A cell of 1.5 V is connected across an inductor of 2 mH in series with a 2 Ω resistor. What is the rate of growth of current…

Q: A cell of 1.5 V is connected across an inductor of 2 mH in series with a 2 Ω resistor. What is the rate of growth of current immediately after the cell is switched on.

Sol: $\large E = L\frac{dI}{dt} + I R $

Therefore , $\large \frac{dI}{dt} = \frac{E-IR}{L}$

E = 1.5 volt , R = 2Ω , L = 2 mH = 2 × 10-3 H when the cell is switched on, I = 0

$\large \frac{dI}{dt} = \frac{E}{L}$

$\large \frac{dI}{dt} = \frac{1.5}{2 \times 10^{-3}}$

= 750 A/s