Q: A coil of resistance 20 Ω and inductance 0.5 henry is switched to dc 200 volt supply. Calculate the rate of increase of current:

(a) At the instant of closing the switch and

(b) After one time constant

(c) Find the steady state current in the circuit.

Sol: (a) This is the case of growth of current in an L – R circuit. Hence, current at time t is given

$\large I = I_0 (1-e^{\frac{-t}{\tau}})$

Rate of increase of current,

$\large \frac{dI}{dt} = \frac{I_0}{\tau} e^{\frac{-t}{\tau}}$

At , t = 0 ;

$\large \frac{dI}{dt} = \frac{I_0}{\tau} = \frac{E/R}{L/R}$

$\large \frac{dI}{dt} = \frac{E}{L} = \frac{200}{0.5}$

= 400 A/s

(b) At , t = τ ;

$\large \frac{dI}{dt} = \frac{I_0}{\tau} e^{-1}$

$\large \frac{dI}{dt} = 400 e^{-1} = 400 \times 0.37 $

= 148 A/s

(c) The steady state current in the circuit is

$\large I_0 = \frac{E}{R} = \frac{200}{20} $

= 10 A