Problem

The maximum current in a $22-\mu \mathrm{F}$ capa…

05:10

Problem 13 Easy Difficulty

The rms current through a $55.5-\mu$ F capacitor is 1.51 A when it is connected to an ac generator. (a) If a second capacitor is connected in series with the first, will the rms current
increase or decrease? (b) Calculate the rms current when a $38.7-\mu \mathrm{F}$
capacitor is connected in series with the $55.5-\mu$ F capacitor.

Answer

a) rms current also decreases.
b) 0.62 $\mathrm{A}$

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Physics 102 Electricity and Magnetism

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Video Transcript

by connecting a second capacitor in Siri'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's these one and he's won the result of capacity off The association is given by C. R. Zico'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

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