V_D) 3V_A - 3V_B = 4V_B - 4V_D 3V_A = 7V_B - 4V_D (Equation 2) KCL at Node C: (V_A - V_C)/(12 \, ) = (V_C - 0 \, V)/(4 \, ) Multiply by 12: V_A - V_C = 3V_C V_A = 4V_C (Equation 3) KCL at Node D: (V_B - V_D)/(6 \, ) = (V_D - 0 \, V)/(3 \, ) Multiply by 6: V_B - V_D = 2V_D V_B = 3V_D (Equation 4) Step 2: Solve the system of equations for node voltages. From Equation 3, V_C = (1)/(4)V_A. From Equation 4, V_D = (1)/(3)V_B. Substitute V_D into Equation 2: 3V_A = 7V_B - 4((1)/(3)V_B) 3V_A = 7V_B - (4)/(3)V_B 3V_A = (21V_B - 4V_B)/(3) 3V_A = (17)/(3)V_B 9V_A = 17V_B V_B = (9)/(17)V_A (Equation 5) Now substitute V_C and V_B (in terms of V_A) into Equation 1: 336 = 17V_A - 3((9)/(17)V_A) - 2((1)/(4)V_A) 336 = 17V_A - (27)/(17)V_A - (1)/(2)V_A Find a common denominator (34) for the coefficients of V_A: 336 = V_A ((17 × 34)/(34) - (27 × 2)/(34) - (1 × 17)/(34)) 336 = V_A ((578 - 54 - 17)/(34)) 336 = V_A ((507)/(34)) V_A = (336 × 34)/(507) = (11424)/(507) \, V Step 3: Calculate the current across the 8 \, resistor. The current across the 8 \, resistor (R_5) is I_R5 = (V_A - V_B)/(8 \, ). First, find the voltage difference V_A - V_B: V_A - V_B = V_A - (9)/(17)V_A = (1 - (9)/(17))V_A = (17 - 9)/(17)V_A = (8)/(17)V_A Now substitute V_A: V_A - V_B = (8)/(17) × (11424)/(507) = (8 × 11424)/(17 × 507) = (91392)/(8619) \, V Finally, calculate the current I_R5: I_R5 = (V_A - V_B)/(8) = (91392)/(8619)8 = (91392)/(8 × 8619) = (11424)/(8619) \, A Simplify the fraction by dividing by common factors. Both numerator and denominator are divisible by 3: (11424 ÷ 3)/(8619 ÷ 3) = (3808)/(2873) Both are divisible by 17: (3808 ÷ 17)/(2873 ÷ 17) = (224)/(169) The fraction cannot be simplified further as 169 = 13^2 and 224 is not divisible by 13. The current across the 8 \, resistor is (224)/(169) \, A. That's 2 down. 3 left today — send the next one.