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A circuit consists of a $2.00-\mathrm{kHz}$ generator and a capacitor. When the rms voltage of the generator is $0.800 \mathrm{V},$ the rms current in thecircuit is 0.515 $\mathrm{mA}$ (a) What is the reactance of the capacitor at2.00 $\mathrm{kHz}$ ? ( b) What is the capacitance of the capacitor?rms voltage is maintained at $0.800 \mathrm{V},$ what is the rms current at 4.00 kHz? At 20.0 kHz?

a) 1.6 $\mathrm{k} \Omega$b) 51 $\mathrm{nF}$c) 5.15 $\mathrm{mA}$

Physics 102 Electricity and Magnetism

Chapter 24

Alternating-Current Circuits

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Alternating Current

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to complete the reactors. We only have to remember that I are. A mass is equal to one over the reactor's times V are a mess. Then so be for the reactor's Get that the reactor'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's on DDE. The initial data given by the problem we have a capacitance that he'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'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'll be equals to 10 times. They need show every mess current and these easy goes to 5.15 million amperes

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