Question

Laboratory experiments with series RLC circuits require some care, as these circuits can produce large voltages at resonance. Suppose you have a 1.00 - \(\mathrm{H}\) inductor (not difficult to obtain) and a variety of resistors and capacitors. Design a series RLC circuit that will resonate at a frequency (not an angular frequency) of \(60.0 \mathrm{~Hz}\) and will produce at resonance a magnification of the voltage across the capacitor or the inductor by a factor of 20.0 times the input voltage or the voltage across the resistor.

Short answer

Answer: To design a series RLC circuit that resonates at 60 Hz with a magnification factor of 20, we need a capacitance of approximately 7.056 x 10^-5 F and a resistance of approximately 37.662 Ohms.

Step-by-step solution

Resonance frequency formula

The resonance frequency, \(f_0\), can be determined using the following formula: $$f_0 = \frac{1}{2 \pi \sqrt{LC}}$$ Here, \(L\) is the inductor value (1 H), \(C\) is the capacitance, and \(f_0\) is the desired resonance frequency (60 Hz).

Q-factor formula

The Q-factor, or quality factor, of a series RLC circuit can be determined using the following formula: $$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$$ Here, \(R\) is the resistance, and \(Q\) is the Q-factor.

Magnification factor formula

The magnification factor, \(M\), for a series RLC circuit is equal to the Q-factor: $$M = Q$$

Calculate the capacitance

Using the resonance frequency formula, we can solve for the capacitance: $$C = \frac{1}{(2 \pi f_0)^2 \cdot L}$$ Substitute the given values: $$C = \frac{1}{(2 \pi (60))^2 \cdot 1}$$ $$C \approx 7.056 \times 10^{-5} \mathrm{~F}$$

Calculate the Q-factor

Next, we will calculate the Q-factor using the magnification factor formula, since \(M = Q\): $$Q = M = 20$$

Calculate the resistance

Now, we can determine the resistance with the Q-factor formula: $$R = \frac{1}{Q} \sqrt{\frac{L}{C}}$$ Substitute the calculated values of \(L\), \(C\), and \(Q\): $$R = \frac{1}{20} \sqrt{\frac{1}{7.056 \times 10^{-5}}}$$ $$R \approx 37.662 \mathrm{~\Omega}$$

Design the circuit

Finally, we have determined the necessary values for the series RLC circuit that resonates at 60 Hz with a magnification factor of 20: - Inductor: \(L = 1.00 \mathrm{~H}\) - Capacitor: \(C \approx 7.056 \times 10^{-5} \mathrm{~F}\) - Resistor: \(R \approx 37.662 \mathrm{~\Omega}\) The student should now create a series RLC circuit using these values to achieve the desired resonance frequency and magnification factor.

Key concepts

Resonance Frequency

Understanding resonance frequency in an RLC circuit is fundamental. It's the frequency at which the reactive components (inductance and capacitance) of the circuit have equal impedance but opposite in nature.
This results in their effects cancelling each other out, leaving only the resistive component. Therefore, the circuit behaves like a pure resistor at this specific frequency.

To find the resonance frequency, we use the formula: \[f_0 = \frac{1}{2\pi\sqrt{LC}}\]where:

• \(f_0\) is the resonance frequency.
• \(L\) is the inductance, given as 1 Henry in this exercise.
• \(C\) is the capacitance, which is what we'll need to calculate.

By achieving resonance, the voltage across either the capacitor or the inductor will be maximized for a given input voltage.

Q-factor

The Q-factor, or Quality factor, is a measure of how underdamped an oscillator or resonator is. In simpler terms, it tells us how sharp or narrow the resonance peak of the circuit is.
The larger the Q-factor, the smaller the bandwidth and the sharper the resonance.

In a series RLC circuit, the Q-factor can be calculated by:\[Q = \frac{1}{R} \sqrt{\frac{L}{C}}\]where:

• \(Q\) is the Quality factor.
• \(R\) is the resistance.
• \(L\) is the inductance.
• \(C\) is the capacitance.

In this exercise, the Q-factor is equated to the magnification factor, due to the requirement that the voltage be magnified by 20 times, thus requiring a Q-factor of 20 at resonance.

Capacitance Calculation

To determine the capacitance that will enable our circuit to resonate at a specified frequency, we can rearrange the resonance frequency formula to solve for \(C\):\[C = \frac{1}{(2\pi f_0)^2 \cdot L}\]Substituting the known values:

• Resonance frequency \(f_0 = 60 \text{ Hz}\)
• Inductance \(L = 1 \text{ H}\)

We calculate:\[C \approx \frac{1}{(2\pi \times 60)^2 \times 1} \approx 7.056 \times 10^{-5} \text{ F}\] This result tells us the specific capacitance needed to achieve resonance at 60 Hz.

Inductor Selection

The choice of the inductor plays a crucial role in determining the resonance frequency and overall behavior of the circuit. For convenience and precision, this problem specifies the inductor value as 1 Henry.
This is a common value for laboratory experiments.

Inductors store energy in the form of a magnetic field, and their opposition to changes in current is critical in defining the time constant and resonance characteristics of the circuit.
In practical circuit design, ensuring the inductor can safely handle the expected current without saturating or significantly altering its inductance due to core material non-linearities is important.

Resistor Calculation

The resistor in an RLC circuit fundamentally influences the overall damping.A properly chosen resistor ensures the circuit meets its specified magnification factor at resonance.
The required resistance can be found via the rearranged Q-factor formula:\[R = \frac{1}{Q} \sqrt{\frac{L}{C}}\]With:

• Q-factor \(Q = 20\)
• Inductance \(L = 1 \text{ H}\)
• Capacitance \(C = 7.056 \times 10^{-5} \text{ F}\)

We calculate:\[R \approx \frac{1}{20} \sqrt{\frac{1}{7.056 \times 10^{-5}}} \approx 37.662 \Omega\]This resistance ensures the circuit resonates correctly and achieves the desired voltage magnification.