LCR series circuits and average power

[physicsStudent00]

Homework Statement

One type of tuning circuit used in radio receivers is a series LCR circuit. You like listening to a station1 that transmits 99.3 MHz in . The government wants to make 100.1 MHz available to station2. Assume that the transmitters of the two stations are equally powerful and that you live the same distance from both stations. You have an inductor with L = 1.00 mH and your goal is to design a circuit that: (1) gives maximum power for station1; and (2) the average power delivered to the resistor in response to statio2 is 1% of the average power in response to station2 (making station2 inaudible when you are tuned to statio1).

Determine the capacitance C, the quality factor Q, and the resistance R that satisfy the design requirements (Hint: you do not need to know V to use condition (2); just take ratios)

Homework Equations

The Attempt at a Solution

i've made a lot of attempts but all the variables and formulas possible i continuously get lost or stuck and don't know where to start.

 

[collinsmark]

Please show your best attempt and maybe we can help you where you get stuck.

If you really just need some help getting started, solving for C should be pretty easy, So I suggest starting by solving for C.

Solving for R is a bit trickier, but before that, can you find a relationship between \langle P_1 \rangle and \langle P_2 \rangle? (Where \langle P_1 \rangle is the average power from station 1 [after the filter] when the receiver is tuned to station 1, and \langle P_2 \rangle is the average power from station 2 [after the filter] when the receiver is still tuned to station 1.)

 

[physicsStudent00]

Please show your best attempt and maybe we can help you where you get stuck.

If you really just need some help getting started, solving for C should be pretty easy, So I suggest starting by solving for C.

Solving for R is a bit trickier, but before that, can you find a relationship between \langle P_1 \rangle and \langle P_2 \rangle? (Where \langle P_1 \rangle is the average power from station 1 [after the filter] when the receiver is tuned to station 1, and \langle P_2 \rangle is the average power from station 2 [after the filter] when the receiver is still tuned to station 1.)

well the ratio of powers would simply be <P_2> = 0.1*<P_1>
however finding C isn't it the same for both of stations as the value of the capacitor in the circuit wouldn't change, which means how i was going about finding C is obviously wrong.

station 1
ω_0 = 1/√LC=2πf
∴C = 2.56*10^-12
station 2
C=2.52*10^-12

im not sure that i have all the of equations needed or if i can use ω_0 in that way. I am probably missing something simple here

 

[collinsmark]

well the ratio of powers would simply be <P_2> = 0.1*<P_1>

I think you mean \langle P_2 \rangle = 0.01 \ \langle P_1 \rangle

however finding C isn't it the same for both of stations as the value of the capacitor in the circuit wouldn't change, which means how i was going about finding C is obviously wrong.

But the receiver is only ever tuned to station 1. You can pick the value of C such that the receiver passes the maximum signal strength of station 1. (In other words, when tuned to station 1, \omega_0 = \omega_1).

station 1
ω_0 = 1/√LC=2πf
∴C = 2.56*10^-12
station 2
C=2.52*10^-12

im not sure that i have all the of equations needed or if i can use ω_0 in that way. I am probably missing something simple here

Is the inductor 1.00 millihenrys or 1.00 microhenrys? The problem statement says it's milli-, but you've calculated it as micro-.

 

[physicsStudent00]

I think you mean \langle P_2 \rangle = 0.01 \ \langle P_1 \rangle

But the receiver is only ever tuned to station 1. You can pick the value of C such that the receiver passes the maximum signal strength of station 1. (In other words, when tuned to station 1, \omega_0 = \omega_1).

Is the inductor 1.00 millihenrys or 1.00 microhenrys? The problem statement says it's milli-, but you've calculated it as micro-.

it is in microhenrys I didn't realize i copied it wrong,
so that's what i was missing that we can have ω_0=ω_1 because its only tuned to one station. so is that how I am supposed to find C? but also by that method doesn't it mean that ω = ω_0

 

[collinsmark]

it is in microhenrys I didn't realize i copied it wrong,
so that's what i was missing that we can have ω_0=ω_1 because its only tuned to one station. so is that how I am supposed to find C? but also by that method doesn't it mean that ω = ω_0

Yes, you may set \omega_0 = \omega_1 in your circuit.

