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Discuss the power in AC circuit with only an inductor. |
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Answer» Solution :The current reaches its maximum value later than the voltage by one-fourth of a PERIOD `(T)/( 4) = ((pi )/( 2))/( omega)` . The instantaneous power supplied to the INDUCTOR is, `P_(L) = IV = I_(m) sin ( omega t - ( pi )/( 2)) xx V_(m) sin (omega t )` `= - I_(m) V_(m) cos ( omega t ) sin ( omega t )` ` = - ( I_(m) V_(m))/( 2) sin ( 2 omega t )` The average power over a complete cycle is, `P_(L) = langle-(I_(2)V_(m) sin ( 2 omega t ))/(2)rangle= (I_(m) V_(m)) /(2) lt sin 2 omega t gt` `= - ( I_(m) V_(m) )/( 2) = 0 ` Since the average of `sin ( 2 omega t )` over a completecycle is zero. Thus, the average power supplied to an inductor over one completecycle is zero. Figures explains it in detail.  0-1 current i through the COIL entering at A increase from zero to a maximum value. Flux lines are setup means the core gets magnetised. With the polarity shown voltage and current are both positive. So their product p is positive. Enbery is ABSORBED from the source.  1-2 current in the coil is still positive but is decreasing. The core gets demagnetised and the net flux becomes zero at the end of a half cycle. The voltage v is negative ( SInce di `//` dt is negative. ) The product of voltage and current is negative and energy is begin returned to source.  2-3 current i becomes negative means it enters at B and comes out of A. Since the direction of current has changed, the polarity of the magnet changes. The current and voltage are both negative. So, their product p is positive . Energy is absorbed.  3-4 current i decreases and reaches its zero value at 4 when core is demagnetised and flux is zero. The voltage is positive therefore, negative energy absorbed during the `1//4` cycle 2-3 is returned to the source. |
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