A circuit consists of a 12 V battery, a switch, and a light bulb--all connected

Question

A circuit consists of a 12 V battery, a switch, and a light bulb--all connected in series. It is known that the light bulb requires a minimum current of 0.15 A in order to produce a visible glow. In the circuit, that particular bulb draws 2.6 W when the switch has been closed for a long time. Now, an inductor is put in series with the bulb and the rest of the circuit. If the light bulb begins to glow 3.0 ms after the switch is closed, how large is the self-inductance of the inductor? Ignore any heating time of the filament and assume the glow is observed as soon as the current in the filament reaches the 0.15 A threshold. Your Answer: Answer units

Explanation / Answer

P = V^2 / R

R = V^2 / P

R = 12^2 / 2.6

R = 55.38 ohm

I after joining inductance = .15 A

So

I = V / X

X = Total resistance = R + inductive reactance (XL)

I = V / (R + XL)

.15 = 12 / (55.38 + XL)

XL = 54.58 ohm

Now another formula of XL

XL = 2.pi.f.L

f = 1/T = 1/(3 x 10^-3) = 333.33

L = XL / 2.pi.f

L = (54.58) / (2 x pi x 333.33)

L = 2.60 x 10^-2 H

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