RLC circuit potential difference

[cupcake]

Homework Statement

An L-R-C series circuit is constructed using a 175{ \Omega} resistor, a 12.5 muF capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V.


At the angular frequency in part A, find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value.

Homework Equations

V= I Z
Z^2 = R^2 + (XL-XC)^2
XL= w.L
XC = 1/ w.C

The Attempt at a Solution

for part A, i get the correct answer for angular frequency = 3160 rad/s
here, i didn't understand what they meant by " the current is equal to one-half its greatest positive value." do they mean i have to find the total current and divide by two??

i found the current through Ir=Ic=Il= 0.143
by doing
I = V/Z = 25/175

 

[rl.bhat]

At resonance the current is maximum, and XL = XC.
You have already found out that current.
Find impedance Z at which the current is half the maximum current.
Then ( Z^2 - R^2 )^1/2 = WL - 1/WC.
Solve the quadratic equation to find the value of W. From that you can find the potential difference across them.

 

[cupcake]

At resonance the current is maximum, and XL = XC.
You have already found out that current.
Find impedance Z at which the current is half the maximum current.
Then ( Z^2 - R^2 )^1/2 = WL - 1/WC.
Solve the quadratic equation to find the value of W. From that you can find the potential difference across them.

then, after i found w??
find the resistance on capacitor and inductor..
then, times with I to get the potential different??

 

[rl.bhat]

To find the potential difference you have to take I/2.

 

[cupcake]

A
Then ( Z^2 - R^2 )^1/2 = WL - 1/WC.
.

hey..but here..
Z=R
so
0= wL - wc..

what did you mean by this equation??

 

[cupcake]

ahh..ic
you meant
Z = V/ (I half of maximum)

rite??

 

[rl.bhat]

Yes.

 

[cupcake]

i think i made a mistake :)
thanks

 

[rl.bhat]

Any one will do. When you take the other value, voltage across L and C will interchange.

 

[cupcake]

how about potential different on ac source and resistor?

 

[rl.bhat]

VR = I/2*R, VC = I/2*XC and VL = I/2*XL
Here V is not equal to VR + VC + VL
So the potential difference on ac source is V itself.

 

[cupcake]

hi, why can't i get the correct answer for this??
though i have done accordingly..
after find the quadratic equation,
i get w=38150 and w=(-261.875)
i choose the positive one,
and do the calculation to find XL and XC,
after that multiply by the current(half of max)

but, my answer is not correct..
please advise!

 

[rl.bhat]

How did you get 3160 rad/s?

 

[cupcake]

for part A the question is
At what angular frequency will the impedance be smallest?

i did
f= 1/2*pi*sqr(LC)
after i got the frequency, then calculated the angular speed w=2*pi*f
ended up with 3160 rad/s

 

[cupcake]

help!

 

[rl.bhat]

To get the current I/2, (WL - 1/WC) must be equal to (3)^1/2*R.
Further simplification shows that W1 = WR - (3)^1/2*R./2L and W2 = WR + (3)^1/2*R./2L

 

[cupcake]

To get the current I/2, (WL - 1/WC) must be equal to (3)^1/2*R.
Further simplification shows that W1 = WR - (3)^1/2*R./2L and W2 = WR + (3)^1/2*R./2L

why it should be equal to (3)^1/2*R. ??
and what is WR?? omega*R?
i don't understand why it is WR??

 

[rl.bhat]

Sorry. I should have explained the symbols. Here omega*R is the resonance frequency. At any other frequencies, I = E/Z, where Z^2 = R^2 + (XL - XC)^2. For I/2, Z' = 2Z. It is possible if (XL - XC)^2 = 3R^2.
If you see the graph of current vs frequency, current will be I/2 for two frequencies. So W1 and W2 must be positive. In the above problem you are getting one positive and other negative. The reason may be either R value is wrong or instead of half the maximum current they might have asked half the maximum power. Just check it.

 

[cupcake]

Sorry. I should have explained the symbols. Here omega*R is the resonance frequency. At any other frequencies, I = E/Z, where Z^2 = R^2 + (XL - XC)^2. For I/2, Z' = 2Z. It is possible if (XL - XC)^2 = 3R^2.
If you see the graph of current vs frequency, current will be I/2 for two frequencies. So W1 and W2 must be positive. In the above problem you are getting one positive and other negative. The reason may be either R value is wrong or instead of half the maximum current they might have asked half the maximum power. Just check it.

hm...
for this one, since we know that Z'=2Z, can i just find the current by I= V/ 2Z, then after find the current, multiply by each of the resistance to get the voltage., without finding omega first.?

 

[rl.bhat]

If you don't know the omega, how can you find XL and XC?

 

[cupcake]

ahh..ya i forgot,, that XL and XC is changed... LOL
thanks rl.bhat.. i'll work on it...

 

[cupcake]

hi..i found the value for w= 10^7 and w=0
is it strange rite??
LOL

why it should (XL-XC)^2=3R^2 ??
shouldn't it..
Z'=2Z
R^2+(XL-XC)^2= 4[R^2+(XL-XC)^2]
(XL-XC)^2= 3R^2 + 4(XL-XC)^2

 

[cupcake]

i can't get the correct answer :( :(