Respuesta :
Answer:
time=4s
Explanation:
we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion
[tex]i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})[/tex], where [tex]i[/tex]max=[tex]\frac{E}{R}[/tex]
Given that, at i(2)=[tex]\frac{imax}{2} =\frac{E}{2R}[/tex]
⇒[tex]\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})[/tex]
⇒[tex]\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}[/tex]
⇒[tex]\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}[/tex]
Applying logarithm on both sides,
⇒[tex]log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}[/tex]
⇒[tex]log(2)=\frac{2}{\frac{L}{R}}[/tex]
⇒[tex]\frac{L}{R}=\frac{2}{log2}[/tex]
Now substitute [tex]i(t)=\frac{3}{4}imax=\frac{3E}{4R}[/tex]
⇒[tex]\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})[/tex]
⇒[tex]\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}[/tex]
⇒[tex]\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}[/tex]
Applying logarithm on both sides,
⇒[tex]log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}[/tex]
⇒[tex]log(4)=\frac{t}{\frac{L}{R}}[/tex]
⇒[tex]t=log4\frac{L}{R}[/tex]
now subs. [tex]\frac{L}{R}=\frac{2}{log2}[/tex]
⇒[tex]t=log4\frac{2}{log2}[/tex]
also [tex]log4=log2^{2}=2log2[/tex]
⇒[tex]t=2log2\frac{2}{log2}[/tex]
⇒[tex]t=4[/tex]
