A solenoid with resistance 4 Ω and inductance 6 mH is used in an automobile ignition circuit similar to that in Fig. 7.78. If the battery supplies 12 V, determine: the final current through the solenoid when the switch is closed, the energy stored in the coil, and the voltage across the air gap, assuming that the switch takes 1 μ s to open.
The final current through the coil is
I = \frac{V_{s}}{R}= \frac{12}{4} = 3 A
The energy stored in the coil is
W = \frac{1}{2} = L I² = \frac{1}{2} × 6 × 10^{-3} × 3 ² = 27 mJ
The voltage across the gap is
V = L \frac{ΔI}{Δt} = 6 × 10^{-3} × \frac{3}{1 × 10^{-6}} = 18 kV