Question

10) A camera flash consists of a 120 μF capacitor charged to 220V which discharges through a flashbulb with 5 Ω resistance. W

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Answer #1

In the solution below, at first i found out time constant then i calculated the discharged energy and in the end i derived a formula to find the amount of discharged energy for any time t .

As there is only resistor apart from capacitor whatever energy discharged must be dissipated across the resistor according to law of conservation of energy.

Power dissipated across resistor = voltage across resistor * resistance.

Solution

ven o= Voltage acoss the eapacitPY voltage drop acs caP vs,time is given b Energy dischargedi within 2t is dissipated coss d七

2て-- 2R 妾[i-r] (2 2.0)(600メー) [0.8, 2 x 5 2,85丁 g,尘xcxYo- I 0+ial stred ener .2.35 -times 2.s o-981 inhal stored energy -

The below graph represents the relation between voltage vs time.Where Vs = 220V and T = 600 microseconds.After 5T we may consider capacitor is fully discharged as almost all the charge is discharged(but remember that it doesnot gets completly dicharged).

Vs 0.5Vs Capacitor Dischaiging Voltage C apacitor Fully Discharged VC I 1T 0.7TI 2T 3T 4T 5T 6T Tirre Constant. (T) Time, t

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