Input & Output Explained: What Does It Mean & Why Is It Relevant?

[Jaccobtw]

Homework Statement

A resonant RLC circuit can be used as an amplifier for a certain band of frequencies around the resonant frequency. Consider a series RLC circuit as an audio band amplifier with an AC voltage source as the input, and the voltage across the 8.0Ω resistor as the output. The amplifier should have a gain (=output/input) of 0.5 at 200Hz and 4000Hz. What is the required value of the inductor in Henry's? What is the required value of the capacitor in Farad's?

Relevant Equations

$$\omega = \sqrt{\frac{1}{LC}}$$

Can someone explain the question so I can attempt a solution? What does it mean by input and output and why is that relevant?

 

[kuruman]

Homework Statement:: A resonant RLC circuit can be used as an amplifier for a certain band of frequencies around the resonant frequency. Consider a series RLC circuit as an audio band amplifier with an AC voltage source as the input, and the voltage across the 8.0Ω resistor as the output. The amplifier should have a gain (=output/input) of 0.5 at 200Hz and 4000Hz. What is the required value of the inductor in Henry's?
Relevant Equations:: $$\omega = \sqrt{\frac{1}{LC}}$$

Can someone explain the question so I can attempt a solution? What does it mean by input and output and why is that relevant?

It is relevant for answering the question. Stated differently: if the voltage at the AC voltage source (input) is $V_0$, what value should the inductor have so that the voltage across the 8.0Ω resistor (output) is $0.5*V_0$ at two frequencies, 200 Hz and 4000 Hz?

 

[Jaccobtw]

It is relevant for answering the question. Stated differently: if the voltage at the AC voltage source (input) is $V_0$, what value should the inductor have so that the voltage across the 8.0Ω resistor (output) is $0.5*V_0$ at two frequencies, 200 Hz and 4000 Hz?

So we know that half the voltage is taken up by the resistor which means the other half has to be taken up by the inductor and the capacitor so I came up with this:

$$L\frac{dI}{dt} + \frac{Q}{C} = 0.5V_{0}sin\omega t = iR$$

The question says the the circuit is at resonance so that means $X_L = X_C$

Where I'm confused is that in using this equation,

$$I_{0} = \frac{V_0}{\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}}$$

wouldn't the total resistance/impedance be $8\Omega$ because the inductive reactance and capacitive reactance cancel each other out? How can that be if there is only half the voltage across the resistor? Wouldn't there have to be resistance somewhere else?

 

[cnh1995]

The question says the the circuit is at resonance

Where does it say that?

Where I'm confused is that in using this equation,

I0=V0R2+(ωL−1ωC)2

Hint: You have two unknowns find, so you need two equations.

Can you go through your post #3 and write those two equations?
(Another hint: No calculus is required in this problem!)

 

[Jaccobtw]

Where does it say that?Hint: You have two unknowns find, so you need two equations.

Can you go through your post #3 and write those two equations?
(Another hint: No calculus is required in this problem!)

$$L = \frac{\sqrt{(\frac{V_0}{I_0})^{2} - R^{2}} + \frac{1}{\omega C}}{\omega}$$

$$C = \frac{1}{\omega (\sqrt{(\frac{V_0}{I_0})^{2}-R^{2}}+\omega L)}$$

 

[sophiecentaur]

Homework Statement:: A resonant RLC circuit can be used as an amplifier for a certain band of frequencies around the resonant frequency.

Can someone explain the question so I can attempt a solution? What does it mean by input and output and why is that relevant?

A passive circuit shouldn't really be described as an 'amplifier' because amplifiers use a power source (battery etc.) and an active device to increase the power of an input signal. This may or may not involve increasing the volts; increasing the available current without changing the volts can be useful too. An RLC circuit near resonance can 'magnify the voltage' of an input signal and that is more accurately described as a transformer. You never get more actual power with a transformer but you can produce a higher voltage. You may say I'm being picky but it's best to start off as you mean to go on and get things right.

 

[Steve4Physics]

Homework Statement:: A resonant RLC circuit can be used as an amplifier for a certain band of frequencies around the resonant frequency. Consider a series RLC circuit as an audio band amplifier with

Ignore the above (misleading) waffle except the part that tells you it's a series RLC circuit!

You are NOT being told that the circuit is operating at resonance (as already noted by @cnh1995).

an AC voltage source as the input, and the voltage across the 8.0Ω resistor as the output. The amplifier should have a gain (=output/input) of 0.5 at 200Hz and 4000Hz. What is the required value of the inductor in Henry's? What is the required value of the capacitor in Farad's?

Take the circuit to be series RLC.
Total applied voltage is the ‘input voltage’.
Voltage across the resistor is the ‘output voltage’.

Since it’s a series circuit, the current through each component is the same. Think AC potential divider.

 

[Jaccobtw]

Ignore the above (misleading) waffle except the part that tells you it's a series RLC circuit!

