Question

A coil of inductance 300mh and resistance 2 ohm is connected to a source of voltage 2v . the current reaches half of its steady state value in

Answer 1

Answer: 0.104 seconds


We are given a Series L-R Circuit. Our data is:


$L = 300 \, \, mH = 300 \times 10^{-3} \, \, H = 0.3 \, \, H \\ \\ R = 2 \,\, \Omega \\ \\ V = 2 \, \, V$


The current through an inductor while it is charging is given by:


$\boxed{I = I_{\circ}\left(1-e^{-\frac{Rt}{L}}\right)}$

Here, $I_{\circ}$ represents the Current in Steady State. 


We are asked the time when the current becomes half of that in steady state. 

So we will put $I=\frac{I_{\circ}}{2}$ in the equation. 


$I = I_{\circ}\left(1-e^{-\frac{Rt}{L}}\right) \\ \\ \\ \implies \frac{I_{\circ}}{2} = I_{\circ}\left(1-e^{-\frac{Rt}{L}}\right) \\ \\ \\ \implies 1-e^{-\frac{Rt}{L}} =\frac{1}{2} \\ \\ \\ \implies 1-\frac{1}{2} = e^{-\frac{Rt}{L}} \\ \\ \\ \implies e^{-\frac{Rt}{L}} = \frac{1}{2} \\ \\ \\ \implies \frac{1}{e^{\frac{Rt}{L}}} = \frac{1}{2} \\ \\ \\ \implies e^{\frac{Rt}{L}} = 2 \\ \\ \\ \implies \frac{Rt}{L} = \ln 2 \\ \\ \\ \implies t = \frac{L}{R} \ln 2$


$\implies t = \frac{0.3}{2} \times \ln 2 \\ \\ \\ \implies \boxed{\bold{t \approx 0.104 \, \, s}}$



Thus, The Current reaches half of its steady state value in approximately 0.104 seconds.

Answer 2

given that ....
then The value of I/2 occurs in .1sec