Question 2.52: A coil having an inductance of 1.4 H and a resistance of 1Ω ......
A coil having an inductance of 1.4 H and a resistance of 1Ω is connected to a 12 V dc source through a switch. What will be the value of current after 400 m sec of switching on the supply? How much time will it take for the steady-state current to drop to half its value after the switch is turned on?
Step-by-Step
We have,
\begin{aligned} i & =I_0\left(1-e^{\frac{-t}{\tau}}\right) \\ I_0 & =\frac{V}{R}=\frac{12}{1}=12 \mathrm{~A} \\ \tau & =\frac{L}{R}=\frac{1.4}{1}=1.4 \\ t & =400 \mathrm{~ms}=0.4 \mathrm{sec} \end{aligned}Therefore,
\begin{aligned} i & =12\left(1-e^{\frac{-0.4}{1.4}}\right) \\ & =12\left(1-e^{-.285}\right) \\ & =3 \mathrm{~A} \end{aligned}For the second part of the problem we will use the expression for decaying current.
i=I_0 e^{-t / \tau}We have to calculate t for i to become I_0 / 2
Therefore, 6=12 e^{\frac{-t}{1.4}}
or, 0.5=e^{-t / 1.4}
or, -\frac{t}{1.4}=\log 0.5
Therefore t = 0 75. seconds
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