Which of the following is the admittance Y of a load whose impedance is Z = 3 + j4 [Ω]?

Answer: A. 0.12 − j0.16 S

Method 1 (rationalizing the denominator): multiplying the numerator and denominator of Y = 1/(3 + j4) by the complex conjugate 3 − j4 gives Y = (3 − j4)/(3² + 4²) = (3 − j4)/25 = 0.12 − j0.16 S. Method 2 (polar form): |Z| = √(3² + 4²) = 5 Ω and the angle is about 53.1°, so Y = (1/5)∠−53.1° = 0.2∠−53.1° S, which in rectangular form is the same 0.12 − j0.16 S. When the impedance is inductive (positive imaginary part), the imaginary part of the admittance (the susceptance) is negative.

• A. Correct. (3 − j4)/25 = 0.12 − j0.16 S.
• B. The sign of the imaginary part is reversed. For 1/(a + jb) you multiply by the complex conjugate, so the sign of the imaginary part flips.
• C. This is 1/3 − j(1/4), taking the reciprocals of the real and imaginary parts separately, which is not how the reciprocal of a complex number is calculated.