Admittance is the inverse of impedance. So, in polar coordinates, the impedance of a circuit that has an admittance of 7.09 millisiemens at 45 degrees is **141 ohms at an angle of -45 degrees**. (E5C16) You calculate it this way:

|Z| = 1/7.09×10^{-3} = 141 ohms

The angle is the mirror image about the x axis:

θ = 0 – -45 degrees = 45 degrees

Let’s look at another example. In rectangular coordinates, the impedance of a circuit that has an admittance of 5 millisiemens at -30 degrees is **173 +j100 ohms**. (E5C17)

|Z| = 1/5×10^{-3} = 200 ohms

θ = 0 – -30 degrees = 30 degrees

R = |Z| × cos 30 degrees = 200 × .866 = 173 ohms

X (the reactance part of the impedance) = |Z| × sin 30 degrees = 200 × .5 = +j100

Now, let’s take a look at some actual circuits.

On Figure E5-2, the point that best represents the impedance of a series circuit consisting of a 400 ohm resistor and a 38 picofarad capacitor at 14 MHz is **Point 4**. (E5C19) Right off the bat, we know that the only choices are really Points 2, 4, and 6 because the resistance is 400 ohms. Next, we calculate the capacitive reactance:

X_{C} = 1/2πfC = 1/(2 × 3.14 × 14×10^{6} × 38×10^{-12}) ≈ 300 ohms

Because the reactance is capacitive, it’s plotted as a negative value.

On Figure E5-2, the point that best represents the impedance of a series circuit consisting of a 300 ohm resistor and an 18 microhenry inductor at 3.505 MHz is **Point 3**. (E5C20) The resistance is 300 ohms and the reactance is:

X_{L} = 2πfL = 2 × 3.14 × 3.505×10^{6} × 18×10^{-6}) ≈ 400 ohms

And, since the reactance is inductive, it’s plotted as a postive value.

On Figure E5-2, the point that best represents the impedance of a series circuit consisting of a 300 ohm resistor and a 19 picofarad capacitor at 21.200 MHz is **Point 1**. (E5C21) The resistance is 300 ohms, and the reactance is:

X_{C} = 1/2πfC = 1/(2 × 3.14 × 21.2×10^{6} × 19×10^{-12}) ≈ 400 ohms

Because the reactance is capacitive, it’s plotted as a negative value.

On Figure E5-2, the point that best represents the impedance of a series circuit consisting of a 300-ohm resistor, a 0.64-microhenry inductor and an 85-picofarad capacitor at 24.900 MHz is **Point 8**. (E5C23) This problem is a little tougher because it has both capacitive and inductive reactance.

X_{C} = 1/2πfC = 1/(2 × 3.14 × 29.4×10^{6} × 85×10^{-12}) ≈ 63.7 ohms

X_{L} = 2πfL = 2 × 3.14 × 29.4×10^{6} × 0.64×10^{-6}) ≈ 118.2 ohms

X = X_{L} – X_{C} = 118.2 – 63.7 = 55.5 ohms

Because the net reactance is inductive, it is plotted as a positive value, and because the resistance is 300 ohms, the answer is Point 8.

I passed my Extra exam in 1976 without having to study the material because I knew it but looking at the information you just gave tells me that I have forgotten much that I once knew. Good post.

It may not be that you’ve forgotten it. There are a lot of questions on the test now that were not on the test back in the 1970s.