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 × 10⁻³ 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
$$\begin{gathered} R\frac{di}{dt}=R\frac{i_0}{\tau } \\ =10\times \frac{0.5}{2\times {10}^{-3}} \\ =2.5\times {10}^3\mathrm{V}/\mathrm{s} \end{gathered}$$

(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