The capacitor in an RC circuit is discharged with a time const...

Question

The capacitor in an RC circuit is discharged with a time constant of 10 ms. At what time after the discharge begins are (a) the charge on the capacitor reduced to half its initial value and (b) the energy stored in the capacitor reduced to half its initial value?

Accepted Answer

Hello, fellow physicists today we're to solve for the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. How much time does it take for the charge on the capacitor in an RC circuit with a time constant of 20 milliseconds to I decrease to 1/4 its initial value I I additionally, after the discharge begins, how long does it take for the energy stored in the capacitor to decrease to 75% of its initial value? OK. So that's our end goal. Our end goal is to solve for two different answers. So we need to solve for an answer for part I and for part I I. So for part I, we're trying to figure out how long will it take time wise to decrease to 1/4 of its initial value for the charge to decrease to 1/4 of its value? We should be specific. And then for our second part, for part I I, we're trying to find out how long does it take for the energy stored in the capacitor to decrease to 75% of its initial value. Awesome. We're also given some multiple choice answers for the different answers. For part I and I, I, and let's note that for parts I and I, I, they're all in the same units of milliseconds. So let's read them off to see what our final answer pair might be. For a for part I is 28 and I, I is 1.2 for B for part I is 12 and I, I is 1.2 for C for part I is 28 and I, I is 2.9. And finally, for D for part I is 12 and I I is 2.9. Awesome. So first off, let us recall that the capacitor discharges through the resistor, which is denoted as capital R as Q equals Q subscript zero multiplied by E to the power of negative lowercase T divided by capital T. Now to solve for part I, we need to note that for Q equals Q subscript zero divided by four, we can then go on to write that Q subscript zero divided by four is equal to Q subscript zero multiplied by E to the power of negative lower case T divided by 20 milliseconds. OK. So now since for part I, we're looking for it to, we're using the information that we're decreasing to 1/4 of its initial value. We need to take the natural log of 1/4 and we need to set it equal to negative lowercase T divided by 20 milliseconds, but we need to convert milliseconds to seconds. So we just need to multiply it by 10 to the power of negative three seconds. Awesome. So then what we're trying to do is we're trying to solve for the amount of time because that's our end goal for how long it takes. So we need to isolate and solve for tea. So when we do that, we will get that Ln the natural log of 1/4 of so natural log of 1/4 multiplied by our 20 multiplied by 10 to the negative three seconds divided by negative one is equal to T. So then when we plug that into a calculator, we'll get that T is equal to 0.0 277 seconds. But we need to convert seconds, two milliseconds. So let us recall using dimensional analysis that in one second there is 1000 milliseconds. So our seconds will cancel out leaving us with milliseconds. So when we plug it into a calculator, we will get 27.7 milliseconds which rounding to one like rounding to the next whole number, we will get 28 milliseconds. And that is our final answer. For part I hooray, we did it. So one down one more to go. So moving right along to solve. For part I I let us recall that the energy equation is U is equal to a note that we're doing, we're focusing on the capacitor or the energy stored in the capacitor decreasing to 75% of its initial values. So we need to use 3/4 is at 75 multiplied by U subscript zero. So let's call this equation A and I'm gonna put it in blue inside blue parenthesis. So this is A and let's go on, we can then take it even further to write that U subscript zero is equal to Q subscript zero squared divided by two multiplied by C. Thus considering this, we can state that C which is the capacitance is equal to Q subscript zero squared divided by two multiplied by U subscript zero. Also, we can write that U is equal to Q squared divided by two multiplied by C. Thus, we can go on to write that see is equal to Q squared divided by two multiplied by U let's rewrite that that's a little sloppy. So to you. So two multiplied by you. OK, let's take a breath. Thus, we can then go on to write that Q subscript zero squared divided by two multiplied by U subscript zero is equal to Q squared divided by two multiplied by U. OK. So then using equation A, we could write that U subscript zero squared divided by two multiplied by U subscript zero is equal to Q squared divided by bracket. Two multiplied by three sports multiplied by U subscript zero bracket. OK. So we can simplify and write that Q squared is equal to 3/4 multiplied by Q subscript zero square. So isolating and solving for cue, we can simplify and write that Q is equal to plant the see square root of three divided by two parentheses multiplied by Q subscript zero. But we need to recall that. And let's put a little star here. We need to recall that Q equals Q subscript zero multiplied by E to the power of negative lowercase T divided by 20 milliseconds. Thus, parentheses square root of three divided by two is equal to E to the power of negative lower case T divided by 20 milly seconds. Thus, we can go on to write since we're trying to isolate and solve for T. Now that's what we're trying to do because we're trying to solve for the time because that's our angle. So we can go on to write that the natural log of the square root of three divided by two is equal to negative lowercase T divided by. And then we have to remember that we have to convert milliseconds to seconds. So 20 multiplied by 10 to the power of negative three seconds. Awesome. So then we can isolate and sulfur T now, so when we do that, we'll get that T is equal to the natural log of the square root of three divided by two multiplied by 20 multiplied by 10 to the power of negative three seconds all divided by negative one, which is equal to 0.00 28, 76 seconds. So like before we need to use dimensional analysis to convert seconds to milliseconds. So you just need to multiply it by 1000 milliseconds in one second, the seconds cancel out, leaving us with milliseconds. So we will get two point 87 milliseconds, but we need to round to one decimal place like the multiple choice answers are. So when we do that, we will get 2.9 milliseconds. And that is our final answer. For part. I I hooray, we did it. So looking at our multiple choice answers, that means our correct answer has to be the letter C which states that part I is 28 milliseconds and for part I I it is 2.9 milliseconds. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

The capacitor in an RC circuit is discharged with a time constant of 10 ms. At what time after the discharge begins are (a) the charge on the capacitor reduced to half its initial value and (b) the energy stored in the capacitor reduced to half its initial value?

Key Concepts

Time Constant

Exponential Decay

Energy Stored in a Capacitor

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