The flash unit in a camera uses a special circuit to “step up” the 3.4 V from th

Question

The flash unit in a camera uses a special circuit to “step up” the 3.4 V from the batteries to 300 V , which charges a capacitor The capacitor is then discharged through a flashlamp. The discharge takes 12s , and the average power dissipated in the flashlamp is 10W. What is the capacitance of the capacitor (in Farads)? Thanks!!! The flash unit in a camera uses a special circuit to “step up” the 3.4 V from the batteries to 300 V , which charges a capacitor The capacitor is then discharged through a flashlamp. The discharge takes 12s , and the average power dissipated in the flashlamp is 10W. What is the capacitance of the capacitor (in Farads)? Thanks!!!

Explanation / Answer

The capacitor is charged to a potential difference of V = 300 V
The time taken by the capacitor to discharge is t = 12 µs = 12 * 10-6 s The average power dissipated in the capacitor is P = 10 W We know from the relation P = (E/t) = [(1/2)CV2/t] or C = (2P * t/V2) where C is the capacitance of the capacitor

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