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The duration of a photographic flash is related to an $RC$ time constant, which is $0.100 \mu \textrm{s}$ for a certain camera. (a) If the resistance of the flash lamp is $0.0400 \Omega$ during discharge, what is the size of the capacitor supplying its energy? (b) What is the time constant for charging the capacitor, if the charging resistance is $800 \textrm{ k}\Omega$?
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Final Answer
  1. $2.50 \mu \textrm{F}$
  2. $2.00 \textrm{ s}$
Solution Video

OpenStax College Physics Solution, Chapter 21, Problem 65 (Problems & Exercises) (0:55)

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Video Transcript

This is College Physics Answers with Shaun Dychko. We know that the time constant for this photographic flash is 0.1 micro seconds which is 0.1 times ten to the minus six seconds and the resistance of a circuit during discharge is 0.04 Ohms. So the time constant tau is equals resistance multiplied by capacitance. And we can solve for the capacitance by dividing both sides by <i>R</i>. And so we have a time constant of 0.10 times ten to the minus six seconds and a circuit with a resistance of 0.04 Ohms. The capacitance must be 2.5 micro Farads. Now when this circuit is charging, because a flash has to charge, a capacitor has to charge in order to discharge to flash again later. The time comes that will be this resistance while charging multiplied by the capacitance. So that’s 800 times ten to the three Ohms times 2.5 times ten to the minus six Farads. And that gives 2.00 seconds.


Submitted by austin004 on Mon, 03/25/2019 - 21:48

The answer for part b is wrong, it should be 0.02s

Submitted by ShaunDychko on Tue, 04/02/2019 - 10:00

Hello, and thanks for sharing your feedback. I'm still getting 2.00s here. Perhaps there's a difference in how you plugged the 800 kOhms into your calculator, by maybe typing 8.00E3 instead of 800E3? The difference is in the decimal point in the first scenario. If you still think I've overlooked something, please let me know.

In reply to by austin004