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.

Bill WA6OHP says

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.

Dan KB6NU says

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.

John says

The MHZ numbers of 24.9mhz are transposed to 29.4 mhz in the last question of this section.

Regardless this guide is the best format for study I’ve found.

73’s

John