Question 2.51: A coil of resistance 5 Ω and inductance 0.1H is switched on ......
A coil of resistance 5 Ω and inductance 0.1H is switched on to a 230 V, 50 Hz supply. Calculate the rate of rise of current at t = 0 and at t = 2τ where τ = L/R. What is the steady-state value of the circuit current?
For the rise in current
i=\frac{V}{R}\left(1-e^{\frac{-R}{L} t}\right)Substituting values, i=\frac{230}{5}\left(1-e^{\frac{-5}{0.1} t}\right)
or, i=46\left(1-e^{-50 t}\right)
The rate of change of current,
\begin{aligned} \frac{d i}{d t} & =46 \times 50 e^{-50 t} \\ & =2300 e^{-50 t} \end{aligned}at t = 0
\frac{d i}{d t}=2300 \mathrm{~A} / \mathrm{sec}at t equal to two times the time constant, \tau=2 \frac{L}{R}.
\begin{gathered} t=2 \frac{L}{R}=2 \frac{0.1}{5}=0.04 \mathrm{sec} \\ \frac{d i}{d t}=2300 e^{-50 \times 0.04}=2300 e^{-2} \mathrm{~A} / \mathrm{sec} \end{gathered}The steady-state value, I_0=\frac{V}{R}=\frac{230}{5}=46 \mathrm{~A}
The larger the time constant is, the more is the time taken by the current to rise or decay in a L–R circuit during transients. Current almost decays to zero at five times the time constant. Beyond this time the magnitude of current is less than one per cent of its steady-state value. When the circuit is switched on, energy is stored in the inductor in the form of a magnetic field and during switching off, the stored energy gets dissipated.