The spark plug system in a car stores energy by a steady current in an inductor

Question

The spark plug system in a car stores energy by a steady current in an inductor coil of 300 turns and inductance of 9.4 mH. The current is driven by a 12V battery and The resistance of The coil is 2 Ohm between The battery and ground. How much energy is stored by The inductor? How long does it take for The current to get to 80 % of its maximum value after all The switches are closed? This current is changed when an engine cam opens The switch in The path to ground. Current now flows into The capacitor (80 micro F), and this charges up in about time RC. If The current stops in The time Delta t = RC, how much back emf is induced across The coil? The coil is one side of a transformer that steps up this coil voltage to a spark plug voltage of 47.1 kV. How many turns does The spark plug side of The transformer have?

Explanation / Answer

1. Energy across inductor U = N* 0.5 LI^2

where L is inductance and i is current

so using ohms V = IR

current I = 12/2 = 6 Amsp

U = (300 * 0.5 * 9.4 e-3* 6*6)

U = 50.76 Joules

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b. Use the growth of Current in inductance as

I = Io ( 1- e-RT/L)

so here

1-e^-Rt/L = I/I0 = 80 % = 0.8

e^-Rt/L = 1-0.8 = 0.2

-Rt/L = ln (0.2)

t = 1.609 * L/R = 1.609 * 9.4 e-3/2

t = 7.56 ms

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c. induced emf e = Q/C

emf e = it./C

emf e = 6 * RC/C = 6*2 = -12 V

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d.Apply in a transforner Ns/Np = Vs/Vp = Ip/Is

where NS ans Np are seconda dr and primary tunrs

Vs and Vp are secondary and primary volatges

Ns = (47100 * 300/12)

Ns = 1.17 e 6 turns

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