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ScanSolve AI Answer
Step 1: Simplify the circuit using source transformation. First, convert the 5A current source on the left with its parallel 6 resistor into an equivalent voltage source. The voltage source value will be V_L = I × R = 5A × 6 = 30V. Since the current source points downwards, the positive terminal of the voltage source will be at the bottom (towards ground), and the negative terminal will be towards node N_L. So, it contributes -30V to node N_L. This 30V source will be in series with the 6 resistor. Similarly, convert the 3A current source on the right with its parallel 4 resistor into an equivalent voltage source. The voltage source value will be V_R = I × R = 3A × 4 = 12V. Since the current source points upwards, the positive terminal of the voltage source will be towards node N_R, and the negative terminal will be at the bottom (towards ground). So, it contributes +12V to node N_R. This 12V source will be in series with the 4 resistor. The circuit now looks like this, with two main nodes N_L and N_R (assuming the bottom wire is ground 0V): From node N_L to ground: A 3 resistor in parallel with a series combination of a 6 resistor and a -30V voltage source. From node N_R to ground: A 3 resistor in parallel with a series combination of a 4 resistor and a +12V voltage source. Between nodes N_L and N_R: A 7 resistor (through which i_o flows), and a series combination of a 1 resistor and a 5V voltage source. The 5V source has its positive terminal towards N_L (through the 1 resistor) and its negative terminal towards N_R. Step 2: Apply Kirchhoff's Current Law (KCL) at nodes N_L and N_R using nodal analysis. KCL at node N_L (sum of currents leaving the node is zero): (N_L - N_R)/(7) + N_L - (N_R + 5V)1 + (N_L)/(3) + N_L - (-30V)6 = 0 (N_L - N_R)/(7) + (N_L - N_R - 5) + (N_L)/(3) + (N_L + 30)/(6) = 0 Multiply the equation by 42 (the least common multiple of 7, 1, 3, 6) to eliminate fractions: 6(N_L - N_R) + 42(N_L - N_R - 5) + 14N_L + 7(N_L + 30) = 0 6N_L - 6N_R + 42N_L - 42N_R - 210 + 14N_L + 7N_L + 210 = 0 (6 + 42 + 14 + 7)N_L + (-6 - 42)N_R = 0 69N_L - 48N_R = 0 (Equation A) KCL at node N_R (sum of currents leaving the node is zero): (N_R - N_L)/(7) + N_R - (N_L - 5V)1 + (N_R)/(3) + N_R - 12V4 = 0 (N_R - N_L)/(7) + (N_R - N_L + 5) + (N_R)/(3) + (N_R - 12)/(4) = 0 Multiply the equation by 84 (the least common multiple of 7, 1, 3, 4) to eliminate fractions: 12(N_R - N_L) + 84(N_R - N_L + 5) + 28N_R + 21(N_R - 12) = 0 12N_R - 12N_L + 84N_R - 84N_L + 420 + 28N_R + 21N_R - 252 = 0 (-12 - 84)N_L + (12 + 84 + 28 + 21)N_R + 420 - 252 = 0 -96N_L + 145N_R + 168 = 0 -96N_L + 145N_R = -168 (Equation B) Step 3: Solve the system of equations A and B. From Equation A: 69N_L = 48N_R N_L = (48)/(69)N_R = (16)/(23)N_R. Substitute the expression for N_L into Equation B: -96((16)/(23)N_R) + 145N_R = -168 -(1536)/(23)N_R + 145N_R = -168 To combine the terms with N_R, find a common denominator: (-1536N_R + 145 × 23 N_R)/(23) = -168 (-1536N_R + 3335N_R)/(23) = -168 (1799N_R)/(23) = -168 1799N_R = -168 × 23 1799N_R = -3864 N_R = -(3864)/(1799)V Now, calculate N_L: N_L = (16)/(23)N_R = (16)/(23) × (-(3864)/(1799)) N_L = -(16 × 3864)/(23 × 1799) = -(61824)/(41377)V We can simplify N_R and N_L by noting that 1799 = 7 × 257. N_R = -(3864)/(7 × 257) = -(552)/(257)V N_L = (16)/(23) × (-(552)/(257)) = -(16 × 24)/(257) = -(384)/(257)V Step 4: Calculate the current i_o through the 7 resistor. The current i_o flows from node N_L to node N_R. i_o = (N_L - N_R)/(7) i_o = (-384)/(257)V - (-(552)/(257)V)7 i_o = (-384)/(257) + (552)/(257)7A i_o = (552 - 384)/(257)7A i_o = (168)/(257)7A i_o = (168)/(257 × 7)A i_o = (24)/(257)A The current i_o is 24/257 A. 3 done, 2 left today. You're making progress.