Answer
given,
battery voltage = 9.9 V
resistance of the circuit = 15 Ω
Inductor = 0.9 H
a) current
[tex]i = \dfrac{V}{R}[/tex]
[tex]i = \dfrac{9.9}{15}[/tex]
I = 0.66 A
b) Time taken to reach current to 50%
[tex]I = I_o (1 - e^{-\dfrac{t}{\tau}})[/tex]
[tex]0.5 I_0 = I_o (1 - e^{-\dfrac{t}{\tau}})[/tex]
[tex] e^{-\dfrac{t}{\tau}} = \dfrac{1}{2}[/tex]
taking log both side
t = τ ln 2
[tex]t= \dfrac{L}{R}\ ln 2[/tex]
[tex]t = \dfrac{0.9}{15}\ ln 2[/tex]
t = 0.0416 s
