The figure below shows a simplified circuit for a photographic flash unit of Bob
ID: 1282392 • Letter: T
Question
The figure below shows a simplified circuit for a photographic flash unit of Bob's new camera. This circuit consists of a 9.0 V battery, an R = 40.0 k resistor, a 140 µF capacitor, a 40 ? flashbulb, and two switches. Take the capacitor to be uncharged initially. When Bob is not taking a picture, the charging side of the circuit is enabled (switch S1 only is closed). Whenever Bob does press the camera button to take a picture (that requires a flash), this closes switch S2 (and opens S1).
a.) How long does it take to charge the capacitor to 5.4 V? in seconds
b.) Now, suppose Bob decides to take a photograph.
How much energy is stored in the capacitor at this moment? Joules
What will be the initial current flowing through the bulb? Amps
How long will the capacitor take to lose 99% of its charge? s
(How does this time compare with your answer from part a?)
Please provide step-by-step explanation. Thank You!
The figure below shows a simplified circuit for a photographic flash unit of Bob's new camera. This circuit consists of a 9.0 V battery, an R = 40.0 k resistor, a 140 MuF capacitor, a 40 flashbulb, and two switches. Take the capacitor to be uncharged initially. When Bob is not taking a picture, the charging side of the circuit is enabled (switch S1 only is closed). Whenever Bob does press the camera button to take a picture (that requires a flash), this closes switch S2 (and opens S1). a.) How long does it take to charge the capacitor to 5.4 V? in seconds b.) Now, suppose Bob decides to take a photograph. How much energy is stored in the capacitor at this moment? Joules What will be the initial current flowing through the bulb? Amps How long will the capacitor take to lose 99% of its charge? s (How does this time compare with your answer from part a?) Please provide step-by-step explanation. Thank You!Explanation / Answer
a) Time constant, T = R*C = 40000*140*10^-6 = 5.6 s
we know,
V = Vo*(1 - e^(-t/T)
5.4 = 9*(1-e^(-t/T)
5.4/9 = 1 - e^(-t/T)
e^(-t/T) = 1 - 5.4/9
e^(-t/T) = 0.4
t = -T*ln(0.4)
= -5.6*ln(0.4)
= 5.13 s
b) U = 0.5*C*v^2
= 0.5*140*10^-6*5.4^2
= 2.041*10^-3 J
c) I = 5.4/40 = 0.135 A (when s2 is closed)
d) now time constant, T = R*c = 40*140*10^-6 = 5.6*10^-3 s
q = Qmax*e^(-t/T)
0.01*Qmax = Qmax*e^-t/T)
0.01 = e^(-t/T)
t = -T*ln(0.01)
= -5.6*10^-3*ln(0.01)
= 25.79*10^-3 s
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