Some camera flashes use flash tubes that require a high...

Some camera flashes use flash tubes that require a high voltage. They obtain a high voltage by charging capacitors in parallel and then internally changing the connections of the capacitors to place them in series. Consider a circuit that uses four AAA batteries connected in series to charge six $10-\mathrm{mF}$ capacitors through an equivalent resistance of $100 \Omega .$ The connections are then switched internally to place the capacitors in series. The capacitors discharge through a lamp with a resistance of $100 \Omega .$ (a) What is the $R C$ time constant and the initial current out of the batteries while they are connected in parallel? (b) How long does it take for the capacitors to charge to $90 \%$ of the terminal voltages of the batteries? (c) What is the $R C$ time constant and the initial current of the capacitors connected in series assuming it discharges at $90\%$ of full charge? (d) How long does it take the current to decrease to $10 \%$ of the initial value?

Transcript

00:01 Hi, here in this given problem, capacitance of each capacitor that is given as 10 millifarrid each and these are 6 overall in number.

00:26 So initially when they are in parallel, so their net capacitance is given as cp is equal to n times of c so it comes out to be equal to 60 milly ferret resistance put across this combination that is 100 oom now in the first part of the problem the time constant of such as circuit r c circuit which is known as r c time constant tau that will be given by the product of resistance with the equivalent capacitance of this circuit cp so here it is 100 om for the resistance and 60 into 10 to 1 minus 3 ferret for the capacitance so this time constant comes out to be equal to 6 .0 second one of the answer for the first part of this given problem now in the second part of the problem as there are four cells each providing 1 .5 volt and these cells are in series so net emf let it be e0 initial emf that will be 4 times of 1 .5 volt means this is 6 .0 volt so initial current because later on as time passes the current will start dropping down so initial current passing through the circuit, i .0, that will be given by e0 by r.

02:42 Means this is 6 volt by 100 or you can say i .0 will be equal to 0 .0 6 ampere.

02:52 Another answer for the first part of this given problem here.

02:55 Now in the second part of the problem, equation of the charging capacitor, equation of the potential difference, potential difference developed across a charging capacitor as these capacitors are being charged in parallel.

03:37 So at any instant potential across them will be given by v is equal to v .0 bracket 1 minus e rish bar minus t by tau.

03:53 Here t is the time at which we are finding this expression for potential.

03:58 T is the time constant.

04:00 Here for v0 it will be e0 means the maximum potential and for v it is given as 90 % of the maximum voltage.

04:10 Means here it will be 0 .9 times of e0.

04:15 1 minus e rach the bar minus t by tau.

04:20 So canceling this e not, finally we get e raced bar minus t by 6 for the time constant is equal to 1 minus 0 .9 which comes out to be equal to 0 .1...

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