Question

An inductor-coil of inductance 20 mH having resistance 10 Ω is joined to an ideal battery of emf 5.0 V. Find the rate of change of the induced emf at (a) t = 0,  (b) t = 10 ms and (c) t = 1.0 s.

Answer 1

Given:-

Self-inductance, L = 20 mH

Emf of the battery, e = 5.0 V

Resistance, R = 10 Ω

Now,

Time constant of the coil:-

\[\tau = \frac{L}{R} = \frac{20 \times {10}^{- 3}}{10}=2\times10^{-3} s\]

Steady-state current:-

\[i_0 = \frac{e}{R} = \frac{5}{10} = 0 . 5\]

The current in the LR circuit at time t is given by

i = i₀(1 − e^−t/τ)

⇒ i = i₀ − i₀e^−t/τ

On differentiating both sides, we get

\[\frac{di}{dt} = \frac{i_0}{\tau} e^{- t/\tau}\]

The rate of change of the induced emf is given by

\[R\frac{di}{dt} = R\frac{i_0}{\tau} \times e^{- t/\tau}\]

(a) At time t = 0 s, the rate of change of the induced emf is given by

\[R\frac{di}{dt} = R\frac{i_0}{\tau}\]

\[ = 10 \times \frac{0 . 5}{2 \times {10}^{- 3}}\]

\[ = 2 . 5 \times  {10}^3   V/s\]

(b) At time t = 10 ms, the rate of change of the induced emf is given by

\[R\frac{di}{dt} = R\frac{i_0}{\tau} \times e^{- t/\tau}\]

Now,

For t = 10 ms = 10 × 10⁻³ s = 10⁻² s,

\[\frac{dE}{dt} = 10 \times \frac{5}{10} \times \frac{1}{2 \times {10}^{- 3}} \times e^{- 0 . 01/(2\times 10^{- 3})} \]

= 16.844 = 17 V/s

(c) At time t = 1 s, the rate of change of the induced emf is given by

\[\frac{dE}{dt} = \frac{Rdi}{dt} = R\frac{i_0}{\tau} \times e^{- t/\tau}\]

\[= 10 \times \frac{5 \times {10}^{- 1}}{2 \times {10}^{- 3}} \times e^{- 1/(2\times 10^{- 3})}\]

= 0.00 V/s