Question

When the switch in the figure is closed, the current takes 3.00 msto reach 98.0% of its final value. If R = 10.0Ω, what is theinductance?

OR

Answer 1

An inductor and a resistor are connected to a battery through a switch. The current in the LR circuit after the switch is closed is, \(i=i_{\max }\left(1-e^{-t / x}\right)\)

Here, \(\tau\) is the time constant and it is given as \(\tau=L / R\). The maximum current in the circuit is,

\(i_{\max }=\frac{\varepsilon}{R}\) here, \(\varepsilon\) is the emf of the battery and \(R\) is the resistance. The current in the circuit reaches \(98 \%\) of maximum value after \(3.0 \mathrm{~ms}\). \((0.98) i_{\max }=i_{\max }\left(1-e^{-t(L / R)}\right)\)

\(0.98=1-e^{-t / R C}\)

\(e^{-t R I L}=1-0.98\)

\(\frac{-t R}{L}=\ln (0.02)\)

Therefore, the inductance of the circuit is, \(L=-\frac{t R}{\ln (0.02)}=-\frac{\left(3 \times 10^{-3} \mathrm{~s}\right)(10 \Omega)}{\ln (0.02)}=7.66 \times 10^{-3} \mathrm{H}(\) or \() 7.6 \mathrm{mH}\)