Question

Consider the two-loop circuit shown in the figure below. The currents I1 and I2 (in amp) satisfy the following system of equations: 16I1 - 9I2 = 100 20I2 - 9I1 + 110 = 0 Calculate I1 and I2. The unit of current is Amp (A). I1 = - 2.25A and I2 = 1.60A I1 = 3.22A and I2 = -2.60A I1 = 4.22A and I2 = -3.60A I1 = -5.55A and I2 = 4.60A

          Consider the two-loop circuit shown in the figure below. The currents I1 and I2 (in amp) satisfy the following system of equations:
16I1 - 9I2 = 100
20I2 - 9I1 + 110 = 0
Calculate I1 and I2. The unit of current is Amp (A).
I1 = - 2.25A and I2 = 1.60A
I1 = 3.22A and I2 = -2.60A
I1 = 4.22A and I2 = -3.60A
I1 = -5.55A and I2 = 4.60A
        
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Consider the two-loop circuit shown in the figure below. The currents I1 and I2 (in amp) satisfy the following system of equations:
16I1 - 9I2 = 100
20I2 - 9I1 + 110 = 0
Calculate I1 and I2. The unit of current is Amp (A).
I1 = - 2.25A and I2 = 1.60A
I1 = 3.22A and I2 = -2.60A
I1 = 4.22A and I2 = -3.60A
I1 = -5.55A and I2 = 4.60A

Added by Emmanuel J.

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Algebra and Trigonometry Real Mathematics, Real People
Algebra and Trigonometry Real Mathematics, Real People
Ron Larson 7th Edition
Chapter 8
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Transcript

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0:00 Solution.
00:01 So the system of equation that we have is 16i1 minus 9i2 is equal to 100.
00:11 12i2 minus 9i1 plus 110 is equal to 0.
00:19 And so 16i1, 9i2 is 100...
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