图中电感内阻为零,开关闭合前(t0-):
i = 3 / (10+20) = 0.1 A

u1 = 0.1 * 10 = 1 V

u2 = 0 V
u3 = 0.1 * 20 = 2 V?
u4 = 0 V
开关闭合瞬间(t0+),电感是电流不会突变,即电流源性质:
u1 = 0.1 * 10 = 1 V
R2 // R3 = 4Ω ? ? ?
u3 = u4 = ?0.1 * 4 = 0.4 V
u2 = Us - u1 - u3 = 3 - 1 - 0.4 = 1.6 V


图中电感内阻为零,开关闭合前(t0-):
i = 3 / (10+20) = 0.1 A

u1 = 0.1 * 10 = 1 V

u2 = 0 V
u3 = 0.1 * 20 = 2 V?
u4 = 0 V
开关闭合瞬间(t0+),电感是电流不会突变,即电流源性质:
u1 = 0.1 * 10 = 1 V
R2 // R3 = 4Ω ? ? ?
u3 = u4 = ?0.1 * 4 = 0.4 V
u2 = Us - u1 - u3 = 3 - 1 - 0.4 = 1.6 V
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