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_Bbe 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 = `L/R`

(a) At t = 0.1 s, time constants of the coils A and B are τ_A and τ_B, respectively.

Now,

\[\tau_A  = \frac{1}{10} = 0 . 1  s\]

\[ \tau_B  = \frac{2}{10} = 0 . 2  s\]

Currents in the coils can be calculated as follows:-

\[i_A = i_0 (1 - e^{- t/\tau} ), \]

\[ = \frac{2}{10}\left( 1 - e^\frac{0 . 1 \times 10}{1} \right) = 0 . 2 (1 - e^{- 1} )\]

\[ = 0 . 126424111\]

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

\[ = \frac{2}{10}(1 - e^{0 . 1 \times 10/2} )\]

\[ = 0 . 2 (1 - e^{- 1/2} ) = 0 . 078693\]

\[\therefore \frac{i_A}{i_B} = \frac{0 . 126411}{0 . 78693} = 1 . 6\]

(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

\[\therefore \frac{i_A}{i_B} = \frac{0 . 172932343}{0 . 126424111} = 1 . 36 = 1 . 4\]

(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

\[\therefore \frac{i_A}{i_B} = \frac{0 . 19999092}{0 . 19999092} \approx 1 . 0\]