Question

Two coils A and B have inductances 1.0 H and 2.0 H respectively. The resistance of each coil is 10 Ω. Each coil is connected to an ideal battery of emf 2.0 V at t = 0. Let i_A and i_B be the currents in the two circuit at time t. Find the ratio i_A / i_B at (a) t = 100 ms, (b) t = 200 ms and (c) t = 1 s.

Answer 1

Given:
Inductance of the coil A, L_A = 1.0 H
Inductance of the coil B, L_B = 2.0 H
Resistance in each coil, R = 10 Ω
The current in the LR circuit after t seconds after connecting the battery is given by
i = i₀ (1 − e^−t/τ)
Here,
i₀ = Steady state current
τ = Time constant = $\frac{L}{R}$

(a) At t = 0.1 s, time constants of the coils A and B are τ_A and τ_B, respectively.
Now,
$$\begin{gathered} {\tau }_{\mathrm{A}}=\frac{1}{10}=0.1\mathrm{s} \\ {\mathrm{\tau }}_{\mathrm{B}}=\frac{2}{10}=0.2\mathrm{s} \end{gathered}$$

Currents in the coils can be calculated as follows:
$$\begin{gathered} i_{\mathrm{A}}=i_0(1-e^{-t/\tau }), \\ =\frac{2}{10}\left(1-e^{\frac{0.1\times 10}{1}}\right)=0.2(1-e^{-1}) \\ =0.126424111 \\ {i}_{\mathrm{B}}=i_0(1-e^{-t/\tau }) \\ =\frac{2}{10}(1-e^{0.1\times 10/2}) \\ =0.2(1-e^{-1/2})=0.078693 \end{gathered}$$

$\therefore \frac{i_{\mathrm{A}}}{i_{\mathrm{B}}}=\frac{0.126411}{0.78693}=1.6$

(b) At t = 200 ms = 0.2 s,
i_A = 0.2 (1 − e^{−0.2 × 10.1})
i_A = 0.2 × 0.864664716
i_A = 0.1729329943
i_B = 0.2 (1 − e^{−0.2 × 10.2})
i_B = 0.2 × 0.632120 = 0.126424111
$\therefore \frac{i_{\mathrm{A}}}{i_{\mathrm{B}}}=\frac{0.172932343}{0.126424111}=1.36=1.4$

(c) At time t = 1 s,
i_A = 0.2 (1 − e^{−1 × 10.1})
= 0.2 − 0.9999549
= 0.19999092
i_B = 0.2 (1 − e^{−1 × 10.2})
= 0.2 × 0.99326 = 0.19865241
$\therefore \frac{i_{\mathrm{A}}}{i_{\mathrm{B}}}=\frac{0.19999092}{0.19999092}\approx 1.0$