flash camera

A flash camera has a 2.5 V bulb with a constant resistance of 112 ohms and its in parallel with a 0.0077 Farad capacitor. The capacitor is in a loop with a 215 ohm resistor and a battery 2.50 V with internal resistance 2.15 ohms. The resistor loop and the bulb loop are have separate switches such that when the resistor switch is closed the capacitor charges up. From the total resistance ( 216.2) and the capacitance ( .0077F) we can calculate the time constant as 1.66 s. We can work out the current from Kirchoff's second law( closed loop V = 0) so 2.50 = 1.20xI + 215xI. This gives a current of 0.0116 amps.
Once the capacitor is fully charged(2.50V) we can open the resistor switch, close the bulb switch and the capacitor discharges through the bulb giving us a short flash as 63% discharges in 1.66 s and a further 63% in another 1.66s ( i/e1 is ~63%).
If we now close both switches the capacitor will recharge but it can be shown that the current is 0.00739 and therefore voltage across the 215 resistor is 1.64 V. Thus a voltage of only 0.85 across the capacitor means it cant make the bulb flash if you try to discharge it to the bulb. So the capacitor switch has to be set to close to get another flash
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Kiwiman

Author:Kiwiman
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