let R=20kΩ, L=10mH, and C=40μF. Which angular frequency ω is attenuated most by this filter? (It would help you to first identify what type of filter this is.) Give your answer in terms of rad/sec, and omit units from your answer.
Well, let's put on our clown noses and figure this out! 🤡
Based on the values given, we can see that the circuit consists of a resistor (R), an inductor (L), and a capacitor (C). This combination is known as an RLC circuit.
Now, to identify the angular frequency (ω) that is attenuated the most by this filter, we need to find the resonant frequency of the circuit.
The resonant frequency (ω₀) of an RLC circuit is given by the formula:
ω₀ = 1 / √(LC)
Plugging in the given values:
L = 10mH = 10 * 10^(-3) H
C = 40µF = 40 * 10^(-6) F
ω₀ = 1 / √(10 * 10^(-3) * 40 * 10^(-6))
Now, let's do some math:
ω₀ = 1 / √(400 * 10^(-9))
Simplifying further:
ω₀ = 1 / (20 * 10^(-3))
Finally, we can calculate:
ω₀ = 50,000 rad/sec
So, the angular frequency (ω) that is attenuated the most by this filter is 50,000 rad/sec. Keep in mind that this answer is theoretical and based on the given values.
Hope this funny explanation brightened up your day! 🌞🎉
To determine the angular frequency that is attenuated the most by the filter, we first need to identify the type of filter.
Based on the given values of R, L, and C, we can see that these components form an RLC circuit.
An RLC circuit can be used as a low-pass, high-pass, or band-pass filter.
In a low-pass filter, lower frequencies are allowed to pass through relatively unaffected, while higher frequencies are attenuated.
To determine the angular frequency at which the filter attenuates the most, we can calculate the cutoff frequency (ωc) using the formula:
ωc = 1 / √(LC)
Given that L = 10mH (10 * 10^(-3) H) and C = 40μF (40 * 10^(-6) F), we can substitute the values:
ωc = 1 / √(10 * 10^(-3) H * 40 * 10^(-6) F)
Simplifying the expression:
ωc = 1 / √(400 * 10^(-9) Hz² * 10^(-6) F)
ωc = 1 / √(400 * 10^(-15) Hz² * F)
ωc = 1 / (√(400) * √(10^(-15)) * √(Hz² * F))
ωc = 1 / (20 * 10^(-8) Hz * sqrt(F))
Now, substituting the value of F = 1 / (2π * ω):
ωc = 1 / (20 * 10^(-8) Hz * sqrt(1 / (2π * ω)))
To find the ω that is attenuated the most by the filter, we need to find the value at which the cutoff frequency is maximum. This occurs when the denominator is minimum.
For the denominator to be minimum, the value inside the square root (√(1 / (2π * ω))) needs to be maximum.
The value inside the square root is (1 / (2π * ω)), so to make it maximum, we need to minimize the value of ω.
Since the numerator of the fraction is constant, minimizing ω is equivalent to maximizing the denominator. This implies that the ω that is attenuated the most is the smallest possible value of ω.
Therefore, the answer is ω = 0 (in rad/sec).
To identify the type of filter, we need to look at the values of the components in the circuit. In this case, we have a resistor (R), an inductor (L), and a capacitor (C). These three components form a series RLC circuit.
The behavior of an RLC circuit depends on the relationship between the resistance, inductance, and capacitance. There are three possible types of filters based on this relationship: low-pass, high-pass, and band-pass.
In a low-pass filter, low-frequency signals are passed through the circuit while high-frequency signals are attenuated. In a high-pass filter, high-frequency signals are passed through while low-frequency signals are attenuated. A band-pass filter allows a specific range of frequencies to pass through while attenuating frequencies outside that range.
In our case, the values of the components R, L, and C determine the type of filter. Comparing the values, we see that R is much larger than both L and C. This indicates that we have a low-pass filter.
Now, to determine the angular frequency (ω) that is attenuated the most by this filter, we need to find the resonance frequency of the circuit. At resonance, the impedance of the capacitor and inductor cancel each other out, resulting in a minimum impedance and maximum attenuation.
The resonance angular frequency can be calculated using the formula:
ω = 1 / √(LC)
Plugging in the given values of L and C:
ω = 1 / √(10mH * 40μF)
Before we can proceed, we need to convert the units to the same base unit. Converting 10mH to Henries (H) and 40μF to Farads (F):
ω = 1 / √(0.01H * 0.00004F)
ω = 1 / √(4 * 10^(-5)) rad/sec
ω = 1 / 0.002 rad/sec
ω = 500 rad/sec
Therefore, the angular frequency (ω) that is attenuated the most by this filter is 500 rad/sec.
Most attenuation will occur when the current is highest.
For an LCR circuit, as is given, this occurs at the condition of resonant frequency.
For the required condition,
ω = 1/(√LC)
Plug in the values of L and C.