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Suppose the circuit inhas a resistance of 15 W. Obtain the average power transferred to each element of the circuit, and the total power absorbed. |
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Answer» Solution :We known that, `I_(rms) = ( V_(rms))/(|Z|)` `= (V_(rms))/( sqrt( R^(2) + ( X_(L ) - X_(C))^(2)))` `= ( V_(rms))/( sqrt( R^(2) +( 2pi fL - ( 1)/( 2pi fC))^(2)))` `= ( 230)/(sqrt( (15)^(2) + ( 2 XX 3.14 xx 50 xx 80 xx 10^(-3) - ( 1)/( 2 xx 3.14 xx 50 xx 60 xx 10^(-6)))^(2)))` `:. I_(rms) = 7.256A` (i) Average power transferred to resistor, `P _(R ) = I_(rms)^(2) R ` `= ( 7.256)^(2) ( 15)` `= 789.7 W` (Watt) (ii) Average power transferred to an inductor, `P_(L) = 0 ``( :. phi = ( pi )/(2) rad )` (iii) Average power transferred to a capacitor, `P_(C )= 0( :. phi= - (pi)/(2) rad)` `rArr` Total power absorbed (or consumed ) by the circuit, `P = P_(R ) + P_(L) + P_(C 0` `= P_(R ) + 0+0` `= P_(R 0` `= 789.7 W` |
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