Answer:i=300 mA
Explanation:
Given
inductance(L)=40 mH
Resistor(R)=[tex]50 \Omega [/tex]
Voltage(V)=15 V
Time constant([tex]\tau [/tex])=[tex]\frac{L}{R}[/tex]
[tex]\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}[/tex]
current [tex]i_0=\frac{V}{R}[/tex]
[tex]i_0=\frac{15}{50}=0.3 A[/tex]
Current as a function of time is given by
[tex]i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )[/tex]
[tex]i=0.3\times 0.9998[/tex]
i= 299.95 mA
