使用戴维南定律求解电路中的4欧电阻的电流
断开4Ω电阻;
Ucd = 6v + 3Ω电阻压降 = 6 + 3 * ( 12 - 6 )/( 3 + 6 ) = 8v;
Uac = 2A * 1Ω = 2v;
Uab = Ucd + Uac = 8 + 2 = 10v;
Rab = 3//6 + 1 + 1 = 4Ω;
故接入4Ω电阻后,I = Uab/( Rab + 4Ω ) = 10/( 4 + 4 ) = 1.25A 。
上一篇:初中物理,急等
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断开4Ω电阻;
Ucd = 6v + 3Ω电阻压降 = 6 + 3 * ( 12 - 6 )/( 3 + 6 ) = 8v;
Uac = 2A * 1Ω = 2v;
Uab = Ucd + Uac = 8 + 2 = 10v;
Rab = 3//6 + 1 + 1 = 4Ω;
故接入4Ω电阻后,I = Uab/( Rab + 4Ω ) = 10/( 4 + 4 ) = 1.25A 。
上一篇:初中物理,急等