Studying General

Discussion in 'Becoming a Ham - Q&A' started by KB3SKU, Mar 17, 2009.

Not open for further replies. 1. Hi all ive started to study for General,my question is to the people without Electrical Engineering backgrounds or Math Majors how in the world did yas pass your general class exams?

Ive noticed alot of formulas and a lot of schematics and diagrams,a lot of it is hard,i never had algerbra or some of the advanced division etc.

My worrie is if ill ever pass the real test. i mean how did yas do it? I am trying but i know ill have a hard time with the formulas and stuff. What did yas do to help?

Ive been studying at HamTestOnline.com alot,i know i may have problems but i will keep studying but some of the stuff i might not understand. I wont quit cause i do want HF priviliges,its just going to be harder than tech hehe. I dont have any books ive just been useing the study mode at hamtestonline.com.hehe i may think this is hard but i wonder what Extra will be like lol.

So any tips or anything? is there any study material online that goes in depth in the subjects of General Class?

i hope i made sense up there Thank you inadvance for the help.

2. Math is your friend! Let's look at the equation for the resonant frequency of a parallel LC circuit.

At resonance, inductive reactance (a.k.a. XL) minus capacitive reactance (a.k.a. Xc) = 0

The formula for XL is:

XL = 2 * pi * f * L, where "*" is the symbol for multiplication, pi ~= 22/7, f = frequency, and L is inductance in Henries.

The formula for Xc is:

Xc = 1 / (2 * pi * f * C), where pi ~= 22/7, f = frequency, and C is capacitance in Farads.

By definition, resonance is the condition where XL = Xc; therefore, at resonance:

2 * pi * f * L = 1 / (2 * pi * f * C)

We need to rearrange this equation to solve for f (the part that requires algebra).

Let's start by dividing both sides of the equation by 2 * pi * L.

(2 * pi * f * L) / (2 * pi * L) = 1 / (2 * pi * f * C) / (2 * pi * L)

In the above equation, dividing ( 2 * pi * f * L) by (2 * pi * L) leaves us with just f on the left-hand side of the equation.

The reduction on the right-hand side is a little less intuitive; however, you were taught how to solve this kind of equation with integer fractions back in elementary school. Like most school-age children, you were taught to invert and multiply.

For example, dividing 1/5 by 2 is the same as multiplying 1/5 by 1/2.

Thus, in the above equation, we can reduce the right-hand side by multiplying 1 / ( 2 * pi * f * C) by 1 / (2 * pi * L), which gives us 1 / ((2 * pi) ^2 * f * L * C) (where “^” means raised to the power of )

Now, we have the equation f = 1 / ((2 * pi) ^2 * f * L * C) ; however, we still need to get all of the f’s on the left-hand side. To do this, we multiply both sides by f.

f * f = 1 / ((2 * pi) ^2 * f * L * C ) * f, which gives us, f^2 =1 / ((2 * pi) ^2 * L * C )

We are interested in f, not f^2; therefore, we must perform the inverse operation of raising a value to the 2nd power, which means taking the square root of both sides of the equation (the symbol “SQRT” is used to denote the square root function here).

SQRT(f^2) = SQRT(1 / ((2 * pi) ^2 * L * C )), which reduces to f = 1 / (2 * pi * SQRT(L * C))

How did were reduce the right-hand side of the equation to 1 / (2 * pi * SQRT(L* C))? Well, the square root of 1 is 1, the square root of (2 * pi)^2 is 2 * pi, and the square root of L * C is SQRT(L* C).

f = 1 / (2 * pi * SQRT(L * C)) is the formula for determining the resonant frequency of a parallel LC circuit. You will see this equation on the Extra exam. Understanding how is it is derived will help you solve other parallel resonance-related equations.

Last edited: Apr 12, 2009
3. Holy smokes that went over my head i never had that math before. Sorry not trying to be dumb. That's what im talking about the math up there,wow ill be studying a long time for General lol. The math up there looks like a foreign language to me.

4. Thank you.

5. You lack confidence in yourself! There are not that many questions involving math on the General Exam. STUDY till you get it!
Gordon West has some fine study tools as well. (W5YI)

Study and you will pass the test!

6. ok sir i will 7. Just curious,when i do pass how will i say my call sign? Will it be "KB3SKU forward slash AG" or "KB3SKU AG"

8. KB3SKU "stroke" AG

9. oh ok thank you.