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ENGR 1990 Engineering Mathematics Applications of Systems of Linear, Algebraic Equations in Electrical Engineering Consider the double-loop DC circuit shown in the diagram. Application of Kirchhoff’s Current Law tells us that the current in the middle path (common to both loops) is the sum of the currents flowing in the outer paths. Using this result, we apply Kirchhoff’s Voltage Law to each loop separately. R1 1 + V1 I1 + I 2 + R3 − − I2 I1 1 1 1 V2 K 2 + R3 ( I1 + I 2 ) loop 1 ∑ ( voltage rises ) = V = ∑ ( voltage drops ) = R 2 loop 2 or I2 I1 ∑ ( voltage rises ) = V = ∑ ( voltage drops ) = R I loop 1 R2 J 2 I 2 + R3 ( I1 + I 2 ) loop 2 ( R1 + R3 ) I1 + ( R3 ) I 2 = V1 ( R3 ) I1 + ( R2 + R3 ) I 2 = V2 (1) Given the resistances and the applied voltages, Eqs. (1) can be solved for the two currents. Example Given: R1 = 6(Ω) , R2 = 10(Ω) , R3 = 5(Ω) , V1 = 6 (volts) , and V2 = 12 (volts) Find: the currents I1 and I 2 using a) substitution, and b) Cramer’s rule. Solution: Using the given values, we have the two simultaneous equations: 11 I1 + 5 I 2 = 6 5 I1 + 15 I 2 = 12 . a) Substitution: 6 − 11 I1 . Substituting this 5 result into the second equation for I 2 gives a single equation that we can solve for I1 . Solving the first equation for I 2 in terms of I1 , gives I2 = ⎛ 6 − 11 I1 ⎞ 12 = 5 I1 + 15 ⎜ ⎟ = 5 I1 + 3 ( 6 − 11 I1 ) = −28 I1 + 18 ⎝ 5 ⎠ 18 − 12 ⎛ 6 − 11 I1 ⎞ = 0.2143 (amps) and I 2 = ⎜ ⇒ I1 = ⎟ = 0.7286 (amps) 28 5 ⎝ ⎠ 1/2 b) Cramer’s Rule: To use Cramer’s rule, we first write the equations as a matrix equation. ⎡11 5 ⎤ ⎧ I1 ⎫ ⎧ 6 ⎫ ⎢ 5 15⎥ ⎨ I ⎬ = ⎨12 ⎬ ⎣ ⎦⎩ 2⎭ ⎩ ⎭ Then, we solve. ⎡6 5⎤ det ⎢ 12 15⎥⎦ (6 × 15) − (5 × 12) 30 ⎣ I1 = = = = 0.2143 (amps) ⎡11 5 ⎤ (11 × 15) − (5 × 5) 140 det ⎢ ⎥ ⎣ 5 15⎦ ⎡11 6 ⎤ det ⎢ 5 12 ⎥⎦ (11 × 12) − (6 × 5) 102 ⎣ = = = 0.7286 (amps) I2 = ⎡11 5 ⎤ (11 × 15) − (5 × 5) 140 det ⎢ ⎥ ⎣ 5 15⎦ 2/2