A capacitor of capacitance `10(mu)F` is connected to a battery of emf 2V.It is found that it takes 50 ms for the charge on the capacitor to bacome `12.6(mu)C.`find the resistance of the circuit.

Explanation

`C = 10 mu f = (10^-5) f, E_emf = 2V, `
` t = 50 ms = 5 xx (10^-2) S `
` Since , q = Qe^(-t/RC)`
` Q = CV = (10^-5) xx 2 `
` q = 12.6 xx (10^-6)C . `
` Hence 12.6 xx (10^-6) = 2 xx (10^-5) e^((5 xx (10^-2))/(R xx (10^-5)))`
` rArr ( (12.6 xx (10^-6))/ (2 xx (10^-5))) = e^ - (5 xx (10^3)/R )`
` rArr 0.63 = e^- (5 xx (10^3)/R )`
` rArr log 0.63 = (5 xx (10^3)/R) `
` rArr R = (5 xx (10^3)/0.79 - 2 ) = 4 k Omega `.