Parallel Impedance and a Meter Tolerance Check
A lab technician is combining two AC impedance measurements in parallel. The measured impedances are Z1 = 6 + 8i ohms and Z2 = 3 − 4i ohms, where i is the imaginary unit. The parallel equivalent is Z = Z1Z2/(Z1+Z2). A bench meter reports the magnitude of Z as 5.0 ohms, and a colleague says the exact equivalent impedance should have an imaginary part of zero because the two imaginary parts, 8i and −4i, partly cancel. The technician needs a written check before filing the calibration record. The meter's phase reading is not available, so only the magnitude can be compared. All arithmetic must be kept exact until the final comparison, and any rounding is limited to two decimal places. The technician also wants to know whether the meter's 5.0-ohm reading is consistent with the computed value within a 0.1-ohm tolerance.
1. Compute the exact equivalent impedance Z in the form a + bi, showing the multiplication and division steps.
Read the worked answer
First multiply the impedances: Z1Z2 = (6 + 8i)(3 − 4i) = 18 − 24i + 24i − 32i² = 18 + 32 = 50. Add the impedances: Z1 + Z2 = (6 + 3) + (8i − 4i) = 9 + 4i. Therefore Z = 50/(9 + 4i). Multiply numerator and denominator by the conjugate 9 − 4i: Z = 50(9 − 4i)/[(9 + 4i)(9 − 4i)] = 50(9 − 4i)/(81 + 16) = (450 − 200i)/97. So Z = 450/97 − 200/97 i.
2. Compute |Z| to two decimal places and decide whether the meter's 5.0-ohm reading is within the stated 0.1-ohm tolerance.
Read the worked answer
Use |a + bi| = sqrt(a² + b²). Here |Z| = sqrt[(450/97)² + (200/97)²] = sqrt(450² + 200²)/97 = sqrt(202500 + 40000)/97 = sqrt(242500)/97. Since sqrt(242500) ≈ 492.44, |Z| ≈ 492.44/97 ≈ 5.08 ohms. The difference from the meter reading is |5.08 − 5.0| = 0.08 ohm, which is less than the 0.1-ohm tolerance. The reading is consistent with the computed magnitude.
3. Respond to the colleague's claim that the imaginary part should be zero because 8i and −4i partly cancel. Why is the complex conjugate used in the division?
Read the worked answer
The imaginary parts 8i and −4i do not simply cancel because the impedances are multiplied and then divided by a complex sum; division by 9 + 4i changes both the real and imaginary components. The conjugate 9 − 4i is used because (9 + 4i)(9 − 4i) = 81 + 16 = 97, a positive real number. Multiplying by 9 − 4i over itself preserves the value and removes i from the denominator. The resulting imaginary part is −200/97 ≈ −2.06, not zero, so the colleague's claim is incorrect.