In Fig. 13.8, the load impedance is Z=(15+j12)Ω at the frequency of the fixed amplitude source voltage Vrms=120 V∠0. The load requires real power P=500 W . What rms line current would be required if the load were Z=15 Ω ? What is the actual rms line current?

Question

In Fig. 13.8, the load impedance is [latex] Z=(15+j 12) \Omega[/latex] at the frequency of the fixed amplitude source voltage [latex] ilde{V}_{r m s}=120 V \angle 0[/latex]. The load requires real power P=500 W . What rms line current would be required if the load were [latex] Z=15 \Omega[/latex] ? What is the actual rms line current?

Accepted Answer

The rms current required to deliver the specified power to the load is

[latex] I_{r m s}{}^2 R=P \Rightarrow I_{r m s}=\sqrt{\frac{P}{R}}=\sqrt{\frac{500 W }{15 \Omega}} \cong 5.77 A . [/latex]

The actual line current is

[latex] I_{r m s}=\frac{|S|}{V_{r m s}}=\frac{V_{r m s}}{|Z|}=\frac{120 V }{\sqrt{15^2+12^2} \Omega} \cong 6.25 A . [/latex]

We find

[latex] 100 imes \frac{6.25 A -5.77 A }{5.77 A } \cong 8.3 \%, [/latex]

which means that the current is about 8 % larger than necessary to deliver the required power.

Also

[latex] 100 imes \frac{(6.25)^2 A ^2-(5.77)^2 A ^2}{(5.77)^2 A ^2} \cong 17.3 \%, [/latex]

which means that the losses in the line are about 17 % greater than they would be if only the required current were delivered. We show in Section 13.8 how the load can be modified such that the reactive power required is reduced, thus reducing the line current and attendant losses.

Question 13.3: In Fig. 13.8, the load impedance is Z=(15+j12)Ω at the freq...

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