Looking for some circuits help

[Fiveohhh]

Got a late night question before class tomorrow, on a circuit I'm trying to solve. It seems simple enough, but I'm not getting an answer that makes sense. Not sure if I'll get much success here, but figured I'd try. There's always one of us waiting to show of his/her circuit solving abilities!

Here is the circuit in questionText

I'm trying to find a value for C that will put the current through the entire circuit in phase with the voltage source. I've done this problem 20 times and I keep getting 40nF and 160nf. The way the question is worded it should be in the uF range and as far as I know there should only be one result. I'm assuming I'm making some mathematical mistake.

 

[frostedflakes]

To keep the current in phase you'd want the reactance of the inductor to equal that of the capacitor. You can use the relationship ?L=1/?C to solve for C. So 200nF is the answer I believe.

 

[Born2bwire]

I'm not getting a valid answer for the given circuit.

 

[Gibson486]

I'll give you a hint.....


current will lag voltage in that purely reactive source by pi/2.

Frostedflakes solution will not work because of the resistor.

phase is arctan(resistance/reactance).

 

[JohnCU]

Originally posted by: Gibson486
I'll give you a hint.....


current will lag voltage in that purely reactive source by pi/2.

Frostedflakes solution will not work because of the resistor.

phase is arctan(resistance/reactance).

isn't it arctan(reactance/resistance)? ie 1+j0 = arctan(0/1) = arctan(0) = 0

 

[frostedflakes]

My bad, yeah I didn't take into account that pesky little resistor.

Also you're right John, probably just a typo.

 

[Gibson486]

Originally posted by: JohnCU

Originally posted by: Gibson486
I'll give you a hint.....


current will lag voltage in that purely reactive source by pi/2.

Frostedflakes solution will not work because of the resistor.

phase is arctan(resistance/reactance).

isn't it arctan(reactance/resistance)? ie 1+j0 = arctan(0/1) = arctan(0) = 0

could be, it's been 2 years since I used this stuff. All I have to know on my current job is X mW + Y mW = Z mW. It's been a while

 

[JohnCU]

Originally posted by: Gibson486

Originally posted by: JohnCU

Originally posted by: Gibson486
I'll give you a hint.....


current will lag voltage in that purely reactive source by pi/2.

Frostedflakes solution will not work because of the resistor.

phase is arctan(resistance/reactance).

isn't it arctan(reactance/resistance)? ie 1+j0 = arctan(0/1) = arctan(0) = 0

could be, it's been 2 years since I used this stuff. All I have to know on my current job is X mW + Y mW = Z mW. It's been a while

lies, X mW + Y mVAR = Z mVA

 

[Fiveohhh]

What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

 

[Gibson486]

Originally posted by: JohnCU

Originally posted by: Gibson486

Originally posted by: JohnCU

Originally posted by: Gibson486
I'll give you a hint.....


current will lag voltage in that purely reactive source by pi/2.

Frostedflakes solution will not work because of the resistor.

phase is arctan(resistance/reactance).

isn't it arctan(reactance/resistance)? ie 1+j0 = arctan(0/1) = arctan(0) = 0

could be, it's been 2 years since I used this stuff. All I have to know on my current job is X mW + Y mW = Z mW. It's been a while

lies, X mW + Y mVAR = Z mVA

Ah man....don't remind me

 

[Born2bwire]

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

 

[Fiveohhh]

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it. So I may very well be overlooking something obvious. Which is why I need some fresh eyes

 

[Fiveohhh]

oops

 

[Born2bwire]

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes

You're taking the s-transform to find the impedances in phasor space. s is j*\omega, the angular velocity not frequency. \omega = 2*pi*frequency.

 

[Gibson486]

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes

why are you setting it to zero? Are you assuming that if reactance and reistance is zero that this will lead to max power and means that they are in phase?

 

[Born2bwire]

Originally posted by: Gibson486

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes

why are you setting it to zero? Are you assuming that if reactance and reistance is zero that this will lead to max power and means that they are in phase?

I believe he's setting the imaginary part of the total impedance to be zero, thus making the current and voltage in phase... At least he should be.

 

[Gibson486]

Originally posted by: Born2bwire

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes

You're taking the s-transform to find the impedances in phasor space. s is j*\omega, the angular velocity not frequency. \omega = 2*pi*frequency.

he may have not been introduced to laplace yet.....if he has been, it would make this so much easier

 

[Fiveohhh]

Originally posted by: Born2bwire

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes

You're taking the s-transform to find the impedances in phasor space. s is j*\omega, the angular velocity not frequency. \omega = 2*pi*frequency.

