Question

Show in details how we can mathematically modify the two equations below to get a straight-line graph from each of these equations. Compare the 2 new equations to the general equation of s straight line and show what the slope (m) represents. How can we calculate the value of capacity from the slope of the line?

Equation 1: V = V₀e^-t/RC

Equation 2: V = V₀(1 - e^-t/RC)

Answer 1

Okay, First Lets do for discharging ( equation 1)

V = V₀e^-t/RC

V / V_o = e^-t/RC

take ln on both sides, we get

- ln ( V / V_o ) = t / RC

Note that whenever there is time involved, we always plot it on x - axis.

straight line equation is

y = mx + c

so, let's try to write our equation in this format

- ln ( V / V_o ) = (1/RC) t + c

therefore, slope of this graph will be 1/RC and we will get a straight line for this

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Now for charging equation

V = V₀(1 - e^-t/RC)

V_o - V / V_o = e^-t/RC

take 'ln' on both sides, we have

- ln ( V_o - V / V_o ) = t / RC

in same way as above, we can write as

- ln ( V_o - V / V_o ) = (1/RC) t + c

so,

this graph will also give a straight line with slope as 1/RC

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Here RC is called as time constant