请帮忙用戴维宁定理计算支路电流I5.
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请帮忙用戴维宁定理计算支路电流I5.
![](http://img.wesiedu.com/upload/e/75/e752274acb9fbd9000a3f7bbcbdfc3e5.jpg)
U=104V,电阻R1=3Ω、R2=6Ω、R3=5Ω、R4=20Ω、R5=1.5Ω,用戴维南定理计算支路电流I5.我做的过程中,将R5断开求解开路电压和内阻R0.开路电压为U0=15V,R0为R1和R3并联,R0=15/8Ω,等效电路是E=U0和电阻R0、R5的串联,计算结果为I5=40/9A,但是答案是15/8A,求高人指点做错的地方,
![](http://img.wesiedu.com/upload/e/75/e752274acb9fbd9000a3f7bbcbdfc3e5.jpg)
U=104V,电阻R1=3Ω、R2=6Ω、R3=5Ω、R4=20Ω、R5=1.5Ω,用戴维南定理计算支路电流I5.我做的过程中,将R5断开求解开路电压和内阻R0.开路电压为U0=15V,R0为R1和R3并联,R0=15/8Ω,等效电路是E=U0和电阻R0、R5的串联,计算结果为I5=40/9A,但是答案是15/8A,求高人指点做错的地方,
![请帮忙用戴维宁定理计算支路电流I5.](/uploads/image/z/1801723-67-3.jpg?t=%E8%AF%B7%E5%B8%AE%E5%BF%99%E7%94%A8%E6%88%B4%E7%BB%B4%E5%AE%81%E5%AE%9A%E7%90%86%E8%AE%A1%E7%AE%97%E6%94%AF%E8%B7%AF%E7%94%B5%E6%B5%81I5.)
![](http://img.wesiedu.com/upload/7/9f/79f7e5b9b9202d1ca6f140735fbdc5e7.jpg)
① 断开R5电阻,则开路电压: Uabo = - I2 * R2 + I1 * R1 = 15 V
其中:I1 = U / (R1 + R3) = 15 / (3 + 5) = 15/8 A
I2 = U / (R2 + R4) = 15 / (6 + 20) = 15/26 A
② 除源的等效内阻(电压源短路):
Rabo = (R2//R4) + (R1//R3) = 6//20 + 3//5 = 675/104 Ω ≈ 6.5 Ω
③ I5 = Uabo / (Rabo + R5) = 15 / (675/104 + 1.5) = 1.8773 A
也可以近似为 = 15 / (6.5 + 1.5) = 15/8 A