Sample Electrical FE Question

[wvgirl14]

Okay it is getting down to the wire and I am working some practice exams. This is a problem I have encountered and cannot come up with the correct answer. Any help would be appreciated!

THREE 33-OHM RESISTORS CONNECTED IN DELTA ARE SUPPLIED THROUGH THREE LINES OF 1.0-OHM RESISTANCE EACH FROM 208-VOLT BALANCED THREE-PHASE SOURCE. tHE MAGNITUDE OF THE ROOT-MEAN SQUARE, LINE-TO-LINE VOLTAGE ACROSS EACH 33-OHM RESISTOR IS MOST NEARLY

(A) 110 V

(B) 120 V

© 190 V

(D) 208 V

(E) 360 V

Question.pdf

 

[Flyer_PE]

I'll take a stab at this. Since it's the FE and you're a CE, I'm not sure what you have for references.

Step 1 is to convert the delta connected load to an equivalent wye circuit:

Zwye = Zdelta / 3 = 33/3 = 11

The equivalent wye circuit is then three 11 ohm resistors connected in a wye configuration.

Step 2 is to convert the line-to-line voltage into a line-to-neutral equivalent. This is done simply by dividing the line voltage by sqrt3.

VL-N = VL-L / sqrt3 = 208/sqrt3 = 120 V.

Now that you have a wye circuit, you can analyze the circuit on a single-phase basis.

It is now a cicuit with a 120 V source in series with a 1 ohm resistor and an 11 ohm resistor. The current in this circuit is then 120 V / 12 ohm = 10 amps.

The line-neutral voltage at the load (11 ohm resistor) is then 120V - (10Amps*1 ohm) = 110V

The Line-to-Line voltage is then the line-neutral voltage multiplied by sqrt 3: VL-L = 110V*sqrt3 = 190.5 Vac

The answer is C

Hope this helps. I can scan a hand written solution in later if that would also help.

 

[wvgirl14]

I'll take a stab at this. Since it's the FE and you're a CE, I'm not sure what you have for references.
Step 1 is to convert the delta connected load to an equivalent wye circuit:

Zwye = Zdelta / 3 = 33/3 = 11

The equivalent wye circuit is then three 11 ohm resistors connected in a wye configuration.

Step 2 is to convert the line-to-line voltage into a line-to-neutral equivalent. This is done simply by dividing the line voltage by sqrt3.

VL-N = VL-L / sqrt3 = 208/sqrt3 = 120 V.

Now that you have a wye circuit, you can analyze the circuit on a single-phase basis.

It is now a cicuit with a 120 V source in series with a 1 ohm resistor and an 11 ohm resistor. The current in this circuit is then 120 V / 12 ohm = 10 amps.

The line-neutral voltage at the load (11 ohm resistor) is then 120V - (10Amps*1 ohm) = 110V

The Line-to-Line voltage is then the line-neutral voltage multiplied by sqrt 3: VL-L = 110V*sqrt3 = 190.5 Vac

The answer is C

Hope this helps. I can scan a hand written solution in later if that would also help.

Thanks, Flyer that helps alot! I looked everywhere trying to solve that darn thing. I really appreciate it!