The inductance of the coil and time is mathematically given as
Generally, the equation for Current is mathematically given as
[tex]\begin{aligned}I(t) &=I_{0} e^{\frac{-t}{\tau}} \\&I(t) =I_{0} e^{\frac{-R t}{L}}\end{aligned}[/tex]
Therefore
[tex]L=\frac{-t R}{\ln \left(\frac{I}{I_{0}}\right)}[/tex]
Hence, the Maximum current is,
[tex]\begin{aligned}I_{0} &=\frac{\varepsilon}{R}=\frac{6.30}{15} \\&I_{0} =0.42 \end{aligned}[/tex]
Thus,
[tex]L &=\frac{-\left(0.280 * 10^{-3})(13.0)}{\ln \left(0.280}{0.42})} \\[/tex]
L=0.1035H
b)
In conclusion, Time taken after S1 is opened is,
[tex](0.01) I_{0}=I_{0} e^{\frac{-t}{\tau}}[/tex]
Time constant is,
[tex]\tau=\frac{L}{R}\\\\\tau=\frac{0.042}{15} \\[/tex]
[tex]\tau=0.0028s[/tex]
[tex](0.01) I_{o} &=I_{o} e^{\frac{-t}{(0.00028)}} \\\\t=-\ln (0.01)(0.0028) \\\\t=-(-4.605)(0.0033) \\\\t=0.01289s[/tex]
t=12.8 ms
Read more about inductance
https://brainly.com/question/18575018
#SPJ1
