a-220-muf-capacitor-is-connected-to-an-ac-generator-with-an-rms-voltage-of-118-mathrmv-and-a-frequen

Question

A 22.0 -\muF capacitor is connected to an ac generator with an rms voltage of 118 $\mathrm{V}$ and a frequency of 60.0 $\mathrm{Hz}$ . (a) What is the rms current in the circuit? (b) If you wish to increase
the rms current, should you add a second capacitor in series or
in parallel? (c) Find the value of the capacitance that should be
added to increase the rms current to 1.50 $\mathrm{A} .$

João Gabriel Alencar Caribé · Accepted Answer

they are the most current off. This circuit is given by by you&#x27;re a mask is equals to two pi times the frequency times the capacitance times our mass voltage and using the given values off the problem. We got anonymous current off to pipe times 60 times 22 my crew times 118 And these these equals to 0.9786 which is approximately 0.979 m peers in the second item. We wish to increase the amass current. So no, stat, the R. M s current is proportional to the capacitance. So in order to increase the R. M s current, we must increase the capacitance and to increase the capacitance, we must add the next capacitor in parallel with the existing one. So in part, a little in the last item, we first discovered the value off the result in capacities and only then we discover their value off the new capacity. So to discover their value off the result and capacitance, we use the fact that I prime Ari Mass Zico&#x27;s truth to pipe times the frequency of times the resulting capacitance times they are amassed. Voltage solving for C prime, we get that the capacitance is equals two I prime every mass divided by to pi times the frequency times that Ari mass voltage And these these equals truth 1.5 divided by to pie times 60 times 118. And these easy close to 0.337 which is approximately 34 me grow for its off resulting capacitance. No, we used the fact that we are that this new capacitor in parallel with the existing one So we&#x27;ll see prime is equals two. 20 to me grow far It&#x27;s off the existing capacitor plus the new capacitor therefore the new capacity or is he goes to the resulting capacitance My nose 22 kids equals two authority four minus 22. Then we should include the new capacitor Wi fi capacitance off approximately 12 me grow parents

João Gabriel Alencar Caribé · Answer

to complete the reactors. We only have to remember that I are. A mass is equal to one over the reactor&#x27;s times V are a mess. Then so be for the reactor&#x27;s Get that the reactor&#x27;s Izzy close to V ROMs divided by I a mess. Then, using the data that was given by the problem, we get that the reactors is opposed to 0.8, divided by 0.515 mealy and piers, Um thes is approximately 155 treat homes, and we can write it as being approximately equals to 1.6 kilo Holmes. To copulate the capacitance, we have to remember that the reactant is given by one divided by two pi times the frequency times the capacitance. Then the capacitance is given by to pipe times. He frequents times the reactors. No using our intermediary result for the reactor&#x27;s on DDE. The initial data given by the problem we have a capacitance that he&#x27;s given by one divided by to pie times the frequency where is too kilo hurt times the capacity of the reactors, which is 0.8 divided by 0.515 I&#x27;m Stan put in my industry at the Z Z equals true um, zero 0.515 time stand to the minus tree divided by two pi times two times stand to the third times 0.8 and this is approximately 5.1 times 10 to the minus eight for its and it can be reading us 51. No, no for its Finally, for the last item, we just have to remember that there are a mess. Current is proportional to the frequency of the sewers. So in the first case there frequents off the source is more to play it by a factor of two. Then the R. M s current in this case will be the R. M s current in the previous case to play it by two. So the R. M s current here is two times 0.515 which is equals true. One point. This is the road tree, really m peers. And finally, in the last case, we have a frequency that is 10 times bigger. Then they need show frequency. So the r m s current here, we&#x27;ll be equals to 10 times. They need show every mess current and these easy goes to 5.15 million amperes

Mayukh Banik · Answer

impotence is given by one over Omega Psi. It was one of our Dubai FC. So putting the values in Dubai times the frequency 50 hearts times the cab assistance is quartered micro fallowed so quarter micro fatted in words of that solving it comes out to be 12 points. Seven key, though Oh, no. The art amiss garden with B the our team is full just divided by the impotence It was Omega Kappa students times the RMS voltage equals to buy f c the are you miss? It was to buy times 50. Actually, we don&#x27;t have to do any of those because we have found ah xsi So we can just put the numbers in right now the enemy&#x27;s voltage is do 20 village divided by 12.7 give the old equals 17 me the ampere.

