- #1
Learnphysics
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Hi guys,
This is a basic circuit that you'd commonly see in a flash Camera.
Hambley says (roughly paraphrased):
When the electronic switch is toggled rapidly (1000s of times a second) the inductor charges due to a change in current. v=L(di/dt).
The diode allows current to flow into the capacitor to create a build up of electrons on one of it's plates creating a growing voltage cross the capacitor. Once it reaches 100s of volts on the positive plate of the capacitor, The Shutter switch closes discharging the capacitor into the flash tube.
Fair enough.
But I have some questions:
Firstly, Why do we even need an inductor? What if we were to simply attach the battery (DC voltage) to a capacitor (in parralell) and let it charge to 100V. I understand this is dependent on the size of the capacitor by: v= 1/C(integral(idt)).
(Speaking of which, and not to get too distracted but, the above formula seems to suggest that voltage across a capacitor is inversely proportional to it's capacitance...Why is this the case I'd figure that the voltage across a capacitor would only INCREASE if we had large capacitors.)
I know I'm probably missing something fundamental when asking this question.
This is a basic circuit that you'd commonly see in a flash Camera.
Hambley says (roughly paraphrased):
When the electronic switch is toggled rapidly (1000s of times a second) the inductor charges due to a change in current. v=L(di/dt).
The diode allows current to flow into the capacitor to create a build up of electrons on one of it's plates creating a growing voltage cross the capacitor. Once it reaches 100s of volts on the positive plate of the capacitor, The Shutter switch closes discharging the capacitor into the flash tube.
Fair enough.
But I have some questions:
Firstly, Why do we even need an inductor? What if we were to simply attach the battery (DC voltage) to a capacitor (in parralell) and let it charge to 100V. I understand this is dependent on the size of the capacitor by: v= 1/C(integral(idt)).
(Speaking of which, and not to get too distracted but, the above formula seems to suggest that voltage across a capacitor is inversely proportional to it's capacitance...Why is this the case I'd figure that the voltage across a capacitor would only INCREASE if we had large capacitors.)
I know I'm probably missing something fundamental when asking this question.