Question

In a series LR growth circuit, the maximum current and maximum induced emf in an inductor of 6 mH are 3A and 8V respectively. In how much time the current in the circuit grows to 63.2% of its final value?

• A. $\frac{9}{4}ms$
• B. $\frac{4}{9}ms$
• C. $16ms$
• D. $\frac{1}{10}ms$

Answer 1

The correct option is A $\frac{9}{4}ms$


In LR growth circuit current grows to 63.2% of its aximum value, in one time constant, it means required time is
$t=T=\frac{L}{R}$
If the circuit is switched on at t = 0, then
$i\left(t\right)=\frac{E}{R}[1-e^{\frac{-t}{T}}][T=\frac{L}{R}]$
$\therefore I_{max}=\frac{E}{R}$
$\Rightarrow 3=\frac{8}{R}\Rightarrow R=\frac{8}{3}\Omega$
So, $t=T=\frac{L}{R}=\frac{6\times {10}^{-3}}{\frac{8}{3}}=\frac{9}{4}ms$