Breanna Kloczkowski · Answer

you guys so to yourself. This problem, you&#x27;re going to start as always, by defining our variables that we&#x27;re gonna be using in the problem. So we&#x27;re gonna play frequency of 750 years, a peak output voltage of 140 and then our RMS current in the circuit is money. Three. There you go. So we&#x27;re gonna be funny. Two things in this problem. So first of all, we&#x27;re going to be finding the capacitance of the capacitor, and then we&#x27;re going to finding the magnitude of the maximum charge on one plate A B capacitor. So to find this, we are going to sir, um by choosing which Christians to you. So we know there are people. Titch is 140 b and we&#x27;re sorry, guys, and we&#x27;re going to refer Teoh that as Visa P. And so with our our mass voltage, we&#x27;re gonna use the question to be R m s. It goes gp over the skirt of to and then that work it US 98.99 And that was our mess. Voltage. So now we can plug it in it Teoh recreation that uses a visa our mess. So we get this, um, the justice employed version of the first part of the equation that he wrote down. So now I can plug in our variables, so But so this is our answer. Purport a I&#x27;m ask you, you&#x27;re in like this. Um, through this is a three. But I think about closet in this format, so no, we&#x27;re going todo park feet. Everybody fined the maximum charge. So the maximum charge is huge. Chicken equals thes of you, etc. So now that way have, like, region, we literally just not Teoh in Tom. The numbers we&#x27;re gonna using and we get our answer, which is nine times something I get 1/4.

Averell Hause · Answer

so far today we have the capacitive reactor. It&#x27;s this is gonna be equaling one over to pie times the frequency time. See the capacitance and this is gonna be equaling one over to pi multiplied by the frequency of 60.0 hurts times the capacitance of 12 Micro Farid Or we can say 12.0 times 10 to the negative Sixth Farage&#x27;s and we have the capacitive reactor. It&#x27;s 221 homes. This would be our final answer for part A. For part B, we have the root means squared current. This would be equaling the root means squared voltage subsea and the capacitive reactant ce. So this would be equaling 36 0.0 votes, divided by 221 homes giving us a point 163 amps, four parts. See, Then we find that the maximum current is equaling radical two times the army&#x27;s current and this would be radical too multiplied by 0.163 amps and this is giving us point 231 amps. So this would be our final answer for part C and then for part D, we can say no. And this is because we can say the charge is a max when bolted JJ is a minimum. And then we can say that. But the voltage cross a capacitor is 90 degrees out of phase with the current. That is the end of the solution. Thank you for watching.

João Gabriel Alencar Caribé · Answer

by connecting a second capacitor in Siri&#x27;s with the 1st 1 we will end up with a result and capacitance that is smaller done. They need show capacitors and also notice that the IRA mass current is equals two to pi times the frequents times, the capacitance times they are a match voltage. Notice that by increasing the capacitance we will increase the armas voltage and by decreasing the capacitance, we will decrease. They are a master politician. So by including a new capacitor, we are decreasing the capacitance. So we are decreasing the R. M s correct that will flow through that circuit. In the second item, we are connecting the two Capasa pers in Siri&#x27;s these one and he&#x27;s won the result of capacity off The association is given by C. R. Zico&#x27;s too see time see two divided by C plus C two and these years equals two 55.5 times 38.7, divided by 55.5. Close 38.7 and it is approximately 22 0.8 me crew forwards of capacitance. And now we can use the following relation between the ROMs, current and the capacitance before and after, including the new capacitor, The relation is the following I, Ari Mass divided by seat disease before including the new capacitor easy close to I ROMs prime divided by CR disease after including the new capacitor. So by using this relation, we get that the Larry Mize current after including the new capacitor easy goes to CR divided by C times I. R s and these is equals two 22 0.8 divided by 55 0.5 times 1.21 and it is approximately 0.6 to 2. Amperes noticed that these relation holds because we are on Lee, including a new capacity. So we are only changing the capacitance off the circuit. We are not changing the voltage, nor the frequency to change the voltage or the frequency. We have to change the soldiers and we are not doing that. So these is why we can use these relations

Problem

Find the impedance of a 60.0 -Hz circuit with a $…

Problem 19 Easy Difficulty

Answer

a) 0.979 $\mathrm{A}$
b) parallel
c) 11.7$\mu \mathrm{F}$

Related Courses

Physics

Related Topics

Discussion

Top Physics 102 Electricity and Magnetism Educators

Christina K.

Aspen F.

Jared E.

Meghan M.

Physics 102 Electricity and Magnetism Courses

Electric Charge

The Electric Force

Recommended Videos

Watch More Solved Questions in Chapter 24

Video Transcript

James S. Walker

Physics

Christina K.

Aspen F.

Jared E.

Meghan M.

Electric Charge

The Electric Force

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

a) 0.979 $\mathrm{A}$b) parallelc) 11.7$\mu \mathrm{F}$

Physics

Christina K.

Aspen F.

Jared E.

Meghan M.

Electric Charge

The Electric Force

James S. WalkerPhysics

Christina K.

Aspen F.

Jared E.

Meghan M.

a) 0.979 $\mathrm{A}$
b) parallel
c) 11.7$\mu \mathrm{F}$

James S. Walker

Physics