When a capacitor discharges through a resistor, the voltage drops exponentially

Question

When a capacitor discharges through a resistor, the voltage drops exponentially over time. A graph of voltage versus time will therefore not be a straight line with a meaningful slope. However, a simple manipulation of the data permits a straight-line slope analysis. Begin with the equation for V (t) across the capacitor during discharge:

V (t) = Viet/RC

(i) Starting with V (t), prove that lnV (t) = lnVi t/RC, where the ln symbol stands for natural
logarithm, and Vi is the voltage of the fully-charged capacitor.

(ii) Note that the result of (i) has the format y = mx + b. If we were to graph lnV (t) versus t, please identify the slope and the y-intercept in terms of the variables Vi, R, and C.

Explanation / Answer

Natural log and e are inverse functions Ln(e^a) = a If we take the natural log of both sides we get Ln{V(t)} = Ln{Vi e^-t/RC} Ln of a product is the sum of the Ln's LnV(t) = LnVi + Lne^-t/RC remembering the inverse realtionship LnV(t) = LnVi + (-t/RC) this is what we were to prove LnV(t) = LnVi - t/RC y = mx + b is called the slope intercept form of a line with y = LnV(t) m = -1/RC =slope x = t b = LnVi = y intercept

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