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.

\[i = i_0 \left( 1 - e^{- t/\tau} \right)\]

\[ \Rightarrow 1 = 2\left( 1 - e^{- t/\tau} \right)\]

\[ \Rightarrow \frac{1}{2} = 1 - e^{- t/\tau} \]

\[ \Rightarrow e^{- t/\tau} = 1 - \frac{1}{2}\]

\[ \Rightarrow e^{- t/\tau} = \frac{1}{2}\]

\[ \Rightarrow \ln\left( e^{- t/\tau} \right) = \ln\left( \frac{1}{2} \right) = \]

\[ \Rightarrow - \frac{0 . 1}{\tau} = - 0 . 693\]

The time constant is given by

\[\tau = \frac{0 . 1}{0 . 693} = 0 . 144 = 0 . 14\]

(b) Given:-

Inductance in the circuit, L = 4 H

Let the resistance in the circuit be R.

The time constant is given by

\[\tau = \frac{L}{R}\]

From the above relation, we have

\[0 . 14 = \frac{4}{R}\]

\[ \Rightarrow R = \frac{4}{0 . 14}\]

\[ \Rightarrow R = 28 . 5728 \Omega\]