Finding R,L,C at resonant state

[MissP.25_5]

Hello again. I hope someone could check if I got the answers correct.

The magnitude of the current I in resonant state was 3[A]. When the angular frequency of the power supply was 1 [rad/s] and 2[rad/s], the magnitude of current I became 1/√2 of the value when resonance occurred. Find R, L, C.

 

[gneill]

When I solve your equations as you've given them I'm seeing -3/8 for the capacitance... which shouldn't be.

You have to be a bit careful about 'losing' solutions when you eliminate squares.

 

[rude man]

Hello again. I hope someone could check if I got the answers correct.

The magnitude of the current I in resonant state was 3[A]. When the angular frequency of the power supply was 1 [rad/s] and 2[rad/s], the magnitude of current I became 1/√2 of the value when resonance occurred. Find R, L, C.

Your approach looks right, but I got a different result. Check your equations against mine & recheck your math:

(L - 1/C)^2 = 16,
(2L - 1/2C)^2 = 16

 

[MissP.25_5]

When I solve your equations as you've given them I'm seeing -3/8 for the capacitance... which shouldn't be.

You have to be a bit careful about 'losing' solutions when you eliminate squares.

Yes, at first I got -3/8 for the Capacitance but since it's an absolute value, I made it positive. Well, looks like I have to redo my calculation again :(

 

[MissP.25_5]

When I solve your equations as you've given them I'm seeing -3/8 for the capacitance... which shouldn't be.

You have to be a bit careful about 'losing' solutions when you eliminate squares.

I redid it and still got the same simultaneous equations. Can you tell me where I am wrong?

 

[rude man]

I redid it and still got the same simultaneous equations. Can you tell me where I am wrong?

Do you agree with my equations or not? If so, what solutions do you get?

Hint: there are multiple solutions but only one where L and C are positive, as they must be.

 

[MissP.25_5]

Do you agree with my equations or not? If so, what solutions do you get?

Hint: there are multiple solutions but only one where L and C are positive, as they must be.

Yes, yours seems to be the same as mine. But can we cancel out the squares by just adding a root to both sides?

 

[rude man]

Yes, yours seems to be the same as mine. But can we cancel out the squares by just adding a root to both sides?

When you take a square root you have to realize there are two polarities involved. Here we are taking two roots so be prepared to get four sets of answers for L and C.

But only one of them has L > 0 and C > 0.

 

[MissP.25_5]

When you take a square root you have to realize there are two polarities involved. Here we are taking two roots so be prepared to get four sets of answers for L and C.

But only one of them has L > 0 and C > 0.

Oh, I totally forgot about that. So that's what "losing solutions" means. So C=1/8 and L=4.
Thank you soooo much! Could you take a look at my new thread? I need help.

 

[rude man]

Oh, I totally forgot about that. So that's what "losing solutions" means. So C=1/8 and L=4.
Thank you soooo much! Could you take a look at my new thread? I need help.

Yes, you have got it!
Will check your new thread.