Moving forward, now that you know what \omega_0 = \omega_1 is, what is the circuit's power output when the signal's frequency is not at the circuit's resonant frequency, but rather at the frequency of station 2?

 

[physicsStudent00]

Yes, you may set \omega_0 = \omega_1 in your circuit.

Moving forward, now that you know what \omega_0 = \omega_1 is, what is the circuit's power output when the signal's frequency is not at the circuit's resonant frequency, but rather at the frequency of station 2?

im sorry i don't quite understand that, so what you are saying is that its not at resonant frequency anymore but at a frequency of station 2. so we are saying that the value i got for C at station2's frequency would be correct. however what is then the difference between ω_1 and ω?

 

[collinsmark]

im sorry i don't quite understand that, so what you are saying is that its not at resonant frequency anymore but at a frequency of station 2. so we are saying that the value i got for C at station2's frequency would be correct. however what is then the difference between ω_1 and ω?

The resonant frequency of receiver [circuit] is \omega_0. And you're designing the circuit such that it's tuned to the frequency of station 1, i.e., \omega_0 = \omega_1. That's fixed and does not change (for this problem).

The frequency of some particular signal is \omega. This can be at any frequency. But for the purposes of this problem, you'll need to ask yourself what the power output of the circuit is when \omega = \omega_2.

 

[physicsStudent00]

The resonant frequency of receiver [circuit] is \omega_0. And you're designing the circuit such that it's tuned to the frequency of station 1, i.e., \omega_0 = \omega_1. That's fixed and does not change (for this problem).

The frequency of some particular signal is \omega. This can be at any frequency. But for the purposes of this problem, you'll need to ask yourself what the power output of the circuit is when \omega = \omega_2.

ahhh ok so if I am able to get the ratio of power which i am, and assume that voltage is the same, resistance is the same and quality factor is the same all because its the same circuit the only thing that would be changing is ω, which is set to ω=ω_2 and ω_0 =ω_1. from that I am able to solve for Q,R and C. I've already solved for C using the relationship 1/√LC = ω_0 = ω_1 = 2πf
ω_0=ω_1, because the radio/circuit is set to station one as its resonant frequency because that's when its at max power,
ω=ω_2 because that's what you are changing.
so when on station1 ω=ω_1=ω_0, when on station2 ω=ω2 and ω_0=ω_1 still.
is that right or am i reading all of this wrong

 

[collinsmark]

ahhh ok so if I am able to get the ratio of power which i am, and assume that voltage is the same, resistance is the same and quality factor is the same all because its the same circuit the only thing that would be changing is ω, which is set to ω=ω_2 and ω_0 =ω_1. from that I am able to solve for Q,R and C. I've already solved for C using the relationship 1/√LC = ω_0 = ω_1 = 2πf
ω_0=ω_1, because the radio/circuit is set to station one as its resonant frequency because that's when its at max power,
ω=ω_2 because that's what you are changing.
so when on station1 ω=ω_1=ω_0, when on station2 ω=ω2 and ω_0=ω_1 still.
is that right or am i reading all of this wrong

Correct (if I'm understanding you correctly).

The receiver is tuned to station 1. Still, some power from station 2 still gets through the filter even though it is tuned to station 1. Specifically, the amount of power from station 2 that gets through is 0.01, in reference to the power of station 1.

Using that you should be able to solve for R and Q.

 

[physicsStudent00]

Correct (if I'm understanding you correctly).

The receiver is tuned to station 1. Still, some power from station 2 still gets through the filter even though it is tuned to station 1. Specifically, the amount of power from station 2 that gets through is 0.01, in reference to the power of station 1.

Using that you should be able to solve for R and Q.

thank you so much for all the help