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Integrated ConceptsA flashing lamp in a Christmas earring is based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. (a) What energy does it dissipate? (b) How much charge moves through the lamp? (c) Find the capacitance. (d) What is the resistance of the lamp?

$\begin{aligned} E &=0.125 \mathrm{J} \\ Q &=0.00417 \mathrm{C} \\ C &=13.9 \mathrm{mF} \\ R &=18 \mathrm{k} \Omega \end{aligned}$

Physics 102 Electricity and Magnetism

Chapter 21

Circuits, Bioelectricity, and DC Instruments

Direct-Current Circuits

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

Lectures

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So before solving this, I would like to say that answers given on the new page are wrong for Q. So let's force first solve for the energy energy is power multiplied by t the time the power is 0.5. At the time, it was honest 0.25 So if you multiply this, you will get answer here as 0.1 to 5 jewels. So I guess this is part a. Yes. For part B. L'm a charge most threw the lamp we need to find the current first current will be equal to power, divided by the potential difference. So power is on fire again divided by the potential of three Vote. If you do this calculation, you'll find the answer is 30.1666 and it goes on. So we don't need this value in this case because we're supposed to find the value of Q pure is equal to ay times T, which is equal to point 1666 times 0.25 So if you do this calculation, you find the value is zero point 0416666 So for 167 on DDE the value given on the numerous page as extra zero over here, which is wrong. So this is the correct value for part C. We're supposed to find the capacitance. Capacitance is simply charge, divided by the potential difference. The charge, as we found in the first parties 0.4167 divided by also 0.0 and then divided by three, we'll get zero point zero 12389 So if you convert this, this is fair it If you can work this to military it, you will get 13.8 95 men. If I did off this valley was correct on the page. So in the last part we're supposed to find the resistance off a lamp. Resistance is equal to potential, divided by current. We know the potential, which is three current, as we found here is 0.1666 So if you do this one three divided by 30.1 66 if you keep many six in this case you will find the answer is 18. Oh, it's not 18 cologne, but it's 18 over. So this answer is correct. Answered on the page is wrong by the factor of 10 power three

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