Lab 216 Part III

# Lab 216 Part III - Figure 2 Part III(B R0 5600 = R1 5600 =...

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Part III(A) The current in entering node A is ( 29 ( 29 -3 -3 0 4 i +i = 1.071 x 10 0 1.071 x 10 + = And the current leaving node A is -3 2 1.071 x 10 i = The voltage across AB is zero because the current across AB is zero. Figure 1 Given i1 1 = i3 1 = i0 1 = i2 1 = i4 1 = i0 i2 - i4 + 0 i1 i3 - i4 - 0 R0 i0 R1 i1 - R4 i4 - 0 R2 i2 R3 i3 - R4 i4 + 0 R1 i1 R3 i3 + v - 0 current Find i0 i1 , i2 , i3 , i4 , ( ) = current 1.071 10 3 - × 6 10 4 - × 1.071 10 3 - × 6 10 4 - × 0 = v 12 = R0 5600 = R1 10000 = R2 5600 = R3 10000 = R4 1000 =

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The current in entering node A is ( 29 ( 29 -4 -4 0 4 i +i = 7.692 x 10 0 7.692 x 10 + = And the current leaving node A is -4 2 7.692 x 10 i = The voltage across AB is zero because the current across AB is zero.
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Unformatted text preview: Figure 2 Part III(B) R0 5600 = R1 5600 = R2 1000 = R3 1000 = R4 1000 = v 12 = Given current Find i0 i1 , i2 , i3 , i4 , ( ) = i0 i2-i4 + i1 i3-i4-R0 i0 ⋅ R1 i1 ⋅-R4 i4 ⋅-R2 i2 ⋅ R3 i3 ⋅-R4 i4 ⋅ + R1 i1 ⋅ R3 i3 ⋅ + v-current 7.692 10 4-× 7.692 10 4-× 7.692 10 4-× 7.692 10 4-× = i1 1 = i3 1 = i0 1 = i2 1 = i4 1 =...
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## This note was uploaded on 11/26/2010 for the course PHYSICS 121 taught by Professor Fayngold during the Spring '10 term at NJIT.

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Lab 216 Part III - Figure 2 Part III(B R0 5600 = R1 5600 =...

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