E- high side = 480VTurns ratio = 2:1 coilResistance per load = 10 ohmsEfficiency = 100%answer the questions, I can't figure this out... VAH4Figure 6T-2T-31-3T-2T-1X-2meX-XX-3X-4Worksheet 3E-XX-3X4H-4 E) I/resistor on the load isa) 2Ab) 4Ac) 6.93Ad) 12AF) I-line on the low side isa) 2Ab) 4Ac) 6.93Ad) 12AG) I-phase on the low side isa) 2Ab) 4Ac)d)6.93A12AH) I/coil on the low side isa) 2Ab) 4Ac) 6.93Ad) 12AI) I-phase on the high side isa) 1Ab) 1.73Ac) 2Ad) 4AJ) I-line on the high side isa) 1Ab) 1.73Ac) 2Ad) 4A

Answered: E) I/resistor on the load is a) 2A b)…

Introductory Circuit Analysis (13th Edition)

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ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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Question

100%

E- high side = 480V

Turns ratio = 2:1 coil

Resistance per load = 10 ohms

Efficiency = 100%

answer the questions, I can't figure this out...

VA
H4
Figure 6
T-2
T-3
1-3
T-2
T-1
X-2
me
X-X
X-3
X-4
Worksheet 3
E-X
X-3
X4
H-4

E) I/resistor on the load is
a) 2A
b) 4A
c) 6.93A
d) 12A
F) I-line on the low side is
a) 2A
b) 4A
c) 6.93A
d) 12A
G) I-phase on the low side is
a) 2A
b) 4A
c)
d)
6.93A
12A
H) I/coil on the low side is
a) 2A
b) 4A
c) 6.93A
d) 12A
I) I-phase on the high side is
a) 1A
b) 1.73A
c) 2A
d) 4A
J) I-line on the high side is
a) 1A
b) 1.73A
c) 2A
d) 4A

Expert Solution

Step 1

given high side voltage is 480 volts

so coil voltage on the primary side is 240(as they are in series)

so coil voltage in the secondary side is 120 volts which is same as line voltage because two cols are in parallel . so load voltage per phase is $\frac{l i n e v o l t a g e}{\sqrt{3}} = \frac{120}{\sqrt{3}} = 69.3$

per phase load current is $\frac{P h a s e v o l t a g e}{R e s i s \tan c e} = \frac{69.3}{10} = 6.93$

So option C is correct.

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