The diagram below shows a simple camera flash circuit. When you turn the flash u
ID: 2029996 • Letter: T
Question
The diagram below shows a simple camera flash circuit. When you turn the flash unit on, switch S1 is closed and the capacitor fully charges through a battery with voltage V0= 8.6 V and a resistor R1= 91 . When the shutter button of the camera is pressed, switch S1 opens, switch S2 closes for a time t2= 3.8×102 s , after which S2 opens and S1 closes for the capacitor to be charged for the next flash. Your task is to find the values of the capacitance C and resistance R2 such that:
1. The capacitor reaches 92% of the total charge in time t1= 3.02 s from a completely discharged state.
2. The energy delivered to the flash resistor R2 is E = 6.4×102 J before switch S2 opens.
Constants The diagram below shows a simple camera flash circuit. When you turn the flash unit on, switch S1 is closed and the capacitor fully charges through a battery with voltage Vo 8.6 V and a resistor R1 91 When the shutter button of the camera is pressed, switch S1 opens, switch S2 closes for a time t2 3.8x10-2 s after which S2 opens and S1 closes for the capacitor to be charged for the next flash. Your task is to find the values of the capacitance C and resistance R2 such that 1. The capacitor reaches 92% of the total charge in time ti-302 s from a completely discharged state. 2. The energy delivered to the flash resistor R2 is E-6.4x102 J before switch S2 opens. S2Explanation / Answer
1.
charge across capcitor q = qo (1- exp(-t/RC) )
q/qo = 0.92 = 1- exp (-3.02/91C )
C = 13 mF
2. energy stroed across the capacitor during charging = CV2/2
the capacitor is charged to 0.92 of its full charge.
energy = 0.922 * 13.0e-3 * 8.62 /2 = 407 mJ
= Q2/2C
Qo = 103 mC
energy discharged through the resistor = 64 mJ
energy retianed in the capacitor = 407-64 = 343 mJ
charge retained Q = 94.44 mC
while discharge charge across the capacitor = Q0 exp(-t/RC)
Qo = 103 , Q = 94.44
exp(-3.8e-2/R213e-3) = 94.44/103
R2 = 33.69 ohm