in a certain series rlc circuit irms 900 a vrms 245 v and the current leads the voltage by 330 a

Step 1: Calculate the total resistance of the circuit using the formula ( R = frac{V_{rms}^2}{I_

In a certain series rlc circuit irms 900 a vrms 245 v and...

In a certain series RLC circuit, Irms = 9.00 A, ΔVrms = 245 V, and the current leads the voltage by 33.0°. (a) What is the total resistance (in Ω) of the circuit? (b) Calculate the reactance (in Ω) of the circuit (XL - XC). (c) What if? If the voltage in the circuit is maintained by an AC source operating at f = 60.0 Hz and the resonant frequency of the circuit is f0 = 80.0 Hz, what are the values of C (in µF) and L (in mH) in the circuit? C = µF L = mH

Transcript

00:01 Hello students so in this question we have to find the four parts right so first of part that is we have to find the total resistance right so power here this is irms multiplied by brms cos of pi and we can write this as i square r rmsr right so this is power now for finding the resistance here we can write here v rms over cosfx that is divided by your i rms is resistance now we'll pull the values here vrm so this is to 45 and here cost of 33 degree five is 33 over irms i rms is here 9 ampere right now here resistance is coming out to do 20 to 0 .83 over so this is the value of resistance now coming to the b part here we have to find the reactance of the circuit that is xl minus xc so 10th 5 here is equal to xl minus xc over xl minus xc is here r multiplied by 10 of 5 right now this is here are that we have found 22 .83 10 of 33 degree so this is coming out to be here minus of 14 .8 o so value of xl minus x xe is here minus 14 .8 o so value of xl minus xe is here minus 14.

02:04 Point eight of it now let's come to the third part so here frequency f is equal to given as 60 hertz and f -not is here 80 hertz right now the resonate frequency that is 80 or we can write this as one over two by root lc so this is 80 right now will a square both the sides here so this is coming out to be 1 by 4 pi square l c multiplied by 6 4 double 0 right so from here we have to find here the value of c and l so we can write this as 1 over c voting 4 pi square l multiplied by 6 4 double and so this is the relation here now we can write here that 2 pi f l minus 1 over 2 pi f c is here minus of 14 .8 now 2 pi multiplied by 16 l minus 1 pi 2 pi 60 multiplied by c is equal to minus 14 .8 now putting the value of 1 by c in the equation here we'll get this as 2 pi 60 minus of 1 by 2 5 60 and this is multiplied by 4 pi square l multiplied by 6 4g000 right here this is the value of 1 by so it will multiply in the above no we will die in the power this is multiplied by 4 5 square l multiplied by 6400 so this is minus 14 point now here on solving we are getting l common and this year at 376 .992 minus 670 .208 so this is coming out to be minus 14 point 9 now l is here coming out to be 0 .5047 n right so this is here the value of inductor we can say this as 50 .47 millie amn we can say this as 50 .47 mm.

04:43 Now 1 by c so this is 4 multiplied by 3 .14 square and putting the value of failure 50 .47 multiplied by 10 to 4 minus 3 that is multiplied by 6400 so value of 1 by c actually we have to find c so reciprocate this and ultimately we'll find c as 7 .8 419 multiplied by 10 .m...

Question

Please give Ace some feedback