(e) By the loop rule (proceeding clockwise), 30V – i4(1.5 Ω) – i5(2.0 Ω) = 0 readily yields i5= 7.5 A. 82. (a) The four resistors R1, R2, R3and R4on the left reduce to 3412eq12347.03.010RRRRRRR RR=+=+=Ω+Ω=Ω++With 30 Vε=across Reqthe current there is i2= 3.0 A. (b) The three resistors on the right reduce to
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