Question

a low pass π-connected symmetrical filter section has an inductance of 200mH in its series arm and capacitance of 400pF in each of its shunt arms. the cutoff frequency of the filter is? ans:25.16kHz ​

Answer 1

Answer:

Cutoff frequency is 35.61 kHz

Explanation:

Given : L = 200 mH = 200 * $10^{-3}$ H

C = 400 pF = 400 * $10^{-12}$ F

To find : Cutoff frequency

Solution :

We know that,

Cutoff frequency = $\frac{1}{\pi \sqrt[]{LC} }$

By applying formula,

Cutoff frequency =  $\frac{1}{\pi \sqrt[]{200 * 400 * 10^{-3} * 10^{-12} } }$

Cutoff frequency =  $\frac{1}{\pi \sqrt[]{2 * 4 * 10^{-11} } }$

Cutoff frequency =  $\frac{1}{\pi * \sqrt[]{0.8 *10^{-10} } }$

By solving above equation,

we get,

Cutoff frequency =  35.61 * $10^{3}$

∴ Cutoff frequency =  35.61 KHz

∴ The cutoff frequency is 35.61 KHz

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Answer 2

Concept:-

It might resemble a word or a number representation of the quantity's arithmetic value. It could resemble a word or a number that represents the numerical value of the quantity. It could resemble a word or a number that represents the numerical value of the quantity. It could have the appearance of a word or a number that denotes the quantity's numerical value.

Given:-

The given question is that a low pass π-connected symmetrical filter section has an inductance of $200$ mH in its series arm and capacitance of $400$ pF in each of its shunt arms.

Find:-

We have to find that the cutoff frequency of the filter is.

Solution:-

Applying cutoff frequency

Cutoff frequency $=\frac{1}{\pi \sqrt{LC}}$

By applying formula,

Cutoff frequency

$=\frac{1}{\pi \sqrt{LC}}\\=\frac{1}{\pi \sqrt{200 \times400 \times 10^{-3}\times 10^{-12}}}\\=\frac{1}{\pi \sqrt{2 \times4 \times 10^{-11}}}\\=\frac{1}{\pi \sqrt{0.8 \times 10^{-10}}}$

By solving above equation, we get,

Cutoff frequency$=35.61 \times 10^3$

∴ Cutoff frequency $= 35.61$ KHz

∴ The cutoff frequency is $35.61$ KHz.

Hence, the cutoff frequency is $35.61$ KHz.

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