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Individual work: Synthesize and no copy from the handout (Max page) 1) Draw circuit diagram for RLC circuit in AC and explain how you would determine the experiment...

Question

Individual work: Synthesize and no copy from the handout (Max page) 1) Draw circuit diagram for RLC circuit in AC and explain how you would determine the experimentally resonant frequency with the instruments in the laboratory:

Individual work: Synthesize and no copy from the handout (Max page) 1) Draw circuit diagram for RLC circuit in AC and explain how you would determine the experimentally resonant frequency with the instruments in the laboratory:



Answers

Find the resonant frequency in each ac circuit. Find the resonant frequency of a circuit containing a $33.0-\mu \mathrm{F}$ capacitor in series with a $43.5-\mu \mathrm{H}$ inductor.

In this problem I am frustrating the given data. So he has given that C. is equal to 10 New F. Which is equal to 10. Multiplication 10 to the power minus. Except The value of independence is given age 37.5 Multiplication 10 to the Power -6. At I just change the unit The Argentine frequency. Our physical to one by who By route and the L. C. So just putting the value in this X. Person, I can write the value of resonant frequency. Yet one by two pi route under 37.5 multiplication 10 to the power minus six. Edge multiplication 10 multiplication 10 to the power minus six. Have on simplification. I get the value of resonant frequency physical too. 8.22 multiplication 10 to the power three. Health changing the unit. I just multiplied by this value. So finally I get the value of resonant frequency at 8.2. Get get.

To solve this problem. First time writing the given data. So he had given that L. Is equal to one min wage. Changing the unit. I can write it as one multiplication, tend to depart minus six H. And capacitance is given edge. Poor new F. Which is equal to four multiplication 10 to the power minus six app regiment frequency F. Is equal to one by two pi aru tender. Elsie solving it further I can write the value of their physical to one by two pi route under L. That is one multiplication. 10 to the power minus six H. Multiplication for multiplication 10 to the par minus six F. So on further simplification I get the value of F. Is equal to 79.6 multiplication. Um 10 to the power three. H. Dad Changing the unit in Kh dad I just multiplied by this value. So I get the value of resonant frequency age 79.6. Kay at dead.

In this problem. First time writing the given data. So here, given that three is equal to 3.75 commune and which is equal to 3.75 multiplication and 10 to the power minus six app. And the value of independence is given as 300 edge. So she is 300 new edge. So I will change it in edge so I can write it as 300 multi pictures and 10 to depart minus six edge. Resonant frequencies given by a physical to one by to buy a route and Elsie. So putting the value here, I can write it as one by two pies. The route ended 300 multiplication 10 to the power minus six H multiplication three point 75 multiplication 10 to the power minus six L. On simplification I get the value of resonant frequency physical too. 4.75 multiplication 10 to the pole. Three hearts changing the unit. I just multiplied by this value. Which on simplification I get the value of resonant frequency eight 4.75 Kh Dead. This is our final answer.

Have to determine the resonating frequency if not, or omega not angular frequency. Okay and for the problem we have capacitance equals to four point double zero micro ferret. That is 10 to the power minus six forward. And in a tense L equals to five point double zero million Henry. That is 10 to the power minus three Henry. And resistance R equals to one point a double zero kill a room. That is 10 to the power plus three. And also we have maximum Walters. We m equals to 10 volt. And also we have to determine the maximum current in this circuit. I not. OK so first of all calculating graduating frequency omega not and it is given by one by underwrote of L. C. So substituting values, we get one by under root of L. Which is equal to five, multiplied by 10 to the power minus three. Multiplied by C. Which is 4 to 10 to the power minus six. Okay, so from here after solving We get omega not equals two 7070 radiant per second, radiant per second. Okay so from here we can also determine the angular frequency from the angular frequency. The resonating frequency. If not it will be omega not divided by two pi. And omega not is equal to 7070 divided by two pi. So from here f naught comes out to be 1126 hurts. Okay so these are the answer for the frequency. Okay so now uh now we will calculate the maximum current in the circuit and maximum current I note it will be given by maximum voltage divided by uh impudence gel. And at the resonance condition we will have jet equals two. Are so VM develop are so substituting values and we m is equal to 10.0 bowled over by the distance. Are is 1000 homes. So from here, peak current, I note comes out to be 10.0 milli ampere. Okay, so this is the answer for the peak current in the circuit. Okay?


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