Question

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resistance R, an inductance L and a battery of emf E. How long will the current take to reach 75% its maximum value (measured from the moment when I = 0)?a. 10 s
b. 8 s
c. 4 s
d. 2 s
e. 1 s

Answer 1

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

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$, where $i$max=$\frac{E}{R}$

Given that, at i(2)=$\frac{imax}{2} =\frac{E}{2R}$

⇒$\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒$\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒$\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒$log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒$log(2)=\frac{2}{\frac{L}{R}}$

⇒$\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒$\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒$\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒$\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒$log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒$log(4)=\frac{t}{\frac{L}{R}}$

⇒$t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒$t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒$t=2log2\frac{2}{log2}$

⇒$t=4$