Input
output
Capacitor impedance
$$ Z_{C} = \frac{1}{jwC} $$
Voltage Divider(Transfer function)
$$ H(jw) = \frac{V_{out}}{V_{in}} $$
$$ = \frac{Z_{C}}{R + Z_{C}} $$
$$ = \frac{\frac{1}{jwC}}{R + \frac{1}{jwC}} $$
$$ = \frac{1}{jwRC + 1} $$
Magnitude of transfer function
$$ \left|H(jw)\right| = \frac{1}{\sqrt{1 + (wRC)^{2} } } $$
At Cutoff w_{c}, the output magnitude is 1/√2 of low frequency (DC) gain
$$ \frac{1}{\sqrt{1 + (wRC)^{2} } } = \frac{1}{\sqrt{2}} $$
$$ 1 + (wRC)^{2} = 2 $$
$$ (wRC)^{2} = 1 $$
$$ wRC = 1 $$
$$ w = \frac{1}{RC} $$
$$ Angular frequency(w) = 2 \pi f $$
\[ \boxed { f = \frac{1}{2 \pi RC} } \]
The voltage at input of filter circuit,
$$ V = iR + V_{C} $$
Current flowing through the capacitor
\[ \boxed { i = C \frac{dv}{dt} } \]$$ V = C \frac{dv}{dt} R + V_{C} $$
$$ V-V_{C} = RC \frac{dv}{dt} $$
$$ \frac{dt}{RC} = \frac{dv}{V-V_{C}} $$
Multiplying by Minus( - ) on both sides for integration
$$ - \frac{dt}{RC} = - \frac{dv}{V-V_{C}} $$
On integrating
$$ log_{e}\left[ \frac{V-V_{C}} {V} \right] = - \frac{t}{RC} $$
$$ V-V_{C} = V\left( e^{- \frac{t}{RC}} \right) $$
\[ \boxed { V_{C} = V \left( 1 - e^{- \frac{t}{RC}} \right) } \]Here \(\tau \) = RC is the time constant of RC low pass filter