# RC Circuit Lab Question

## Homework Statement

This isn't really a homework question- at the end of my last lab, in which we measured the time required to charge and drain a 10micro farad capacitor with a 100k resistor hooked up to a switch and a 9 volt battery our professor asked us what the natural log of voltage plotted against time would represent. The class kicked it around some but we could never come up with what it represented and it's driving me nuts.

## Answers and Replies

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cepheid
Staff Emeritus
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If, at t = 0, you close a switch that connects a resistor and capacitor to a battery (all in series), then the voltage across the charging capacitor as a function of time is given by the equation:

$$v_C(t) = V_0(1 - e^{-\frac{t}{RC}})~~~~~~~~~~[1]$$​

If, once the capacitor has been charged up to the battery voltage, you then remove the battery from the circuit and close the RC loop by providing a connection to ground, the equation for voltage across the discharging capacitor as a function of time is:

$$v_C(t) = V_0e^{-\frac{t}{RC}}~~~~~~~~~~[2]$$

here, vC is the capacitor voltage and V0 is the battery voltage. Does that give you a better sense of what ln(vC) represents? First of all it doesn't make sense to take the logarithm of a dimensional quanity, so we should divide both sides by V0 to make it dimensionless. Then, if you plot this normalized voltage vs. time on a log plot (i.e. a plot in which the y-axis uses a logarithmic scale) you'll end up with a straight line. That's because, taking the ln of both sides of [2], you can see we end up with a linear function of time:

$$\ln\left(\frac{v_C}{V_0}\right) = -\frac{t}{RC}$$

Now, the t-intercept is 1 and the slope is -1/RC. RC has dimensions of time and is called the "time constant" of the circuit. It gives you a sense of the time scale for the exponential decay (or growth in the charging case). The larger the time constant, the shallower the slope of the discharge line on your log-log plot, and the longer the discharging takes.

To derive equations [1] and [2]:

For the first case (battery in the circuit), KVL says that the sum of the voltages around the loop is equal to zero so that:

$$V_0 = v_c(t) + Ri(t)$$​

Now, since the cap and resistor are in series, it's the same current through both of them, so we can replace i(t) with the expression for the current through a capacitor:

$$V_0 = v_c(t) + RC\frac{dv_C(t)}{dt}$$​

Now you just have to solve this differential equation, given the initial condition that vC(0) = 0 (when the switch that connects to the battery is closed).

For the second case, solve the same differential equation, but replace the V0 term on the left hand side with 0. In this case, the initial condition is that vC(0) = V0 (the capacitor voltage initially equals the battery voltage, before you make the connection to ground to discharge it).

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