Page 3 - 6- اجابات نيوتن

‫إجابات‬
‫الفصل‬
‫الأول‬

2VB1  54 VB1  27 V VB2  3V
VB1  27  9
VB2 3

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Vab = 10 + (42) + 17 = 25 V
(34) - (17) = VB
VB = 5 V

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I1 + I2 - I3 = 0 (1)
30 I1 + 0 I2 + 30 I3 = 40 (2)
0 I1 + 10 I2 + 30 I3 = 20 (3)

I1  2 A I2 = zero I3  2 A
3 3

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12

6

2A 1A

PWX  I2 X  112  12  3
PWY I2 Y 44 16 4

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6 6 , 2

4 8 , 8

I\  VB  20  4A
R\ 5

PW = I2 R = (2)2  6 = 24 watt

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R = 10 

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