Answer:
PartA) resistance R=1.86Ω
PartB) inductnce L=1.26mH
Explanation:
In an RL circuit, [tex]i(t)=\frac{E}{R}(1-e^{\frac{-t}{\frac{L}{R}}})[/tex]
where E is the emf of battery
L is inductance of inductor
R is the internal resistance of inductor
given [tex]i(\infty)=[/tex][tex]i[/tex]max=[tex]\frac{E}{R}[/tex]
⇒[tex]6.45=\frac{E}{R}[/tex]
⇒[tex]6.45=\frac{12}{R}[/tex]
⇒[tex]R=\frac{6.45}{12}=1.86[/tex]Ω
given [tex]i(0.94ms)=4.86[/tex]
⇒[tex]4.86=6.45(1-e^{\frac{-0.94*10^{-3}}{\frac{L}{R}}})[/tex]
⇒[tex]0.75=1-e^{\frac{-0.94*10^{-3}}{\frac{L}{R}}}[/tex]
⇒[tex]0.25=e^{\frac{-0.94*10^{-3}}{\frac{L}{R}}}[/tex]
applying logarithm on both sides,
⇒[tex]log(0.25)=\frac{-0.94*10^{-3}}{\frac{L}{R}}[/tex]
⇒[tex]log(4)=\frac{0.94*10^{-3}}{\frac{L}{1.86}}[/tex]
⇒[tex]L=\frac{0.94*10^{-3}*1.86}{log4}=1.26mH[/tex]
