Xc=1divided by 2pifC solve for C
ib=ibq+ sq.root of 2is solve for is
Q=XLdivided byR solve XL
Lt=L1+L2-2Lmsolve for Lm
R1=i2R2divided by i2 solve for R
Es=E1+E2+E3solve for E2
R1divided by Rx=R2divided by R3 solve for Rx
y=Mx+b solve for b
i2=i1R1divided by R2 solve for i1
2006-11-07 08:53:36 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1. Xc = 1/2πfC
So 2πfC = 1/Xc
So C = 1/2πfXc
2. ib = ibq + √2is
So √ = ib - ibq
So 2is = (ib - ibq)²
ie is = ½(ib - ibq)²
3. Q = XL/R
So XL = QR
4. Lt = L1 + L2 - 2Lm
So 2Lm = L1 + L2 - Lt
ie Lm = ½(L1 + L2 - Lt)
5. R1=i2 R2 / i2
??????? So R1 = R2
6. Es = E1 + E2 + E3
So E2 = Es - E1 - E3
7. R1 / Rx = R2 / R3
So Rx / R1 = R3 / R2
So Rx = R1 * R3/R2
8. y = Mx + b
So Mx = y - b
So M = (y - b)/x
9. i2=i1 R1 / R2
So i1 R1 = i2R2
ie i1 = i2 R2 / R1
I would be extremely concerned about your progress if you find this simple algebra a serious chore!!!!!! (Especially if you are learning the electrical trade)
2006-11-07 09:16:54 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
in #1) Xc=1/(2*pi*f*C) - I hope this is what you meant
first you get the inverse of both sides of the equation
so 1/Xc=2*pi*f*C
Then you divide both sides by 2*pi*f to get C
To divide 1/Xc by 2*pi*f, you can multiply 1/Xc by 1/(2*pi*f)
therefore C=1/(2*pi*f*Xc)
and you can do the rest of your own homework.
Ok, I take back my original comment and apologize for being rude, but seriously, problem #6 is very very elementary. Anyone over age 11 should be able to logically figure that one out.
2006-11-07 17:10:08 · answer #2 · answered by Anonymous · 0⤊ 0⤋
wow!!! what grade math is that?
2006-11-07 17:04:17 · answer #3 · answered by campkid4ever 2 · 0⤊ 0⤋