Example 10.3: The RL circuit

Question: A coil has a resistance of $R=5.0\,\Omega$ and an inductance of $L= 100\,{\rm mH}$. At a particular instant in time after a battery is connected across the coil, the current is $i=2.0\,{\rm A}$, and is increasing at a rate of $d i/dt=20\,{\rm A\,s}^{-1}$. What is the voltage $V$ of the battery? What is the time-constant of the circuit? What is the final value of the current?
 
Answer: Application of Ohm's law around the circuit gives [see Eq. (257)]

$$V = i\,R + L\,\frac{di}{d t} = (2.0)\,(5.0) + (0.1)\,(20) = 12\,{\rm V}.$$

The time-constant of the circuit is simply

$$\tau = \frac{L}{R} = \frac{(0.1)}{(5.0)} = 0.020\,{\rm s}.$$

The final steady-state current $I$ is given by Ohm's law, with the inductor acting like a conducting wire, so

$$I= \frac{V}{R} = \frac{(12)}{(5)} = 2.4\, {\rm A}.$$