In polar coordinates, what is the impedance of a network comprised of a 400-ohm- reactance inductor in parallel with a 300-ohm resistor?

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Multiple Choice

In polar coordinates, what is the impedance of a network comprised of a 400-ohm- reactance inductor in parallel with a 300-ohm resistor?

When you have a resistor in parallel with a pure inductor, use admittance to combine them. The total impedance is the reciprocal of the sum of their admittances. Impedances: resistor is 300 ohms (Z_R = 300), inductor is j400 ohms (Z_L = j400). Their admittances are Y_R = 1/300 ≈ 0.0033333 and Y_L = 1/(j400) = -j/400 = -j0.0025. Add them: Y = 0.0033333 - j0.0025. Now Z = 1/Y. With Y = a + jb where a ≈ 0.0033333 and b ≈ -0.0025, Z = (a - jb)/(a^2 + b^2). This gives Z ≈ (0.0033333 - j(-0.0025)) / (0.0033333^2 + 0.0025^2) ≈ 192 + j144 ohms. The magnitude is sqrt(192^2 + 144^2) ≈ 240 ohms, and the angle is arctan(144/192) ≈ 36.9 degrees. Since the imaginary part is positive, the net impedance is inductive. So the impedance is about 240 ohms at +36.9 degrees.

When you have a resistor in parallel with a pure inductor, use admittance to combine them. The total impedance is the reciprocal of the sum of their admittances.

Impedances: resistor is 300 ohms (Z_R = 300), inductor is j400 ohms (Z_L = j400). Their admittances are Y_R = 1/300 ≈ 0.0033333 and Y_L = 1/(j400) = -j/400 = -j0.0025. Add them: Y = 0.0033333 - j0.0025.

Now Z = 1/Y. With Y = a + jb where a ≈ 0.0033333 and b ≈ -0.0025, Z = (a - jb)/(a^2 + b^2). This gives Z ≈ (0.0033333 - j(-0.0025)) / (0.0033333^2 + 0.0025^2) ≈ 192 + j144 ohms.

The magnitude is sqrt(192^2 + 144^2) ≈ 240 ohms, and the angle is arctan(144/192) ≈ 36.9 degrees. Since the imaginary part is positive, the net impedance is inductive.

So the impedance is about 240 ohms at +36.9 degrees.

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