Question

Write a small report of maximum one page answering the following questions: [8 marks] (a) What is apparent power? (b) What is reactive and real power? (c) What is the power factor? (d) What cause the changes of power factor and how can you improve the power factor in a network

Answer 1

- Apparent power :- . s= Vrms Irms - ② Reaclive power? Q = Vrms Irms Cose © Real power? P- Vrms Irmis coso Power factor (PA) = coso cosine of angle between Voltage and current is called as power factor. changes of power bacter caures Reactive power absåbed sy the system is changes power factor can be improved sy using capacitor banks, ②

1.

Consider the formula for the bandwidth BW. BW= 1 BWRC Substitute 10 uF for C and 1 ks for R. BW= (1 k)(10 u) 0.01 = 100 rad/s Therefore, the value of the capacitor is 100 rad/s.

Consider the formula for the resonant frequency in terms of inductor and capacitor. 0, TIC Substitute 10 uF for C and 100 mH for L. 00 (100 m)(104) 1 1 m = 1000 rad/s Therefore, the value of the resonant frequency is 1000 rad/s.

of the circuit in terms of the resonant frequency Consider the formula for the quality factor and bandwidth. BW Substitute 100 rad/s for BW and 1000 rad/s for 0,- Q=1000 9 100 = 10 Therefore, the value of the quality factor is 10.

Consider the formula for the lower half power frequency. Lo = 0, - Substitute 1000 rad/s for 0, and 10 for Q. 00 = (1000) 2009 (200) =(1000) = (1000) – (0.05)+ J(0.05) +1 = (1000)[-(0.05)+ V2.5 m+1] 010 = (1000)[-(0.05)+ V1.0025] = (1000)[-(0.05)+1.00125] = (1000)[0.95125] =951.25 rad/s Therefore, the value of the lower half power frequency is 951.25 rad/s.

Consider the formula for the higher half power frequency in terms of the lower half power frequency and the bandwidth. 0.1 = BW + 10 Substitute 100 rad/s for BW and 951.25 rad/s for 0.0 0,1 = 100+951.25 = 1051.25 rad/s Therefore, the value of the higher half power frequency is 1051.25 rad/s.

Consider the circuit diagram for the parallel combination of resistor capacitor and inductor as shown in Figure 1. 24 © 2 A Sr Sjol V(jo) Tjo C Figure 1

Consider the expression for the admittance of the network at lower half power frequency. Y(10,0)=171 Rjo.+300C Substitute 951.25 rad/s for 0 , 1 ks for R. 10 uF for C and 100 mH for L. Y(0,0)=15 Ik ;(951.25)(100 m)': +7(951.25)(10 u) = 0.001+ 1 *;(95.125)+ |(9.5125 m) = 0.001- 10.010512+ j(9.5125 m) = 1- jl ms Therefore, the value of the admittance at the lower half power frequency is 1-jl ms.

Consider the expression for the voltage across the network at lower half power frequency. v (0,0) Y(0,0) Substitute 1- jl ms for Y(20.). V(0 ) T-j1 m 2 722-45° m =V22450 kV Therefore, the value of the voltage across the circuit at lower half power frequency is 12245° kV

Consider the expression for the admittance of the network at higher half power frequency. 11 @C Y(j) = jQL Substitute 1051.25 rad/s for 0,4; 1 ks for R. 10 uF for C and 100 mH for L. y (10 ";(1051.25)(100 m) *+,(1051.25)(104) (105.125) + j(10.5125 m) = 0.001 - 19.512485 m+ j(10.5125 m) = 1+ jl ms = 0.001+_ Therefore, the value of the admittance at the higher half power frequency is 1+ jl ms.

Consider the expression for the voltage across the network at higher half power frequency VjYj Substitute 1+ jl ms for Y(jo). V(0:1) = " 1+ jl m 2 2245° m =/22-45° kV Therefore, the value of the voltage across the circuit at higher half power frequency is V22-45° kv

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