Question

No. 2 Find the current throngh cach resistor in the circuit shown below: Note: "R" is the lowest value resistor, and all others are muliples of that value as t indicated.) (25 Points) 15 A

Answer 1

feel free to ask doubts.

Given circuit KCL at Node: - I_1 + I_2 + I_3 + I_4 + I_5 = 15 V/R + V/2R + V/4R + V/6R + V/8R = 15 V/R [1 + 1/2 + 1/4 + 1/6 + 1/8] = 15 V/R [24 + 12 + 6 + 4 + 3/24] = 15 V/R [49/24] = 15 doublerightarrow V/R = 15 times 24/49 V/R = 7.34 doublerightarrow V = 7.34 R I_1 = V/R = 7.34 doublerightarrow I_1 = 7.34 A I_2 = V/2R = 7.34/2 doublerightarrow I_2 = 3.67 A I_3 = V/4R = 7.34/4 doublerightarrow I_3 = 1.83 A I_4 = V/6R doublerightarrow I_4 = 1.22 A I_5 = V/8R = 7.34/8 doublerightarrow I_5 = 0.91 A