3) A 5000-? resistor and a 50-?F capacitor are connected in...

3) A 5000-? resistor and a 50-?F capacitor are connected in series at t = 0 with a 6-V battery. The capacitor is initially uncharged. Find, a. The current in the circuit at t = 0? b. The current at t = 0.5 s? c- The charge on the Capacitor at t=0.5s d. The maximum charge stored on the capacitor? e- if the battary is disconnected afetr a long time ( compare to the time contant of the circuit), find how long does it take the charge to drop to 50% of its maximum value.

Transcript

00:01 Hello, here we have to solve an assignment of five sub -questions and first we have to calculate the current in the circuit at time when time is zero.

00:11 So this current equals to voltage of the battery divided by the resistance of the resistor which is six volts over 5 ,000 tombs.

00:30 Let's calculate it that equals to 0 .00 12 amperes which is 1 .2 millamper.

00:53 Now let's calculate current at time 0 .5 seconds so and current can be calculated as following.

01:14 So first of all we need to write down the expression for for the charge of the capacitor as a function of time or actually we can just use a following expression for the current so current as a function of time equals to i0 times exponent power by negative t over tau where tau is a time constant and that equals to rc therefore at half a second this current equals to 1 .2 milamper times exponent power by negative 0 .5 seconds over 50 times 10 power by negative 6 ferrets times 500 tombs, 5 ,000 tom sorry let's calculate it that equals to 0 .16 milampeer now let's calculate the charge as a capacitor at this time and to do this we have to calculate we have to write down the expression which describes charge as a function of time so charges function of time for charging equals to maximum charge times one power one minus exponent of t or over rc.

03:33 So here q maximum charge and that will be the answer to question d equals to capacitance times voltage of the battery which is 50 microferrets so which is 50 times power by negative 6 ferrets times 6 volts that is 3 times 10 power by negative 4 columns.

04:07 So therefore q at half a second equals to 3.

04:15 Times 10 power by negative 4 columns times 1 minus exponent of negative 0 .5 seconds over tau.

04:31 Let's calculate it.

04:56 That equals to 2 .59 times 10 power by negative 4 columns, which is roughly 2 .6 times 10 power by negative 4 columns...

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