Ask question

Ask question

All categories

2 years ago

6

A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha

t is the resonant frequency of this circuit

1 answer:

ipn [44]2 years ago

4 0

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, $L=400\ \mu H=400\times 10^{-6}\ H$

Capacitance of parallel LCR circuit, $C=0.3\ \mu F=0.3\times 10^{-6}\ F$

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

$f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }$

$f=14528.79\ Hz$

or

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

You might be interested in

In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla

Harlamova29_29 [7]

Answer:

$\lambda'=78.086\ nm$

Explanation:

Given:

wavelength of light in the air, $\lambda=530\times 10^{-9}\ m$

time taken to travel from the source to the photocell via air, $t=16.7\ s$

time taken to reach the photocell via air and glass slab, $t'=21.3\times 10^{-9}\ s$

thickness of the glass slab, $x=0.87\ m$

<u>Now we have the relation for time:</u>

$\rm time=\frac{distance}{speed}$

hence,

$t=\frac{d}{c}$

c= speed of light in air

$16.7\times 10^{-9}=\frac{d}{3\times 10^8}$

$d=16.7\times 10^{-9}\times 3\times 10^8$

$d=5.01\ m$

For the case when glass slab is inserted between the path of light:

$\frac{(d-x)}{c} +\frac{x}{v} =t'$ (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

$\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}$

$v=4.42\times 10^7\ m.s^{-1}$

Using Snell's law:

$\frac{\lambda}{\lambda'} =\frac{c}{v}$

$\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}$

$\lambda'=78.086\ nm$

5 0

2 years ago

What is the number of complete oscillations of a wave per second

Ira Lisetskai [31]

I believe the term Frequency is what you are looking for.

8 0

2 years ago

Read 2 more answers

Why do high and low tides happen 1 hour later each day? just answer please and it is a written response

8_murik_8 [283]

With each<span> passing </span>day<span>, the </span>high tides occur<span> about an </span>hour later<span>. The moon rises about an </span>hour later each day<span>, too (actually, 54 minutes </span>later<span>). Since the moon pulls up the </span>tides<span>, these two delays are connected. As the earth rotates through </span>one day<span>, the moon moves in its orbit.</span>

6 0

2 years ago

Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

6 0

3 years ago

How can you predict the action of the element

patriot [66]

Stark contrast to paths on energy surfaces or even mechanistic reactions, rule-based and inductive computational approaches to reaction prediction mostly consider only overall transformations. Overall transformations are general molecular graph rearrangements reflecting only the net change of several successive mechanistic reactions. For example, Figure 1 shows the overall transformation of an alkene interacting with hydrobromic acid to yield the alkyl bromide along with the two elementary reactions which compose the transformation.

8 0

2 years ago