A L-R combination is connected to a battery of emf 4 volt. If L = 0.1 H and R = 4.0 `Omega`, then the time taken to reach a current of 0.631 ampere is:-

Options

A. (1/40) sec
B. 0.4 sec
C. 1.6 sec
D. 0.63 sec

Answer

A

Explanation

Final maximum current `=E/R=4/4=1` amp
Instantaneous current , `I=E/R (1-e^(-Rt//L))`
Here , I=0.6321 ampere
0.6321 = 1(1–e​–Rt/L)
0.6321 = 1–e–Rt/L
e–Rt/L = 1–0.6321 = 0.3679 = `1/e`
or `"Rt"/L=1` or `t=L/R = 0.1/4=1/40` sec.