Q. 16

# (i) When an AC so

Answer :

(i) The source is AC, so, The average power provided by the source over a complete cycle is given as ,

Paverage = Erms.Irms.cos θ

Where,

Erms is the root mean square value of voltage,

Irms is the root mean square value of the current provided.

θ is the phase difference between the current and the voltage.

If the circuit contains an ideal inductor then, θ = π/2

So, cosθ = 0

Hence, Paverage = 0

Therefore, when an AC source is connected to an ideal inductor, the average power supplied by the source over a complete cycle is zero.

(ii) If the key is plugged in and the iron rod is inserted inside the inductor, the inductance is decreased. Hence, the reactance of the inductor (XL = ω0L) decreases. So, the impedance of the circuit increases, which decreases the current in the circuit. Thus, the brightness of the lamp decreases.

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