What Is the Resistance Value in an RC Circuit with a 37V Drop After 2 Seconds?

[huykma]

Homework Statement

A simple RC Circuit consisting of a 100.0 V-battery in series with a 10.0- uF capacitor and a resistor. Initially, the switch is open and the capacitor is uncharged. Two seconds after the switch is closed, the voltage across the resistor is 37V.
Determine the numerical value of the resistance R.
A) 2.0 x 10^5
B) 5.0 x 10^4
C) 0.37
D) 4.3 x 10^5
E) 2.70

Homework Equations

q = qo e^(-T/RC)

The Attempt at a Solution

I'm having trouble starting it at all, could I get a point in the right direction to get it going?

 

[kuruman]

Your equation provides the charge as a function of time for a discharging capacitor. You need the equation that provides the voltage as a function of time for a charging capacitor.

 

[huykma]

Your equation provides the charge as a function of time for a discharging capacitor. You need the equation that provides the voltage as a function of time for a charging capacitor.

so that would be...

q = qo [1-e^(-t/rc)]

 

[kuruman]

so that would be...

q = qo [1-e^(-t/rc)]

The time dependence is correct, but the conventional symbol for voltage is V not q. So replace q with V and q₀ with V₀. At this point the problem is a straightforward plugin. Of course, first you need to solve algebraically for R in terms of the other quantities.

 

[huykma]

ah that makes much more sense, maybe because my professor studied in berlin, so his formula sheet has q and qo...

what i ended up getting is...
37 = 100(1-e^(-2/rc))

37/100 = (1-e^(-2/rc))

ln (63/100) = -2/rc

R = -2 / (ln 63/100)(10^-6)

= 4.3 x 10^6...

which is incorrect...maybe I plugged in something wrong?

 

[Dmytry]

The voltage across the resistor is actually same as if you had capacitor charged to 100v and discharging through that resistor. The differential equation is same, dv/dt=-v/(rc) when the v is voltage across the resistor.
You get
100*e^(-t/rc)=37

The current depends to voltage across resistor, the battery and discharged capacitor in series are here instead of charged capacitor to check if you understood that. Battery and capacitor in series behave exactly the same as just the capacitor charged to same voltage as the sum of capacitor's and battery's voltages.

You mixed up voltage on resistor with voltage across capacitor.
You could of used
63 = 100(1-e^(-2/rc))
where 63 is the voltage across capacitor, which gives
63 = 100(1-e^(-2/rc))
0.63 = 1-e^(-2/rc)
0.37=e^(-2/rc)

 

[kuruman]

ah that makes much more sense, maybe because my professor studied in berlin, so his formula sheet has q and qo...

what i ended up getting is...
37 = 100(1-e^(-2/rc))

37/100 = (1-e^(-2/rc))

ln (63/100) = -2/rc

R = -2 / (ln 63/100)(10^-6)

= 4.3 x 10^6...

which is incorrect...maybe I plugged in something wrong?

10 μF = 10x10^-6F = 10^-5F

*** On edit ***
Also, Dmytry is correct. The 37 Volts is across the resistor which is equal to 100 Volts minus the voltage across the capacitor given by the equation that you used.

 

[Dmytry]

10 μF = 10x10^-6F = 10^-5F

he also mixed up voltage across resistor with voltage across capacitor. I've did such stuff timed on contests. The trick is to realize (or know beforehand) how to simplify the circuit. Suppose you got a circuit of battery and capacitor in series in a black box, that's indistinguishable from just the capacitor - you take out or put in charge q, the voltage changes by q/c .

 

[huykma]

You mixed up voltage on resistor with voltage across capacitor.
You could of used
63 = 100(1-e^(-2/rc))
where 63 is the voltage across capacitor, which gives
63 = 100(1-e^(-2/rc))
0.63 = 1-e^(-2/rc)
0.37=e^(-2/rc)

could you clarify on this point please? I'm confused as to how you progressed. Sorry, its 4am.

 

[Dmytry]

could you clarify on this point please? I'm confused as to how you progressed. Sorry, its 4am.

the voltage across capacitor is 63 volts, not 37 . The formula for 'charging capacitor' is for voltage across capacitor.

The better way to solve is to right away see that voltage on resistor (and it's change over time) is same as if battery wasn't there (was a short circuit), and capacitor was charged to -100 volts. I.e. to simplify circuit first then do the math.

 

[gneill]

A helpful thing to remember with RC and LC circuits is that all of the quantities that are changing over time in the circuit (voltages, currents, charges) do so following an exponential curve of the form X(t) = X₀ e(-t/tau), or X(t) = X₀(1 - e(-t/tau)), where X₀ is some maximum value and tau is the time constant, either RC or L/R.

The first form of exponential function presents a curve that starts at some maximum value (X₀) and decays down to zero, while the second form presents a curve that begins at zero and rises to some final maximum value.

In the RC circuit with the initial charge on the capacitor being zero the charge and voltage on the capacitor starts at zero and rises to the maximum value while the current flowing starts at a maximum value and decays down to zero. In the same circuit, the voltage across the resistor starts at a maximum value and decays down to zero (because the initial current is maximum and decays similarly).

So it should be a simple matter to determine whether the parameter of interest is going to decay or rise, determine its maximum value, and directly write the appropriate form of the exponential function for it.

In this case you are interested in the voltage across the resistor. You know that it will start out with the full voltage of the supply developed across it (100V) because the initial voltage on the capacitor is zero and the current is maximum at time t = 0. So directly you can write:

$$V_R(t) = 100V e^{-\frac{t}{\tau}}$$

with $$\tau = R C$$