Question

The current in a discharging LR circuit without the battery drops from 2.0 A to 1.0 A in 0.10 s. (a) Find the time constant of the circuit. (b) If the inductance of the circuit 4.0 H, what is its resistance?

Answer 1

The current in the discharging LR circuit after t seconds is given by
i = i₀ e^−t/τ
Here,
i₀ = Steady state current = 2 A
Now, let the time constant be τ.
In time t = 0.10 s, the current drops to 1 A.
$$\begin{gathered} i=i_0\left(1-e^{-t/\tau }\right) \\ \Rightarrow 1=2\left(1-e^{-t/\tau }\right) \\ \Rightarrow \frac{1}{2}=1-e^{-t/\tau } \\ \Rightarrow e^{-t/\tau }=1-\frac{1}{2} \\ \Rightarrow e^{-t/\tau }=\frac{1}{2} \\ \Rightarrow \ln \left(e^{-t/\tau }\right)=\ln \left(\frac{1}{2}\right)= \\ \Rightarrow -\frac{0.1}{\tau }=-0.693 \end{gathered}$$
The time constant is given by
$\tau =\frac{0.1}{0.693}=0.144=0.14$ s

(b) Given:
Inductance in the circuit, L = 4 H
Let the resistance in the circuit be R.
The time constant is given by
$$\begin{gathered} \tau =\frac{L}{R} \\ \mathrm{From}\mathrm{the}\mathrm{above}\mathrm{relation},\mathrm{we}\mathrm{have} \\ 0.14=\frac{4}{R} \\ \Rightarrow R=\frac{4}{0.14} \\ \Rightarrow R=28.5728\mathrm{\Omega } \end{gathered}$$