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If you look at an oscilloscope trace of a discharging capacitor, measuring exact values from a smooth curve with a ruler is notoriously inaccurate. How do engineers transform that curved line into a perfectly straight line to extract exact circuit values?

Linearizing the Discharge Curve

A capacitor discharging through a resistor follows the exponential decay equation V(t)=V0eβˆ’t/RCV(t) = V_0 e^{-t/RC}, where V0V_0 is the starting voltage and the time constant Ο„=RC\tau = RC is the time in seconds for the voltage to fall to about 37%37\% of its initial value.
In 1614, Scottish mathematician John Napier invented the natural logarithm (ln\\ln), which serves as the inverse operation to the exponential function exe^x, turning multiplicative powers into simple linear additions.
Taking the natural logarithm of both sides gives ln⁑V=ln⁑(V0β‹…eβˆ’t/RC)=ln⁑V0+ln⁑(eβˆ’t/RC)\ln V = \ln(V_0 \cdot e^{-t/RC}) = \ln V_0 + \ln(e^{-t/RC}), which simplifies directly to ln⁑V=βˆ’1RCt+ln⁑V0\ln V = -\frac{1}{RC}t + \ln V_0.
This matches the standard linear equation y=mx+cy = mx + c, where y=ln⁑Vy = \ln V, the independent variable x=tx = t, the gradient m=βˆ’1RC=βˆ’1Ο„m = -\frac{1}{RC} = -\frac{1}{\tau}, and the y-intercept c=ln⁑V0c = \ln V_0.
How do we pull real circuit parameters like resistance, capacitance, and starting voltage out of actual experimental graph points?

Extracting Values from the Graph

To find the circuit's time constant Ο„\tau, calculate the slope of the line using two points (t1,ln⁑V1)(t_1, \ln V_1) and (t2,ln⁑V2)(t_2, \ln V_2): the gradient is m=Ξ”(ln⁑V)Ξ”t=βˆ’1Ο„m = \frac{\Delta(\ln V)}{\Delta t} = -\frac{1}{\tau}, which means Ο„=βˆ’1m\tau = -\frac{1}{m}.
To find the initial voltage V0V_0, read the vertical intercept cc where the line crosses t=0t = 0, then undo the logarithm using the exponential base ee: V0=ecV_0 = e^c.
Let's apply this to a concrete lab scenario where a capacitor discharges through a known 10Β kΞ©10\text{ k}\Omega resistor.