Parallel Impedance and the Trap of Adding Magnitudes
An intern at an audio equipment company is combining two speaker drivers in parallel. Each driver is modeled as a complex impedance: Z1 = 3 + 4i ohms and Z2 = 5 - 2i ohms, where the imaginary part represents reactance. The intern correctly finds the series combination Z1 + Z2 = 8 + 2i ohms. For the parallel combination the company's design sheet gives Zp = Z1Z2/(Z1 + Z2). Wanting a quick estimate, the intern writes that the magnitude of Zp should be |Z1| + |Z2| = 5 + sqrt(29) ≈ 10.39 ohms and plans to enter that value in a load table. Before signing off, a senior engineer asks for the exact parallel impedance, its magnitude, and a check of the estimate.
1. Compute Zp = Z1Z2/(Z1 + Z2) in the form a + bi. Show the multiplication and the division, and give exact fractions.
Read the worked answer
Multiply first: Z1Z2 = (3 + 4i)(5 - 2i) = 15 - 6i + 20i - 8i^2. Since i^2 = -1, this is 15 + 14i + 8 = 23 + 14i. The sum is Z1 + Z2 = 8 + 2i. Now divide:
Zp = (23 + 14i)/(8 + 2i).
Multiply numerator and denominator by the conjugate 8 - 2i:
Numerator: (23 + 14i)(8 - 2i) = 184 - 46i + 112i - 28i^2 = 184 + 66i + 28 = 212 + 66i.
Denominator: 8^2 + 2^2 = 68.
So Zp = 212/68 + (66/68)i = 53/17 + 33/34 i ohm. Check: 53/17 ≈ 3.12 and 33/34 ≈ 0.97.
2. Compute |Zp| exactly and as a decimal to the nearest hundredth. Compare it with the intern's 10.39-ohm estimate and with |Z1| and |Z2|.
Read the worked answer
Using the magnitude identity for quotients and products:
|Zp| = |Z1||Z2| / |Z1 + Z2|.
Here |Z1| = sqrt(3^2 + 4^2) = 5, |Z2| = sqrt(25 + 4) = sqrt(29) ≈ 5.385, and |Z1 + Z2| = |8 + 2i| = sqrt(64 + 4) = sqrt(68) ≈ 8.246.
Thus |Zp| = 5sqrt(29)/sqrt(68) = (5sqrt(493))/34 ≈ 3.27 ohms.
This is far below the intern's 10.39-ohm estimate. It is also below both |Z1| = 5 and |Z2| ≈ 5.39, so the parallel load is not found by adding the two magnitudes.
3. Explain, using the magnitudes of Z1, Z2, and Z1 + Z2, why adding the magnitudes gives an estimate that is too large. What value should replace 10.39 in the load table?
Read the worked answer
The intern added magnitudes as if the two complex numbers pointed in the same direction. They do not: 3 + 4i has a positive imaginary part and 5 - 2i has a negative imaginary part, so their vector sum partially cancels. Indeed |Z1 + Z2| = |8 + 2i| = sqrt(68) ≈ 8.25, while |Z1| + |Z2| ≈ 10.39. The parallel formula divides by |Z1 + Z2|, not by |Z1| + |Z2|:
|Zp| = (5)(sqrt(29))/sqrt(68) ≈ (5)(5.385)/(8.246) ≈ 3.27 ohms.
The 10.39 value is too large by more than a factor of 3. The load table should be corrected to about 3.27 ohms, with the exact value (5sqrt(493))/34.