When a camera flash goes off, batteries immediately begin to recharge the flash\

Question

When a camera flash goes off, batteries immediately begin to recharge the flash's capacitor, which stores electric charge given by the following Q (t) = Q_0 (1 - e^-t/a) (The maximum charge capacity is Q_0 and f is measured in seconds.) Find the inverse of this function. Explain Its meaning. This gives us the time r with respect to the maximum charge capacity Q_0. This gives us the time: necessary to obtain a given charge Q. This gives us the charge Q obtained within a given time t. How long does it take to recharge the capacitor to of capacitor if a = 5

Explanation / Answer

Q = Q0(1 - e-t/a)

Q/Q0 = 1 - e-t/a

e-t/a = 1 - Q/Q0

-t/a = ln(1 - Q/Q0)

t = (-a) ln(1 - Q/Q0)

t = (-a) ln(1 - Q/Q0) tells us the time taken in seconds to recharge the flash's capacitor till it stores electric charge Q after the camera flash goes off.

(b)

Note: Since the question did not restrict us to using answer from (a) to do (b), we can use the given formula Q = Q0(1 - e-t/a) to solve part (b). You may prefer to do it in this manner, just in case your answer to part (a) is wrong. Alternatively, we can use answer from part (a), which will be a more direct method since we are trying to find the time taken for the recharging process. Anyway, both works fine by substituting Q = 0.9Q0 and a = 2. You can use both to check your answer.

I will use part (a) answer to solve this.

t = (-a) ln(1 - Q/Q0)

= (-5) ln(1 - 0.9Q0/Q0)

= (-5) ln(0.1)

= 11.512 (rounded off to 3 s.f.)

In conclusion, it takes 11.51 s to recharge the capacitor to 90% of capacity if a=5

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