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Define the power for AC circuit. Obtain an equation of average power for L-C-R series AC circuit. |
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Answer» Solution :Electric power is the rate of ENERGY consumption in an electric CIRCUIT. Instantaneous power cannot be measured in AC circuit, hence true power is measured. True power in AC circuit means the value of average power in full period. Let a voltage `V = V_(m) sin omega t ` applied to a AC circuit drives a current in the circuit by `I = I_(m) sin ( omega t +phi )` where `I_(m) = (V_(m))/( Z) ` and `phi= tan^(-1) ((X_(C ) - X_(L))/( R ))` The intantaneous power supplied by the source, P=VI `= ( V_(m) sin omega t ) [ I_(m) sin ( omega t + phi )]` `= V_(m) I_(m) sin omega t. sin ( omega t + phi )` but `2 sin A sin B = cos ( A-B) - cos ( A+B)` `:. P = ( V_(m) I_(m)) /( 2) [ cos phi - cos ( 2 omega t + phi )]`....(1) The average power over a cycle is given by the average of the two terms in R.H.S. of EQUATION. It is only the second term `cos ( 2 omega t + phi )` which is time - dependent . Its average is zero. `:. P = ( V_(m) I_(m))/( 2) cos phi ` `:. P = ( V_(m))/( sqrt(2)). ( I_(m))/(sqrt(2)) cos phi` `:. P = V_("rms") I_("rms") cos phi` but `V_("rms")` is denoted by V and `I_("rms")` is denoted by I. `:. P = VI cos phi` But taking `V = IZ``[ :. V = IR ]` `P = I^(2) Z cos phi` So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle `phi` between them. The quantity `cos phi` is called the power factor. It is UNITLESS. The ratio of resistance and impedence is called power factor it is denoted by `cos phi`. Power factor ` = cos phi = ( P )/( I^(2) Z ) = ( R )/( Z )` |
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