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Simultaneous equation problem as circuit analysis to find unknown currents. Students need to solve the equations and type in the solutions for each variable. Advice is given in terms of solution by elimination.

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We need to solve \\begin{align} \\simplify[unitFactor]{-{a}I_2 + {b}I_1} &= 0\\\\ \\simplify[unitFactor]{{c}I_1 - {d}I_2 + {f}} &=0\\end{align} which can be written as \\begin{align} \\simplify[unitFactor]{{b}I_1-{a}I_2} &= 0\\quad (1)\\\\ \\simplify[unitFactor]{{c}I_1 - {d}I_2} &=\\simplify{-{f}}\\quad (2)\\end{align}

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Using the method of elimination to eliminate \\(I_2\\) we multiply equation \\((1)\\) by \\(\\var{factor2}\\) and equation \\((2)\\) by \\(\\var{factor1}\\) to get \\begin{align}\\simplify[unitFactor]{{b*factor2}I_1-{a*factor2}I_2} &= 0\\quad (3)\\\\ \\simplify[unitFactor]{{c*factor1}I_1 - {d*factor1}I_2} &=\\simplify{-{f*factor1}}\\quad (4)\\end{align}

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Taking \\((3)-(4)\\) we obtain \\[\\simplify{({b*factor2}-{c*factor1})I_1={f*factor1}}\\]

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and hence \\[I_1=\\var{rational(a*f/(b*d-a*c))}\\]

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Substituting this into \\((1)\\) gives us \\[\\simplify[unitFactor]{-{a}I_2 + {b*a*f/(b*d-a*c)}} = 0\\] and hence \\[I_2=\\simplify{{b*f/(b*d-a*c)}}\\]

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Thus the currents are \\(I_1=\\simplify{{a*f/(b*d-a*c)}}\\) amps and \\(I_2=\\simplify{{b*f/(b*d-a*c)}}\\) amps.

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(Note that this problem could also be solved by eliminating \\(I_1\\) instead of \\(I_2\\) or by the method of substitution, however the final answers will be the same.)

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Coefficent of I_1 in equation 2, random integer between 2 and 10

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constant term in equation 2, random value between 5 and 15

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factor used to multiply equation 2 by in advice showing solution by elimiation method

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factor used to multiply equation 1 by in advice showing solution by elimiation method

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Analysis of a particular electronic circuit shows that the currents across two sections of the circuit must satisfy the equations \\begin{align} \\simplify[unitFactor]{-{a}I_2 + {b}I_1} &= 0\\\\ \\simplify[unitFactor]{{c}I_1 - {d}I_2 + {f}} &=0 \\end{align} Find the values of \\(I_1\\) and \\(I_2\\). Write your final answer in the box below and show full working on your handwritten notes. (Note: Leave answers in fraction form. For example to enter the fraction \\(\\frac32\\) type 3/2.)

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\\(I_1=\\)[[0]] amps

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\\(I_2=\\)[[1]] amps

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