Question

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 $800 \textrm{ k}\Omega$?

Final Answer

- $2.50 \mu \textrm{F}$
- $2.00 \textrm{ s}$

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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

*R*. 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.
## Comments

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

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

In reply to The answer for part b is… by austin004