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The flash on a high-tech camera contains a capacitor of 750 μ F. The battery in the camera supplies 330 V. (a) Determine the energy that is used to produce the flash. (b) Assuming that the flash lasts for 5 ms, find the power of the flash.

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Two equations for this one: Q=CV  & PE=(1/2)QV. 
C=750 * 10^-6 F , V=330 V . 
so if you use Q=CV=(750 * 10^-6)(330). Q=0.2475

Then solve for PE. PE=(1/2)QV= (1/2)(0.2475)(330)= 40.8 J
So PE=40.8 J

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