The circuit in Figure P4.59 models the charging circuit of...

The circuit in Figure P4.59 models the charging circuit of an electronic camera flash. The flash should be charged to $v_C \leq 7.425 \mathrm{~V}$ for each use. Assume $C=1.5 \mathrm{mF}, R_1=1 \mathrm{k} \Omega$, and $R_2=1 \Omega$. a. How long does it take the flash to recharge after taking a picture? b. The shutter button stays closed for $1 / 30 \mathrm{~s}$. How much energy is delivered to the flash bulb $R_2$ in that interval? Assume the capacitor is fully charged. c. If the shutter button is pressed 3 s after a flash, how much energy is delivered to the bulb $R_2$ ? (FIGURE CAN'T COPY) Figure P4.59

The circuit in Figure P4.59 models the charging circuit of an electronic camera flash. The flash should be charged to $v_C \leq 7.425 \mathrm{~V}$ for each use. Assume $C=1.5 \mathrm{mF}, R_1=1 \mathrm{k} \Omega$, and $R_2=1 \Omega$.
a. How long does it take the flash to recharge after taking a picture?
b. The shutter button stays closed for $1 / 30 \mathrm{~s}$. How much energy is delivered to the flash bulb $R_2$ in that interval? Assume the capacitor is fully charged.
c. If the shutter button is pressed 3 s after a flash, how much energy is delivered to the bulb $R_2$ ?
(FIGURE CAN'T COPY)
Figure P4.59

Transcript

00:01 So this question we are talking about the capacitors that are in the camera flashes.

00:07 In the start, the capacitors are connected in parallel.

00:12 And so to find the rc constant, we'll need what is the effective resistance as well as the effective capacitance.

00:20 So effective resistance is already given to be 100 oms.

00:25 For the effective capacitance, since they are connected in parallel, then we will just add them up together.

00:30 So 6 of them multiplied by 10 micro ferret for each of the capacitor.

00:40 60 microferrets.

00:43 So the rc time constant is just 100 oms multiplied by 60 mini ferrets.

00:54 It gives us 6 seconds.

01:00 Next, we want to find how long it takes for the capacitor.

01:06 To charge to 90 % of our battery.

01:13 So we're going to use the equation for the voltage, which is equals to v0, 1 minus e -t of negative t over r c.

01:29 When we want v to be 90%, it means that we want v to be equal to 9 over 10 times v0, right, where v0 is the terminal voltage of our battery.

01:44 We substitute this value of v into our equation, and we cancel the v0 on both left hand side and right hand side, we will get 9 over 10 equals to 1 minus e negative t over rc.

02:03 Now all we have to do is to rearrange a little bit.

02:06 So 1 over 10, e negative t over rc, then we take long of both sides.

02:20 Finally we can get our value of t to be close 10 .75 seconds.

02:40 Finally we want to find when it is discharging...

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