 #1
CAF123
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Homework Statement
The power supply for a large flash lamp is shown below where a 500 V DC power supply changes up at 10 μF capacitor that then discharged through the flash lamp.
Such a system is potenitally lethal, so to reduce risk a bleed resistor Rb is connect across the capacitor to discharge it when the DC power supply is disconnected. The safety specification is that the capacitor should be reduced to a safe voltage (30 V), 15 second after the DC power supply is disconnected. Calculate,
1. the required value of Rb,
2. the maximum power dissapated in Rb when the system when(??) the DC power supply is connected.(disconnected?)
The Attempt at a Solution
1) I think this is just a use of the equation $$V(t) = V_o exp\left(\frac{t}{RC}\right),$$with V = 30, V_{0} = 500, t = 15 and C = 10μF and then solve for R.
2) I don't really understand the question, particularly because I don't understand the english used. What I have though is that ##P_R(t) = ##power dissipated in the resistor = ##I_R(t) V_R(t) = I_2^R(t) V_R(t).## Should I first find the time it takes for the capacitor to charge up to 500V. I am presuming that the 'connected' there should be 'disconnected'.
Many thanks.
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