An inductor L of reactance XL is connected in series with a bulb B to an a.c. source .Briefly explain how does the brightness of the bulb changes when a) no. of turns of the inductor is reduced. b)a capacitor of reactance XC=XL is included in series in the same circuit

Asked by sonalchandra | 11th Sep, 2010, 03:00: PM

Expert Answer:

Dear student
In case of AC circuit with a bulb and L, it is similar to LR circuit.
In such a case as L will reduce in case the number of turns are reduced.
As teh instantenous current
as L lowers, the current through the circuit reduces & hence the brightness of the bulb will reduce.
In case of a capacitor being introduced such that Xc= XL,
the resonance condition is achived.
In such secenario,
Io = Vo/Z, where Z = R of the bulb i minimum.
Hence the current in te circuit will be maximum.
bulb will glow the brightest.
 
Hope that clarifies your query.
Regards
Team
Topper learning

Answered by  | 14th Sep, 2010, 10:59: AM

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