Question

QUESTION 23 A single-phase AC power network is shown in the figure below with four different loads. Determine the root mean squared (rms) magnitude of the current in the 20-hp motor load. (Select the one best answer from the choices given below.) $I_S = \frac{100}{15^\circ} A$ (without Fp correction load) Oa-31.08 A Ob-38.85 A Oc-44.4 A Od-55.5 A O e-60 A $E_S = 480\angle 0^\circ V$ Sync machines 10 kW lighting load 15 kVA $F_p = 0.866$ 20 hp $\eta = 80\%$ Capacitive Load power factor correction leading $F_p = 0.7$ logging

          QUESTION 23
A single-phase AC power network is shown in the figure below with four different loads. Determine the root mean squared (rms) magnitude of the current in the 20-hp motor load. (Select the one best answer from the choices given below.)
$I_S = \frac{100}{15^\circ} A$ (without Fp correction load)
Oa-31.08 A
Ob-38.85 A
Oc-44.4 A
Od-55.5 A
O e-60 A
$E_S = 480\angle 0^\circ V$
Sync
machines
10 kW
lighting
load
15 kVA
$F_p = 0.866$
20 hp
$\eta = 80\%$
Capacitive Load
power factor
correction
leading
$F_p = 0.7$
logging

QUESTION 23
A single-phase AC power network is shown in the figure below with four different loads. Determine the root mean squared (rms) magnitude of the current in the 20-hp motor load. (Select the one best answer from the choices given below.)
IS = (100)/(15^∘) A (without Fp correction load)
Oa-31.08 A
Ob-38.85 A
Oc-44.4 A
Od-55.5 A
O e-60 A
ES = 480∠ 0^∘ V
Sync
machines
10 kW
lighting
load
15 kVA
Fp = 0.866
20 hp
η = 80%
Capacitive Load
power factor
correction
leading
Fp = 0.7
logging

Added by Magdalena S.

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