Q: 4μF capacitor and a resistance 2.5 MΩ are in series with 12 V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [given in (2)= 0.693]

Sol: (a) Charging current $\large I = \frac{V_0}{R} e^{\frac{-t}{R C}}$

Potential difference across R is

$\large V_R = I R = V_0 e^{\frac{-t}{R C}} $

Potential difference across ‘C’ is

$\large V_C = V_0 – V_R = V_0 – V_0 e^{\frac{-t}{R C}} $

$\large V_C = V_0 (1 – e^{\frac{-t}{R C}}) $

But given V_{C} = 3 V_{R} , we get

$\large V_0 (1 – e^{\frac{-t}{R C}}) = 3 V_0 e^{\frac{-t}{R C}} $

$\large (1 – e^{\frac{-t}{R C}}) = 3 e^{\frac{-t}{R C}} $

$\large 1 = 4 e^{\frac{-t}{R C}} $

$\large e^{\frac{t}{R C}} = 4 $

$\large \frac{t}{R C} = ln 4 $

$\large t = 2 R C ln2 $

t = 2 × 2.5 × 10^{6} × 4 × 10^{-6} × 0.693

or , t = 13.86 sec.