Solve for I1 from the simultaneous linear equations 4I1 + 4I2 = 2 and 6I1 + 7I2 = 4 in the context of circuit analysis (report the correct sign and in amperes).

Difficulty: Easy

Correct Answer: –0.5 A

Explanation:


Introduction / Context:
Systems of linear equations frequently arise from Kirchhoff's laws in electrical engineering. Here, I1 and I2 represent assumed branch or mesh currents. Solving the pair of equations correctly, including the sign, reveals the true direction of I1 relative to the assumed reference.


Given Data / Assumptions:

  • Equation 1: 4I1 + 4I2 = 2.
  • Equation 2: 6I1 + 7I2 = 4.
  • I1 and I2 are steady-state DC currents; units are amperes.
  • Algebraic signs indicate direction with respect to the chosen current reference.


Concept / Approach:

Use substitution or elimination to solve two linear equations in two unknowns. The negative sign of a current solution means the real current flows opposite to the initially assumed direction—a common outcome in mesh/branch analyses.


Step-by-Step Solution:

From 4I1 + 4I2 = 2, isolate I1: I1 = (2 − 4I2) / 4 = 0.5 − I2.Substitute into 6I1 + 7I2 = 4: 6(0.5 − I2) + 7I2 = 4.Compute: 3 − 6I2 + 7I2 = 4 ⇒ 3 + I2 = 4 ⇒ I2 = 1 A.Back-substitute: I1 = 0.5 − 1 = −0.5 A.


Verification / Alternative check:

Plug (I1, I2) = (−0.5, 1) into both equations: 4(−0.5) + 4(1) = −2 + 4 = 2; 6(−0.5) + 7(1) = −3 + 7 = 4. Both hold, confirming the solution.


Why Other Options Are Wrong:

0.5 A and 50 mA ignore the required negative sign and magnitude. −50 mA has the correct sign but is off by a factor of 10.


Common Pitfalls:

Dropping the negative sign, arithmetic slips while distributing coefficients, or failing to verify the solution in both equations.


Final Answer:

–0.5 A

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