A `4 muF` capacitor, a resistance of `2.5 M Omega` is in series with `12V` 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]

Options

A. 13.86s
B. 6.93s
C. 7s
D. 14s

Answer

A

Explanation

(a) KEY CONCEPT : At any instant of time t during charging process, the transient current in the circuit is
given by
`I = V_0/R e^(-t//RC)`
`:. Potential difference across resistor R is `.
`V_R = [V_0/R e^(-t//RC)] xx R`
`=V_0e^(-t//RC).......(i)`
`:. Potential diff. across C`
`V_c = V_0 - V_0 e^(-t//RC) = V_0 (1-e^(-t//RC)).......(ii)
`:. V_c = 3V_R`
`:. V_0 (1-e^(-t//RC)) = 3V_0. e^(-t//RC)`
` rArr (1-e^(-t//RC)) = 3e^(-t//RC) rArr 1 = 4e^(-t//RC)`
Taking log on both sides
`log_e1 = 2log_e2 + (-t/(RC))`
` log_e 1 = 2 log_e 2 + (t/(RC))`
`rArr 0 = 2 xx 2.303 log_(10) 2- t/(RC)`
`t = [2xx2.303 log_10 2] xx 2.5 xx 10^6 xx 4 xx 10^-6`
`= 13.86 sec.`