Hello,
i thought i would go through this thinking it would help. i was able to work through most of question 1 without too much of a problem.
question 2 is a little confusing.
Through algebraic substitution, write an equation that gives the Q factor of a series resonant LC circuit exclusively in terms of L, C, and R, without reference to reactance (X) or frequency (f).
this is the answer:

im guessing the 1 over R is in parallel with the square root of L(an absolute value?) over C?
rearranging this i think is a big step from question 1.
i think the person/people making the worksheets is doing a great job as i cant find that much elsewhere to help practice with this stuff. i have bought many books. i got one recently "engineering circuit analysis" by william h. hayt. i flicked through it a couple of times and threw it. it is heavy with calculus and i think you need to be an expert at calculus before starting that book.
does question 2 look right? i cant read the answer to well or figure out how to get there.
thanks
simon
i thought i would go through this thinking it would help. i was able to work through most of question 1 without too much of a problem.
question 2 is a little confusing.
Through algebraic substitution, write an equation that gives the Q factor of a series resonant LC circuit exclusively in terms of L, C, and R, without reference to reactance (X) or frequency (f).
this is the answer:

im guessing the 1 over R is in parallel with the square root of L(an absolute value?) over C?
rearranging this i think is a big step from question 1.
i think the person/people making the worksheets is doing a great job as i cant find that much elsewhere to help practice with this stuff. i have bought many books. i got one recently "engineering circuit analysis" by william h. hayt. i flicked through it a couple of times and threw it. it is heavy with calculus and i think you need to be an expert at calculus before starting that book.
does question 2 look right? i cant read the answer to well or figure out how to get there.
thanks
simon