A 230 V RMS source supplies power to two loads connected is paral
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A. (18 + j1.5) kVA
B. (18 – j1.5) kVA
C. (20 + j1.5) kVA
D. (20 – j1.5) kVA
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Analysis:
The circuit for the given information is drawn as:
Complex power Delivered by source = Apparent power of load 1 + Apparent power of load 2
Consider load 1:
Given: P1 = 10 kW
Power factor = 0.8 (lead)
Since the power factor is leading, we can say that capacitive reactance is present in load 1.
Calculation:
Load 1 = R1 – j Xc and cos θ1 = 0.8
Apparent power S1 = P – jQC
θ1 = cos-1 (0.8) = 36.86°
The power triangle for the above is drawn as:
\(\tan {\theta _1} = \frac{{{Q_c}}}{{{P_1}}}\)
\(\tan 36.86 = \frac{{{Q_c}}}{{10}}\)
Reactive power QC = 10 × 0.75
S1 = (10 – j7.5) kVA
Qc = 7.5 VAR
Consider load 2:
Given S2 = 10 kVA
Power factor = cos θ2 = 0.8 (lag)
θ2 = cos-1 (0.8)
θ2 = 36.86°
Lagging power factor means inductive reactance.
The power triangle for the above is drawn as:
P2 = S2 cos θ2 and
Ql = S2 sin θ2
P2 = (10 × 0.8) kW
QL = 10 × 0.6
P2 = 8 kW
QL = 6 kVAR
S2 = P2 + jQL
S2 = (8 + j6) kVA
The equivalent complex power delivered by the source will be:
Sev = S1 + S2
Sev = (10 – j7.5) + (8 + j6) kVA
Sev = (18 – j1.5) kVA
∴ Option B is correct.
Alternate Method:
Apparent power angle is of opposite nature to that of the power factor angle.
Load 1:
P1 = 10 kW, cos θ1 = 0.8 (leading)
θ1 = cos-1 (0.8) = 36.86°
\(\left| {{S_1}} \right| = \frac{{\left| {{P_1}} \right|}}{{\cos \theta }}\)
\(\left| {{S_1}} \right| = \frac{{10\;kW}}{{0.8}} = 12.5\;kVA\)
S1 = 12 kVA ∠-36.86°
Angle is of opposite in nature to that of power factor angle because S = V(I*)
This conjugate makes it of opposite sign.
S1 = 12.5 ∠-36.86° kVA
Or S1 = (10 – j7.5) kVA
Load 2:
|S2| = 10 kVA, P.F. = 0.8 (lag)
S2 = 10 ∠+36.86° kVA
The angle is of opposite nature to that of power factor angle.
S2 = 8 + j6 kVA
Sev = S1 + S2
Sev = 18 – j1.5 kVA