(50 points) iin(t) Given R1 = 4.5kΩ, C1 = 7μ F. a)....

(50 points) $i_{in}(t)$ \\ Given $R_1 = 4.5k\Omega$, $C_1 = 7\mu F$. a). Calculate the frequency response $\tilde{H}(f)$ in terms of $f$: $\tilde{H}(f) = $ $\Omega$ help (formulas). b). Calculate the magnitude and phase of the frequency response as functions of frequency $f$: $|\tilde{H}(f)| = \frac{[4.5*(10^3)]}{[sqrt(1+4*(\pi)^2*(f^2)*([4.5*(10^3)])]}$ $\Omega$ help (formulas). $\theta_{\tilde{H}}(f) = -tan^{-1}(2\pi f*(4.5*10^3)*(7*10^{-6}))$ help (formulas). c). Calculate the half-power frequency of the system. $f_{HP} = 5.053$ Hz help (numbers).

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00:01 The question is taken from physics and from electricity and magnetism.

00:05 And the question is, in an rlc circuit, if the napr frequency is 50 ,000 l and r are also given to us, the resonant frequency will be.

00:17 We know that in case of rlc circuit, generally the value of napa frequency is given as, and its value is zeta is equal to, alpha by omega 0 where zeta is given as 50 ,000 and we know that alpha can be given as r by 2 l okay so alpha is equal to r by 2 l into omega 0 so for the value of r and l are also given to us omega 0 is the resonant frequency so omega 0 is equal to the value of r is already given to us and the value of r is thousand divided by the value of l is equal to 10 millie henry so that will be 2 into 10 into 10 to the power minus 3 divided by 50 ,000 okay so this will give you the result thousand divided by 20 into 50 so that will be one hertz which one radian per second which is the required value of the resonant frequency but they are asking about resonant frequency so a resonant frequency is new 0 that is equal to 1 by 2 pi so that is 1 by 6 .28 so answer is none of the option is correct okay now the second part is during the transient analysis of the rlc circuit if the response vs is given by this so it could be so since the response is like vs is equal to 30 divided by s plus 1 into s plus 3 so you can solve it using the partial fraction so that is 30 into 1 over s plus 1 minus 1 over s plus 3 so this will give you value s plus 3 minus so that is divided by 2 okay so that will be 15 into 1 by s plus 1 minus 1 by s plus 3 so this is a kind of step function which will vary with a step okay so it is a step response of a series rlc circuit so answer is of a series rlc circuit okay and third part is in the rlc circuit if the napr frequency is 25 ,000 c is equal to 1000 nanofarad r is equal to 5 kilo the resonant frequency would be okay so in case of rlc circuit napper frequency is given as 25 ,000 okay so the value of a zeta is equal to alpha by omega 0 okay so omega 0 will be equal to alpha by zeta okay the value of alpha is r by 12 but since it in terms of in terms of a c it can be written as zeta is equal to r by 2 into square root c by l that we want the value of the value of alpha is equal to generally r by 2 l so other value of zeta is equal to r by 2 into square root c by l so that is equal to r by 2 into square root c by l okay so from here we can easily evaluate the value of l and we know that what is the resonant frequency so buying that value we can easily evaluate the value of resonant frequency so first zeta is equal to 25 ,000 and the value of r is 5 kilo which is 5 ,000 divided by 2 square root c is capacitance that is 100 into 10 to the power minus 9 divided by l we need to evaluate the value of l will be so this is 2500 this will be 10...

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