Simple Capacitor charging and discharging question
I am not sure that he has done there. Has he used the capacitor charge or discharge formula and rearranged it of some sort?


The charging equation look like this: 

$$V_C = V_{th}(1 - e^{\frac{-t}{RC}}) $$ 

Now to solve for the "time"\$(t)\$ we need to rearranged this formula. 

First we can divide by \$V_{th}\$  

$$\frac{V_C}{V_{th}} = 1 - e^{\frac{-t}{RC}} $$ 

We subtract 1 from both sides and multiply both sides by -1

$$\frac{V_C}{V_{th}} - 1 = - e^{\frac{-t}{RC}} $$ 

$$1-\frac{V_C}{V_{th}} = e^{\frac{-t}{RC}} $$ 

take the ln() of both sides

$$ln(1-\frac{V_C}{V_{th}}) = {\frac{-t}{RC}} $$ 

Multiply both sides by -RC

$$-RC*ln(1-\frac{V_C}{V_{th}}) = t $$ 

Swap sides

$$t = -RC*ln(1-\frac{V_C}{V_{th}})$$ 


  (iv): Not sure what hes done here?


\$t_d\$ is a discharge time constant \$t_d = R*C \$ 

Where \$R\$ is a resistance seen by the capacitor when discharging.