The spark plug in an automobile engine is an R-L circuit as shown in figure. The circuit that provides the spark uses an inductor as the energy source. Initially switch is closed and allows current to build through the inductor. When the switch is open the current decreases rapidly through inductor and a large emf is induced by inductor. Given ε = 12V, L = 10 mH, Rc = 10Ω, Rp = 7kΩ

The spark plug in an automobile engine is an R-L circuit as shown in figure. The circuit that provides the spark uses an inductor as the energy source. Initially switch is closed and allows current to build through the inductor. When the switch is open the current decreases rapidly through inductor and a large emf is induced by inductor. Given ε = 12V, L = 10 mH, R_c = 10Ω, R_p = 7kΩ

Linked Question 1

What is maximum heat dissipated in the spark plug ?

14.4 mJ

7.2 mJ

6 mJ

3.5 mJ

$U_{S} = \frac{1}{2} {Li}^{2} = \frac{1}{2} (10^{- 2} H) {(\frac{12 V}{10 Ω})}^{2} = 7.2 mJ$

Linked Question 2

The spark occurs in the spark plug because

Spark plug short circuits the inductor.

Spark produced is due to conversion of magnetic field energy of inductor to visible electromagnetic radiations.

Spark plug short circuits the battery.

Due to large induced emf generated by inductor air gap in spark plug gets ionized.

When switch is opened, the inductor opposes change in flux by inducing a current in it. A large potential difference develops across the inductor for change to jump the spark plug gap.

Linked Question 3

If switch must be closed for up to three time constants. Find this time

3 ms

1.5 ms

1/3 ms

6 ms

Time constant $τ_{L} = \frac{L}{R_{C}} = \frac{(1 \times 10^{- 2})}{10} = 1 m / s$

$3 τ_{L} = 3 ms$