- #1

WJSwanson

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## Homework Statement

The RC circuit pictured below begins with capacitor C completely uncharged.

*I:*What is the current at point A immediately after the switch closes?

*II:*What is the current at point A 'at the end of the quaternary period' (I lol'd when he worded it like that) after the switch is closed?

*III:*'After a few proton half-lives,' (again, lol) the switch is opened by alien explorers. Let us call this time t

_{0}. After how much additional time will the charge on capacitor C equal half its charge when the switch was first opened?

*IV:*At that time, what is the current at point B?

[itex]E = 12.00V[/itex]

[itex]C = 15.00\mu F[/itex]

[itex]R_1 = 1000\Omega[/itex]

[itex]R_2 = 2000\Omega[/itex]

[itex]R_3 = 1500\Omega[/itex]

[itex]R_4 = 1000\Omega[/itex]

(NOTE: I will be using "E" to denote the emf because I can't find a script E.)

The RC Circuit:

## Homework Equations

Charging:

[itex] q = CE(1 - e^{(\frac{-t}{RC})})[/itex]

[itex] i = \frac{dq}{dt} = \frac{-CE}{RC} * e^{(\frac{-t}{RC})}[/itex]

Discharging:

[itex] q = q_0 e^{(\frac{-t}{RC})}[/itex]

[itex] i = \frac{dq}{dt} = \frac{-q_0}{RC} * e^{(\frac{-t}{RC})}[/itex]

## The Attempt at a Solution

*I:*Since the capacitor is uncharged at the beginning and point A is connected directly to the voltage source, the current across A is simply given by

[itex]i_A = \frac{E}{R_1} = \frac{12V}{1000\Omega} = 0.0120A = 12.00mA[/itex]

*II:*Because the capacitor is fully charged by the time the next geological period commences, there is no current across point A.

*III:*By the relation noted above for the charge on a discharging capacitor in an open RC circuit,

[itex]q = \frac{1}{2}q_0 = q_0 * e^{(\frac{-t}{RC})} \Rightarrow e^{(\frac{-t}{RC})} = \frac{1}{2}[/itex]

[itex] \frac{-t}{RC} = ln(\frac{1}{2} \Rightarrow t = -RC * -ln(2) = RC * ln(2)[/itex]

[itex]t = 1000\Omega * 1.500 * 10^{-5}F * 0.6931 = 1.040 * 10^{-2}s = 10.40ms[/itex]

*IV:*The current through capacitor C, resistor R

_{1}, and point A (as well as through the equivalent resistance of R

_{2}in parallel with R

_{3}and R

_{4}) goes as the quotient of the potential difference across the capacitor and the equivalent resistance of the circuit:

[itex]R = R_1 + (\frac{1}{R_2} + \frac{1}{R_3 + R_4})^{-1} = 2111\Omega

[itex]i = \frac{dq}{dt} = \frac{q_0}{RC} * e^{(\frac{-t}{RC})} = \frac{q_0}{RC} * e^{ln(1/2)} = \frac{q_0}{2RC} = \frac{CV_0}{2RC} = \frac{V_0}{2R} = \frac{12V}{4222\Omega} = 2.842mA = 2.842 * 10^{-3}A[/itex]

This is the part I'm less sure about:

I think that if you traverse the circuit counterclockwise (starting at C, going through R

_{1}, and then going through the parallel resistors) then the circuit follows the Kirchhoff rules. (It would follow them going clockwise as well, with the opposite signs compared to counterclockwise.)

So the potential across the capacitor is given by [itex]V = \frac{q}{C}[/itex]. This means that the potential across the wire connecting R

_{2}and R

_{4}to R

_{1}-- and thus at point B -- is given by

[itex]V_B = q_{0}e^{ln(1/2)} - iR_1 = \frac{9.000 * 10^{-5}C}{1.500 * 10^{-5}F} - 2.842 * 10^{-3}A * 1.000 * 10^{3}\Omega = 3.158V[/itex].

And since passing through the parallel resistors through either path leads to an equipotential surface with 0 potential, the potential drop across the R

_{3}& R

_{4}series and across R

_{2}(which is in series with point B) is 3.158V. Thus,

[itex]i_B = \frac{V_B}{R_2} = \frac{3.158V}{2000\Omega} = 1.579 * 10^{-3}A = 1.579mA[/itex]

**So to summarize the results:***I:*i

_{A}= 12.00mA = 1.200 * 10

^{-2}A

*II:*i

_{A}= 0

*III:*t = 10.40ms = 1.040 * 10

^{-2}s

*IV:*i

_{B}= 1.579mA = 1.579 * 10

^{-3}A