Question

A coil of inductance 300mH and resistance 2Ω is connected to a source of voltage 2V. The current
reaches half of its steady value in:
(A) 0.05 s
(B) 0.1 s
(C) 0.105 s
(D) 0.3 s

Answer 1

Hint:When a current flows through a coil, there is a change in flux linked with the coil due to which an opposing emf is produced. So it takes some time to reach a steady value. Using Kirchoff’s
second law and differentiation, we can find out the formula for calculating the value of current at
any time t.

Complete step by step answer:
We know that the growth of current in an inductor is given by,
$I = {I_0}(1 - {e^{\dfrac{{ - Rt}}{L}}})$
Where ${I_0} = $ steady value of current, $R = $ resistance connected with the inductor, $L = $ the
measure of inductance and $t = $ the time taken by the current to reach the value $I$ .
Here we have to find the time taken by the current to reach half of its steady value that is, $I =
\dfrac{{{I_0}}}{2}$
So, we get
$
\dfrac{{{I_0}}}{2} = {I_0}(1 - {e^{\dfrac{{ - Rt}}{L}}}) \\
\dfrac{1}{2} = 1 - {e^{\dfrac{{ - Rt}}{L}}} \\
{e^{\dfrac{{ - Rt}}{L}}} = \dfrac{1}{2} \\
{e^{\dfrac{{ - Rt}}{L}}} = {2^{ - 1}} \\
$
Taking log on both sides,
$
\dfrac{{Rt}}{L} = \ln (2) \\
\dfrac{{Rt}}{L} = 0.693 \\

t = 0.693 \times \dfrac{L}{R} \\
t = 0.693 \times \dfrac{{300 \times {{10}^{ - 3}}}}{2} \\
t = 103.95 \times {10^{ - 3}} \\
t = 0.10395 \approx 0.105 \\
t = 0.105 \\
$
Hence, option (C) is the correct answer.

Additional information:
For a coil of N-turns, the magnetic flux linked with all the turns is the same so when the flux through
the coil changes, each turn contributes to the induced emf. Therefore, when flux linked with the coil
is equal to \[N{\phi _B}\]
Then, total flux,
$
N{\phi _B} \propto I \\
N{\phi _B} = LI \\
$
L is a constant of proportionality and is known as inductance.
The S.I. unit of inductance is Henry (H).
Resistance is the opposition offered by a conductor to the flow of current through it. It is
represented by ‘R’. Its S.I. unit is Ohm (Ω).

Note:
In an R-L circuit, when the voltage is applied, it takes some time for the current to reach its final value.
This is because when there is a change of flux in the inductor, the inductor develops an induced emf
that opposes the change in current so after overcoming this opposition, the current reaches a steady
value.