A 0.30 μF capacitor is connected across an AC generator that prod...

Question

A 0.30 μF capacitor is connected across an AC generator that produces a peak voltage of 10 V. What is the peak current to and from the capacitor if the emf frequency is (a) 100 Hz?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem in a domestic cooling system. A 0.40 micro Ferra capacitor is incorporated into a motor operating on a 20 volt ac power source with a frequency of 110 Hertz. What is the peak current flowing through the capacitor? So that's our end goal is we're ultimately trying to figure out what the peak current flowing through the capacitor is for this particular problem. And that will be our final answer. Awesome. We're also given some multiple choice answers and they're all in units of milly amps. So let's read them off to see what our final answer might be. A is 4.0 B is 5.5 C is 7.0 and D is 8.5. Awesome. So first off, let us write down all of our known variables. So we know the capacitance C is equal to 0.40 micro fads, which is equal to 0.40 multiplied by 10 to the power of negative six fa we also know the voltage which is equal to 20 volts. And we also know the frequency which is equal to 110 Hertz. So now our first step to solve this problem is to calculate the angular frequency omega. So to find the angular frequency, we need to note that the angular frequency omega is two pi times the frequency in units of Hertz. Thus, for 110 Hertz, the angular frequency will be found as. So the angular frequency is equal to two multiplied by pi multiplied by the frequency which is equal to when we plug in all of our known variables, two multiplied by pi multiplied by 110 hurts which is equal to 691.2 radians per second when you round to one decimal place. So now our second step is to solve for the capacitive reactance, which is denoted as XC. So let us recall and use the equation to solve for XC. So XC is equal to one divided by the angular frequency multiplied by the capacitance C. So now we can plug in all of our known variables to solve for XC. So one divided by 691.2 radians per second multiplied by our capacitance, which is 0.40 multiplied by 10 to the power of negative six Ferras. So we had to convert micro ferras to ferrets in order to keep our units consistent, which you, you will see in a moment why it's super important. So when we plug that into our calculator, we will get that. The answer for XC, which is the capacitive reactance is equal to 3617 homes. Awesome. So moving right along here, our final step is to calculate the peak current, which is gonna be denoted as IC, which is our final answer. So let us recall and use the equation to help us solver the peak current. So the peak current IC is equal to VC divided by XC. So now we can plug in all of our known variables and sulfur. So I see is equal to 20 volts because the voltage was given to us as 20 volts and we divided by our XC value which we just found together which was 3617 ohms. So let me plug that into our calculator. We will get 5.529 multiplied by 10 to the power of negative three amps. But we have a bit of a problem here. All of our multiple choice answers aren't milliamps. So we need to multiply this by 10 to the power of three to convert it from amps to milliamps. And when we do rounding to one decimal place, we will get 5.5 milliamps. And that will be our final answer for the peak current hooray we did it and this corresponds with multiple choice answer letter B 5.5 milliamps. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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A 0.30 μF capacitor is connected across an AC generator that produces a peak voltage of 10 V. What is the peak current to and from the capacitor if the emf frequency is (a) 100 Hz?

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