Question 7.20: An electronic flashgun has a current-limiting 6-kΩ resistor ......
An electronic flashgun has a current-limiting 6-kΩ resistor and 2000-µF electrolytic capacitor charged to 240 V. If the lamp resistance is 12Ω , find: (a) the peak charging current, (b) the time required for the capacitor to fully charge, (c) the peak discharging current, (d) the total energy stored in the capacitor, and (e) the average power dissipated by the lamp.
(a) The peak charging current is
I_{1} = \frac{V_{s}}{R_{1}} = \frac{240}{6 × 10³} = 40 mA
(b) From Eq. (7.65),
t_{charge} = 5 R_{1}C= 5 × 6 × 10^{3} × 2000 × 10^{-6} = 60 s = 1 minute
(c) The peak discharging current is
I_{2} = \frac{V_{s}}{R_{2}} = \frac{240}{12} = 20 A
(d) The energy stored is
W = \frac{1}{2} CV^{2}_{s} = \frac{1}{2} × 2000 × 10^{-6} × 240² = 57.6 J
(e) The energy stored in the capacitor is dissipated across the lamp during the discharging period. From Eq. (7.66),
t_{discharge} = 5 R_{2}C = 5 ×12 ×2000 × 10^{-6} = 0.12 s
Thus, the average power dissipated is
p = \frac{W}{t_{discharge}} = \frac{57.6}{0.12}= 480 watts