V0 r 4r c 6r the capacitor is uncharged and the switch is...

$V_0$ $R$ $4R$ $C$ $6R$ The capacitor is uncharged and the switch is open. (a) If we close the switch at $t=0$, what is the power supplied by the battery at $t=0$? If the resistors are lightbulbs, are any of them off? A long time passes by. (b) What is the power supplied by the battery now and what is the charge on the capacitor? We now open the switch and reset our clock to $t=0$. (c) What time is the charge on the capacitor 13% of the original value? (d) Find the power being dissipated in the $4R$ resistor as a function of time. Sketch a plot. (e) Now find the total energy dissipated up to some time $t$. Is you answer consistent with what you should expect $t \rightarrow \infty$?

Transcript

00:01 For this problem, we're considering the circuit that is given, and the switch of this rc circuit is closed at t equals zero.

00:13 We want to figure out what is the power dissipated in each of these resistors just after t equals zero right after the circuit is closed, as well as at the limit of t going to infinity, so after the capacitor has been fully charged.

00:32 For part b, we want to figure out what the charge on the capacitor will be after 0 .35 milliseconds.

00:41 For part c, we want to see how much total energy will be stored in the capacitor once it is fully charged at t equals infinity, or as t approaches infinity.

00:53 And for part d, we want to know if we were to double the voltage on the battery, how much does the answer to part c change? so for part a we want to find power, and power is going to be equal either to v squared over r for each resistor, or i squared times r, depending on if we know the voltage or the current going through each resistor.

01:26 The 24 -oom resistor is pretty easy.

01:30 The voltage across that resistor at t equals zero just after the switch is closed will be 15 volts because our capacitor will act like a wire as immediately after the switch has been closed so the power on the 24 -oom resistor is just going to be its voltage of 15 volts squared divided by its resistance of 24 alms for a grand total of 9 .4 watts.

02:11 The other two resistors are not so easy.

02:16 We don't know what the voltage across each of those will be.

02:20 I've chosen to instead use that power will be i squared times are because we can pretty quickly find what the current will be.

02:31 The two resistors there are in series with each other, so they will have the same current, and we can figure out what that current is by using the relationship that v equals ir.

02:50 So v equals ir, if we take the equivalent resistance of those two resistors, they will also experience a voltage or a potential difference of 15 volts.

03:04 And so we can, if we use the equivalent resistance, figure out what the current through that branch of the circuit will be.

03:14 The two resistors are in series, and in series you just add the resistances to find the equivalent resistance.

03:23 So for this case, the equivalent resistance is just going to be 19 .5 oms, which means that the current through that branch is going to be 0 .76923 amps.

03:50 Now that we have the current, it's pretty easy to find the power.

03:55 So for the 6 .5 oom resistor, it's going to be the current that we just found squared times the resistance for a grand total of 3 .85 watts.

04:10 And for the 13 -on resistor, it will again be the current we found squared times 13 to get 7 .69 watts of power.

04:24 At time t equals infinity, there's going to be zero current flowing through the circuit once the capacitor is fully charged.

04:33 Current will no longer flow.

04:35 So as t approaches infinity, power is going to be zero for all of the resistors...

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