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 V vs t will therefore not be a straight line.) However, a simple manipulation of the data permits

a straight-line slope calculation. Begin with the equation for V(t) during discharge:

V(t) = Vi*e^(t/R*C)

From the equation above, prove that lnV(t) = lnVi - t/R*C, where the ln symbol stands for natural

logarithm, and Vi is the voltage of the fully-charged capacitor. Note that the new equation has the

y = mx + b format.

Explanation / Answer

V(t) =Vie(-t/RC)

tanking logarthim on both sides

ln[v(t)] =ln[Vie-t/RC]

since ln(ab) =ln a+ln b

ln[V(t)]=ln(Vi) +ln(e-t/RC)

ln[V(t)]=ln(Vi) +(-t/RC)ln e

since ln e=1

ln(v(t)) =ln(Vi) -(t/RC)

hrnce proved

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