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5) The switch has been closed for a very long time The switch is opened a) What is the frequency of oscillation? 6) What is the maximum voltage on the 1OuF capaci...

Question

5) The switch has been closed for a very long time The switch is opened a) What is the frequency of oscillation? 6) What is the maximum voltage on the 1OuF capacitor?R- 50 0 ~2o_FIzv"mf40.25mh 22 0. Jo_h

5) The switch has been closed for a very long time The switch is opened a) What is the frequency of oscillation? 6) What is the maximum voltage on the 1OuF capacitor? R- 50 0 ~ 2o_F Izv "mf 40.25mh 22 0. Jo_h



Answers

A 12.0$\mu \mathrm{F}$ capacitor and a 5.25 $\mathrm{mH}$ inductor are connected in series with an open switch. The capacitor is initially charged to 6.20$\mu \mathrm{C}$ . What is the angular frequency of the charge oscillations in the capacitor after the switch is closed?

For this problem. On the topic of inductions were shown in Rlc circuit with an M 4 50 volts Resistance 250 owns And capacitance of 0.5 micro carrots. If the switch is closed for a long time and no voltage is measured across the capacitor. We want to know the inductions. L if the Voltage across the capacitor reaches a maximum value of 150 volts when the switch is opened. Now when the switch has been closed for a long time, the battery resistor and coil carry a constant current imax of epsilon over our, which is the maximum voltage divided by the total resistance. When the switch is opened, the current in the battery and resisting dropped zero. But the coil carries the same current for moment as oscillations begin in the L. C. Loop. We interpret the problem to mean that the voltage amplitude of these oscillations is delta v. And so by equating the energy stored in the capacitor half. See delta v squared is equal to the energy in the induct er a half hour I'm X squared. So by rearranging this equation we can solve for the inductions L. And so we get the induction cell to be the capacitance C. And the voltage delta v squared divided by the maximum current squared. And so this is the best and see times delta v squared times are squared over epsilon squared. Now all of these values are known. So we can find this conductance. This is a capacitance of 0.5 Times 10 to the -6 Ferrets, times 100 and 50 volts squared Times 250 home squared, divided by the IMF of 50 volts squared. This gives us the conductance of the conductor, how to be 0.281 Henry's

When the sewage is open. So in the sweeties open the battery is no longer a part of a sark. It so voltage equals zero. No, the finals guarding the circuit. He's fix voltage over five. Home equals one point toe ampere. So the voted across that 10 home register is so from here we can find it. The would be across the 10 ohm resistor is 1.2. Tampere must fell by 10. Home equals to as fault.

It's so first we can calculate the ah angular frequency. And then from angular frequency, we can get the frequency by falling this equation where F is the frequency and omega is the angular frequency. Now, in this problem, since we have the capacitor, the resistor and thean doctor So the frequency will be one over square. Root off one over l time see minus R squared, divided by four other squid. So now if we use the given numbers, we have one divided by l, which is 22,000,000 Henry's or 22 times into the power. Negative three Hendry. See, here is, um, 15 Nana Ferret. So that's 10 to the power negative nine carat. And then we subtract that from, ah, 75. I'm squared with his r squared than four times 22 times 10 to the bar. Negative three Henry. And then we were square there. So combining this we see that the angular frequency is 5.5 times 10 to the power for radiance per second. So now from there we can actually find the frequency. So f is again Omega by to buy. So in our case on my guys 5.5 times 10 to the power for radiance for a second. And we divide that by two fight. It gives us 8.76 times 10 to the bar. Three herds are 8.76 he lords. Now, for part B on amplitude decrees is as a function of time where e not is the amplitude at equal to zero. And then we have the exponential decay. So that's e to the power. Negative are over two l Thanks, t. So now if we saw for L from here, what we can do here is we can take long on both sides and then if we take long onboard sites, we have long base e e by easier. Where is a function of tea is equal to negative. Sorry. This is gonna be positive because we all reconsider negatives in here. So negative are by two l time, Steve. And from which we see that okay, time. So, basically, we need the time. So So we're solving for time from which we see that time is equal to negative twice. L longest e over year zero. And we divide that by our So we used the Given numbers we have. He'II is equal to negative toe 22 times 10 to the power three Hendry Sorry, negative tree because it's in Millie Henry. Then low base e 0.1 and we divide that by 75 homes, which is 1.35 times 10 to the power negative. Three second or 1.35 millisecond. Now for the final part, the critical damping can be calculated using this accretion so ours equal do square root off for l over seeing were seized the capacitance. So again, if we use the given numbers, we have 22 times 10 to the power negative three Hendry and let's do very dead by the capacitance, which is 15 dying. Stand to the bar. Negative nine Farid So that gives us 24 to 0 homes. All right, Thanks for watching. We'll see

On the topic off oscillations and lt circuits. We had given he self induct INTs as well as the capacitance, often oscillating Elsie circuit. And they're asked to calculate the frequency of the oscillations, as well as the maximum current in the circuit if we know the maximum potential difference between the plates of the capacitor. So, firstly, we need to calculate the angular frequency of oscillations. So in an L C circuit, the angular frequency omega is equal to the square root off one over L. C, where Alice the Induct INTs and sees the capacitance. Now, since all these values unknown, we can put them in and find these angular the Anglo frequency. So that's one divided by the self. Inducted is given as 2010 to 10 to the minus three Henry's 20 million Henry's Times the capacitance off one micro Farid one times 10 to the minus six parrots. This gives us the angular frequency to be 7000 and 71 0.7 radiance second, And since we know the angular frequency, we can calculate the frequency f, which is simply this angular frequency omega divided by two pi. Calculating this so we divide the answer by two pi. We get 1100 and 25 huts and that's all for a solution. That's the frequency off oscillations in the circuit. 1125 huts for the next part were given the potential difference between the plates off the capacitor and asked to calculate the maximum current. Now we know the energy stored in the magnetic field off the in dr is a half hour. I squared and this energy is equal to the energy stored in the magnetic electric field across the plates off the capacitor at a different time, half C v squared so we can actually rearrange this equation and soul for the maximum current. I provided that the voltage is at a maximum. So I is equal to the square root off, see V squared over our and again. These values are all known so we can calculate this current. So if you put our values in, this becomes the square root with the capacitors. One times 10 to the minus. Six parrots claims the potential difference across the plates of the capacitor 50 squared over the induct INTs, which is 20 times 10 to the minus three and calculating we get the maximum current in the circuit to be 0.35 and piers, that's


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