用叠加定理求电路的U和I.
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用叠加定理求电路的U和I.
![](http://img.wesiedu.com/upload/c/b2/cb21a838a7f610a4009d4c0adbf4aee1.jpg)
![](http://img.wesiedu.com/upload/c/b2/cb21a838a7f610a4009d4c0adbf4aee1.jpg)
![用叠加定理求电路的U和I.](/uploads/image/z/15423684-60-4.jpg?t=%E7%94%A8%E5%8F%A0%E5%8A%A0%E5%AE%9A%E7%90%86%E6%B1%82%E7%94%B5%E8%B7%AF%E7%9A%84U%E5%92%8CI.)
(1)电压源单独作用,电流源开路,负载是不平衡电桥,两臂电阻按各自的比例对电压源分压:
U‘ = 28 * 4 / 7 - 14 = 2 V
I’ = - 28 / (4 * 7 / (4 + 7)) = - 11 A
(2)电流源单独作用,电压源短路,电路分析见下图,各支路电流是按电阻值反比例分流:
![](http://img.wesiedu.com/upload/c/19/c1972ccc631d887ae33275935be3780c.jpg)
I‘’ = 8 * 4 / 7 - 4 = 0.5714 A
U‘’ = 8 * (3 // 4 + 1) = 8 * 2.7143 =21.714 V
(3)合并:
U = U' + U'' = 23.714 V
I = I' + I'' = - 10.43 A
U‘ = 28 * 4 / 7 - 14 = 2 V
I’ = - 28 / (4 * 7 / (4 + 7)) = - 11 A
(2)电流源单独作用,电压源短路,电路分析见下图,各支路电流是按电阻值反比例分流:
![](http://img.wesiedu.com/upload/c/19/c1972ccc631d887ae33275935be3780c.jpg)
I‘’ = 8 * 4 / 7 - 4 = 0.5714 A
U‘’ = 8 * (3 // 4 + 1) = 8 * 2.7143 =21.714 V
(3)合并:
U = U' + U'' = 23.714 V
I = I' + I'' = - 10.43 A