I think I see what you're saying. It is supposed to be 1000 rads/sec. I just drew that up and didn't change the units.

 

[Born2bwire]

Originally posted by: Gibson486

Originally posted by: Born2bwire

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes

You're taking the s-transform to find the impedances in phasor space. s is j*\omega, the angular velocity not frequency. \omega = 2*pi*frequency.

he may have not been introduced to laplace yet.....if he has been, it would make this so much easier

Can't imagine how he would know to replace C with 1/(j\omega C) and L with j\omega L without knowing Laplace transform.

 

[Fiveohhh]

Originally posted by: Born2bwire

Originally posted by: Gibson486

why are you setting it to zero? Are you assuming that if reactance and reistance is zero that this will lead to max power and means that they are in phase?

I believe he's setting the imaginary part of the total impedance to be zero, thus making the current and voltage in phase... At least he should be.

Yeah that is what I was attempting to do.

 

[Fiveohhh]

Originally posted by: Born2bwire

Originally posted by: Gibson486

he may have not been introduced to laplace yet.....if he has been, it would make this so much easier

Can't imagine how he would know to replace C with 1/(j\omega C) and L with j\omega L without knowing Laplace transform.

This was before we learned Laplace. We just started on Laplace Friday and these need to be done with phasors.

 

[Gibson486]

Originally posted by: Born2bwire

Originally posted by: Gibson486

Originally posted by: Born2bwire

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes

You're taking the s-transform to find the impedances in phasor space. s is j*\omega, the angular velocity not frequency. \omega = 2*pi*frequency.

he may have not been introduced to laplace yet.....if he has been, it would make this so much easier

Can't imagine how he would know to replace C with 1/(j\omega C) and L with j\omega L without knowing Laplace transform.

welll....that is what sinosoidal capacitance looks like. You do not repalce C with 1/(jwC)....1/(jwc) is the defination of impedance for a cap in sinosoidal analysis.

Laplace tells you to replace anything related to the imaginary part with s.

 

[Born2bwire]

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Fiveohhh
What I was doing was ((1/12.5k)+j(1000)(c))^-1

From here I plugged it out by hand and a calculator to get the imaginary part. I than set this imaginary part(which was negative since this portion is capacitive) equal to the reactance of j5000 from the inductor. doing this I get 2 real roots at 160nF and 40nF. Which I would think is not correct since there should only be one value as the circuit moves from inductive to capacitive it should only be purely resistive once.

s = j\omega, not j\nu.

Not following ya.

1/12.5k<---admittance of resistor
j(1000)(c)<--------admittance of cap

Adding them together and taking the inverse gets the impedance of the rc section. add to that the impedance of the inductor(1000)(5). Set to zero and solve for c.

I've been staring at this problem all day, so it's hard for me to get a fresh view on it, which is why I need some fresh eyes

You're taking the s-transform to find the impedances in phasor space. s is j*\omega, the angular velocity not frequency. \omega = 2*pi*frequency.

I think I see what you're saying. It is supposed to be 1000 rads/sec. I just drew that up and didn't change the units.

Well in that case your answer is 0.16 \mu F and 0.04 \mu F.

 

[Gibson486]

Originally posted by: Fiveohhh

Originally posted by: Born2bwire

Originally posted by: Gibson486

he may have not been introduced to laplace yet.....if he has been, it would make this so much easier

Can't imagine how he would know to replace C with 1/(j\omega C) and L with j\omega L without knowing Laplace transform.

This was before we learned Laplace. We just started on Laplace Friday and these need to be done with phasors.

so....this means you have to do it in this form....

Ae^(jwt)....


Remember....current is gonna be pi/2 phase ahead.

hint:

Ae^(jwt)*e^(jwt).

The second part of the equation (after the multiplication sign) is where you take into account your phase change.

 

[Fiveohhh]

Originally posted by: Born2bwire

Originally posted by: Fiveohhh


I think I see what you're saying. It is supposed to be 1000 rads/sec. I just drew that up and didn't change the units.

Well in that case your answer is 0.16 \mu F and 0.04 \mu F.

Does it make sense that there are two solution? I'm still new to most of this AC, but I would think that as the circuit progresses from inductive to capacitive that it would only be in phase once.