In an AC circuit, the voltage E, current I, and impedance Z are related by the

Formula E = I · Z

Find the current in a circuit with voltage 14 – 8j volts and Impedance is 2 – 3j ohms.

E = I · Z

Substitute E and Z values

14 – 8j = i · (2 – 3j)

Divide each side by (2 – 3j) \ \

Multiply the numerator and denominator by the conjugate of the denominator

2 – 3j. Conjugate of 2 – 3j is 2 + 3j.

  (Multiply the numerator using FOIL method)

  (Using multiply formula: )

  (Evaluate powers: ) \ \

               (Multiply: )

                       (Multiply: )

         (Distributive property: )

                      (Subtract: )

               (Substitute the imaginary unit value )

                        (Product of two same signs is positive)

                        (Commutative property: a + b = b + a)

                                 (Add: 28 + 24 = 52 and 4 + 9 = 13)

                              (Write standard division form: )

                                         (Divide: )

The current is 4 + 2j amps.