A coil of inductance 300mh and resistance 2 ohm

Asked by admin @ 17/08/2021 in Physics viewed by 280 People

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

Answered by admin @ 17/08/2021

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.

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