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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 12 Ω, 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.

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(a) The peak charging current is

I_{1}=\frac{V_{s}}{R_{1}}=\frac{240}{6 \times 10^{3}}=40 mA

(b) From Eq. (7.65),

t_{\text {charge }}=5 R_{1} C=5 \times 6 \times 10^{3} \times 2000 \times 10^{-6}=60 s =1 \text { 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} C V_{s}^{2}=\frac{1}{2} \times 2000 \times 10^{-6} \times 240^{2}=57.6 J

(e) The energy stored in the capacitor is dissipated across the lamp during the discharging period. From Eq. (7.66),

t_{\text {discharge }}=5 R_{2} C=5 \times 12 \times 2000 \times 10^{-6}=0.12 s

Thus, the average power dissipated is

p=\frac{W}{t_{\text {discharge }}}=\frac{57.6}{0.12}=480 W

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