Question

用戴维宁定理和诺顿定理分别计算R1上的电流。 R1=9，R2=4，R3=2， R4=2

Answer 1

Uab = 10 - 2 * 4 = 2V

Rab = R2 = 4Ω

I1 = 2 / (4 + 9) = 2/13 A

用线性叠加:

1 电流源开路，U=10V求出 R1 电流 I1U = 1

2 电压源短路，IS=2A 求出 R1 电流 I1I = 1

3 叠加: I1 = I1U + I1I = 2A

Answer 2

Uab = 10 - 2 * 4 = 2V

Rab = R2 = 4Ω

I1 = 2 / (4 + 9) = 2/13 A

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