Question

A coil of resistance $40\Omega$ is connected across a $4.0$V battery. $0.10$s after the battery is connected, the current in the coil is $63$mA. Find the inductance of the coil.

Answer 1

$R=40\Omega ,E=4$V,$t=0.1,i=63$mA

$i=i_{\circ }(1-e^{\frac{tR}{2}})$

$\Rightarrow 63\times {10}^{-3}=\frac{4}{40}(1-e^{-0.1\times \frac{40}{L}})$

$\Rightarrow 63\times {10}^{-2}=1-e^{\frac{-4}{L}}$

$\Rightarrow 1-0.63=e^{\frac{-4}{L}}$

$\Rightarrow e^{\frac{-4}{L}}=0.37$

$\Rightarrow \frac{-4}{L}=\ln \left(0.37\right)=-0.994$

$\Rightarrow L=\frac{-4}{-0.994}=4.024$H$=4$H