Question

For the circuit Figure 5: V = 480220° V, Zline = 0.14+j0.22 2. Zoad = 5+j3 22 Determine the current through the circuit, the voltage across the line, the voltage across the load, the real power of the source, line and load, the reactive power of the source, line and load, and the apparent power of the source, line and load. Is the magnitude of the voltage at the load higher or lower than that of the source? Discuss what can be done to the circuit to make the magnitude of the voltage at the load be the same as that of the source. I NN 1. Figure 5

Answer 1

Ans)

current through the circuit is given as

480220 0.14 + 20.22 + 5+j3 79.142 - 12.06° 4. ne Zline + Zload

Iline = 79.142 - 12.06 A

as it is series circuit
IG = Iline = Iload = 79.142 – 12.06 A

voltage across the line is

Vine = Iline Zline = 79.142 - 12.06º * (0.14 +0.22)

Vline = 20.64245.46% v

voltage across load is

$V_{load}=I_{line}Z_{load}=79.14\angle -12.06^{o}*(5+j3)$

Vload = 461.45218.90 v

Now complex power across source is

$S_{G}=VI_{line}^{*}=480\angle 20^{o}*79.14\angle 12.06^{o}$

$S_{G}=P_{G}+jQ_{G}=32191.37 +j20166.58=37986.51\angle 32.06^{o}\: \: VA$

So real power of source is

$\mathbf{P_{G}=32191.37 \: \: W}$

reactive power is source is

$\mathbf{Q_{G}=20166.58 \: \: VAR}$

apparent power of source is

$\mathbf{\left |S_{G} \right |=37986.51 \: \: VA}$

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Now complex power across line is

$S_{line}=V_{line}I_{line}^{*}=20.64\angle 45.46^{o}*79.14\angle 12.06^{o}$

Sline = Pline + jQline = 876.81+; 1377.84 = 1633.2257.53' VA

So real power of line is

$\mathbf{P_{line}=876.81 \: \: W}$

reactive power is line is

$\mathbf{Q_{line}=1377.84 \: \: VAR}$

apparent power of line is

$\mathbf{\left |S_{line} \right |=1633.2 \: \: VA}$

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Now complex power across load is

SL = Viline = 461.45218.90 * 79.14Z12.06

$S_{L}=P_{L}+jQ_{L}=31314.56+j18788.73=36518.74\angle 30.96^{o}\: \: VA$

So real power of load is

$\mathbf{P_{L}=31314.56 \: \: W}$

reactive power is load is

$\mathbf{Q_{L}=18788.73 \: \: VAR}$

apparent power of load is

$\mathbf{\left |S_{L} \right |=36518.74 \: \: VA}$

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You can see that magnitude of load voltage is lower than the source i.e

Vil = 461.45 < V = 480

To make the voltages equal we need improve the power factor and makes it leading by connecting a capacitor in parallel with the load ,