Question

When an AC source is connected to an ideal

inductor show that the average power supplied

by the source over a complete cycle is zero.

(ii) A lamp is connected in series with an

inductor and an AC source. What happens

to the brightness of the lamp when the key is

plugged in and an iron rod is inserted inside the

inductor? Explain.

Answer 1

Answer:

1) For an idol inductors phase difference between current and applied voltage =Pi/2

Power ,P = Vŕms I rms coso =

V rms I rms Cos pi/2 =0

The power consumed in a pure induction is zero .

2 ) Because of the current flowing across bulb will decrease and the decreasing the brightness of the bulb .

Please fallow me

Answer 2

Explanation:

QUESTION:-

When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.(ii) A lamp is connected in series with an inductor and an AC source. What happens to the brightness of the lamp whplugged in and an iron rod is inserted inside

theinductor? Explain

ANSWER :-

(i) As Pav = Vrms Irms cosf

extra information :-

_____________________________________

In ideal inductor, current Irms lags behind applied voltage V rms by π/2.

_____________________________________

Fi = π/2 so, P av = V rms I rms cos π/2

or Pav = V rms I rms × 0.

or P av = 0.

(ii) Brightness of the lamp decreases. It is because when iron rod is inserted inside the inductor, its inductance L increases, thereby increasing its inductive reactance XL and hence impedance Z of the circuit. As I rms = V

rms / Z

so, this decreases the

so, this decreases the current I rms in the circuit and hence the brightness of

rms in the circuit and hence the brightness of lamp.

____________________________________

@TheStellar♥️

Answer 3

Explanation:

QUESTION:-

When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.(ii) A lamp is connected in series with an inductor and an AC source. What happens to the brightness of the lamp whplugged in and an iron rod is inserted inside

theinductor? Explain

ANSWER :-

(i) As Pav = Vrms Irms cosf

extra information :-

_____________________________________

In ideal inductor, current Irms lags behind applied voltage V rms by π/2.

___________________________________

Fi = π/2 so, P av = V rms I rms cos π/2

or Pav = V rms I rms × 0.

or P av = 0.

(ii) Brightness of the lamp decreases. It is because when iron rod is inserted inside the inductor, its inductance L increases, thereby increasing its inductive reactance XL and hence impedance Z of the circuit. As I rms = V

rms / Z

so, this decreases the

so, this decreases the current I rms in the circuit and hence the brightness of

rms in the circuit and hence the brightness of lamp.

____________________________________

@TheStellar♥️