You are NOT being told that the circuit is operating at resonance (as already noted by @cnh1995).Take the circuit to be series RLC.
Total applied voltage is the ‘input voltage’.
Voltage across the resistor is the ‘output voltage’.

Since it’s a series circuit, the current through each component is the same. Think AC potential divider.

I understand. I'm just confused about how to find the two unknowns. I got 16 ohms for the impedance. Do I need both frequencies to solve for the the two unknowns?

 

[kuruman]

I understand. I'm just confused about how to find the two unknowns. I got 16 ohms for the impedance. Do I need both frequencies to solve for the the two unknowns?

Yes. Find an expression for the voltage across the 8 Ω resistor divided by the source voltage. That's the gain G that depends on frequency. Write equations for the gain at the two given frequencies. You will have a system of two equations and two unknowns, L and C.

 

[cnh1995]

I got 16 ohms for the impedance.

Yes.
Now you can use the impedance equation for the LC combination at the two given frequencies to find the values of L and C.

 

[Jaccobtw]

Yes. Find an expression for the voltage across the 8 Ω resistor divided by the source voltage. That's the gain G that depends on frequency. Write equations for the gain at the two given frequencies. You will have a system of two equations and two unknowns, L and C.

Yes.
Now you can use the impedance equation for the LC combination at the two given frequencies to find the values of L and C.

Not sure what I'm doing wrong. When I set up a system of equations for L and C, I don't see how having two frequencies makes a difference. Am I on the right track at least or no?

$$\sqrt{8^{2} + (X_L - X_C)^{2}} = 16$$
$$X_L - X_C = \sqrt{192}$$
$$\omega L - \frac{1}{\omega C} = \sqrt{192}$$
$$L = \frac{\sqrt{192} + \frac{1}{\omega C}}{\omega}$$
$$C = \frac{1}{\omega(\omega L - \sqrt{192})}$$

 

[cnh1995]

X_L - X_C = \sqrt{192}

Is this true at both the frequencies?
What happens to the reactances as you go from 200Hz to 4000 Hz?

 

[Jaccobtw]

X_L - X_C = \sqrt{192}

Is this true for both the frequencies?

I’d imagine at high frequencies, X_L - X_C is true and X_C -X_L at low frequencies is true

 

[cnh1995]

I’d imagine at high frequencies, X_L - X_C is true and X_C -X_L at low frequencies is true

Yes.
So you can write two equations with two unknowns from this and solve for L and C.

 

[kuruman]

Can you explain where this came from?
$$\sqrt{8^{2} + (X_L - X_C)^{2}} = 16$$

 

[Jaccobtw]

Can you explain where this came from?
$$\sqrt{8^{2} + (X_L - X_C)^{2}} = 16$$

My reasoning is that because half of the voltage goes across the resistor which is 8 ohms the other half must go across the inductor and capacitor making the total impedance 16. The current is constant throughout an entire series circuit

 

[kuruman]

That is faulty reasoning. The current in the inductor and the capacitor is not in phase with the voltage across these element as is the case with the resistor.

 

[cnh1995]

That is faulty reasoning. The current in the inductor and the capacitor is not in phase with the voltage across these element as is the case with the resistor.

Voltage across the resistor V_r= IR
I= Input Voltage/Impedance =V/Z.

Gain as defined in the problem=V_r/V=0.5
So, IR/V=0.5
(V/Z)*R/V=0.5 --> R/Z=0.5
Given that R= 8 ohm, the impedance is 16 ohm.

 

[Jaccobtw]

That is faulty reasoning. The current in the inductor and the capacitor is not in phase with the voltage across these element as is the case with the resistor.

You're right. I explained that poorly.

The gain is the voltage across the resistor over the total voltage which is 0.5

$$\frac{V_R}{V_t} = 0.5$$

$$V_R = iR$$
$$R = 8$$

Rearranging gives,

$$\frac{V_t}{i} = \frac{8}{0.5} = 16$$

 

[Jaccobtw]

Thank you @kuruman and @cnh1995 I finally figured it out

 

[kuruman]

Voltage across the resistor V_r= IR
I= Input Voltage/Impedance =V/Z.

Gain as defined in the problem=V_r/V=0.5
So, IR/V=0.5
(V/Z)*R/V=0.5 --> R/Z=0.5
Given that R= 8 ohm, the impedance is 16 ohm.

Right you are, I didn't immediately recognize the 16 on the right-hand side as 2R. Anyway, it seems OP has figured it out.

 

[cnh1995]

I’d imagine at high frequencies, X_L - X_C is true and X_C -X_L at low frequencies is true

I think this is where the concept of resonance is relevant.
For the given series RLC circuit, you have the same current at two different frequencies. This means the two frequencies are on either side of the resonant frequency (where X_L=X_C). So for the lower frequency, X_C > X_L and for the higher frequency, X_L > X_C.