电路分析基础(总复习)(8)

解：由KCL i1 = iS – i2 = 1A i 1 3Ω 故电压 u = 3 i1 + uS = 3+5 = 8(V) uS u 5V 电阻 R = u / i2 = 8/1 = 8Ω i 产生的功率 P1 = u iS = 8×2 = 16 (W) uS产生的功率 P2 = - u i1 = - 5×1 = - 5 (W)S

i2 iS 2A R