Inductive Reactance - Assignment Example

With 25 F capacitor connected in series,
Current =
(Impedance)2 = (resistance)2 + (net reactance)2
Z2 = R2 + (XL-XC)2
XC = =
= 63.66 Ω
Z2 = 452 + (66.162 - 63.66)2
= 2027.502
Z = 45 Ω
Current =
=
= 2.22 A
(h) Voltage across the resistor IR = 2.22 x 45
= 99.9 V
(i) Voltage across the coil = IXL
= 2.22 x 66.162
= 146. 87 VAR
(j) Voltage across the capacitor = IXC
= 2.22 x 63.66
= 141. 33 VAR

(k) Phase angle between supply voltage and current is given by
tan-1( ) = tan-1( )
= 3.18°

Task 2;
(a) Efficiency =
Apparent input power =
=
= 9.33 KID
The apparent current is drawn by the motor =
=
= 40.56 A
(b) With the power factor improved to 0.925,
Load KVA =
= 7.31 KVA
Supply current =
=
= 31.78 A
(c) Capacitor connection involves the parallel connection of the capacitive reactance and motor impedance.
(Resultant current)2= (sum of active components of current)2 + (sum of reactive components of current)2
31.782 = (IC cos(90) + IM cos )2 + (IC sin(90) - IM sin )2
= (40.56 x 0.725)2 + (IC - (40.56 x 0.689))2
1009.97 = 864.713 + (IC - 27.94)2
(IC - 27.94)2= 145.257
IC - 27.94 = 12.05
IC = 27.94 + 12.05 = 39.99 A

(d) With the power factor at 0.725,
Load KVAR= KVA x sin
= 9.33 x sin (43.53)
= 6.43 KVAR
With the power factor improved to 0.925,
New KVAR = New KVA x sin (22.33)
= 7.31 x sin (43.53)
= 2.78 KVAR
Therefore leading KVAR required to be supplied by the capacitor
= 6.43 - 2.78
= 3.65 KVAR
But leading KVAR= IXC
XC =
=
= 91.27 Ω
XC =
C= =
= 34.87 F
(e) KVAR rating of the capacitor = Load KVAR- New KVAR
= 6.77- 2.78
= 3.65 KVAR leading

Task 3;
(a) The p-n junction diode is a two-terminal semiconductor electronic device that conducts current in one direction only. It allows low resistance to the flow of current in one direction and very high resistance in the reverse direction. A semiconductor material, usually silicon, germanium, or gallium arsenide is added with impurities or doped to form an n-type region with negative charge carries on one side and a p-type region with positive charge carries on another side. The two are attached together resulting in a depletion region with no charge carriers between them and the terminals are connected at the ends of the two regions. The resulting crystal allows the flow of electrons across the junction from the n-type region to the p-type region (forward bias) but not the reverse (reverse bias), thus blocking current flow in one direction.

(b)
(i) In the purely resistive form of circuit, the current is always in phase with the voltage.
For the fundamental wave, R1= 25 Ω. For the second harmonic, R2=2 x 25 = 50 Ω. For the third harmonic, R3= 3 x 25= 75

Current iR = = (1000 t + ) + (2000 t - ) + (3000 t + )
iR = 8 (1000 t + ) + 3 (2000 t - ) + (3000 t + ) A
(ii) In a purely inductive circuit, the current is lags behind the voltage by rads.
For the fundamental wave, X1= 2 x 500 x 0.025 Ω = 78.54 Ω. For the second harmonic, R2=2 x 78.54 = 157.1 Ω. For the third harmonic, R3= 3 x 78.54= 235.6 Ω

Current iL = = (1000 t + - ) + (2000 t - - ) + (3000 t + - )
iL = 2.55 (1000 t - ) + 0.955 (2000 t - ) + 0.32 (3000 t + ) A
(iii) In a purely capacitive circuit, the current is leads the voltage by rads.

For the fundamental wave, X1= Ω = 12.73 Ω. For the second harmonic, R2= 2 x 12.73= 25.46 Ω. For the third harmonic, R3= 3 x 12.73= 38.2 Ω
Current iC = = (1000 t + + ) + (2000 t - + ) + (3000 t + + )
iC = 15.71 (1000 t - ) + 5.89 (2000 t + ) + (3000 t + ) A

Task 4;
(a) Given the turn ratio 25: 1, secondary voltage= 5000 x
= 200 V
Full load secondary rating = VSIS = 10 kVA
IS =
= 50 A

(b) Given secondary voltage= 200 V,
minimum load resistance, RL = = = 4 

(c) = for which primary current = I1 = x I2 = = 2 A

(d) Secondary impedance referred to the primary = ZS’ = ZS x ( )2
Impedance seen at the input = 4 x 252= 2500 

(e) Power dissipated across load resistance = I2R = 502 x 4
= 10 kW Read More