please explain the steps. somehow, i am getting power factor wrong. 2.5 S2 j4.0

Question

please explain the steps. somehow, i am getting power factor wrong.

2.5 S2 j4.0 2 Ib 80.0 S2 250/0°v -j 150 ? j25.0 2 Given: The load voltage shown in the above circuit is an rms phasor voltage Required a. Calculate the rms magnitude and angle of the voltage source, Vs b. Calculate the rms magnitude and angle of the phasor current, Ib c. Calculate the real and reactive power delivered by the voltage source, Vs d. Calculate the apparent power and power factor (in percent) of the voltage source, Vs Solution: Your answer for the power factor bpe is incorrect. Please try again. Number of tries: 25! Hint: First, determine the total current drawn from the source. Second, find the voltage drop across the feeder and add it to the given load voltage. Now you can calculate the complex power delivered by the source a. V 254.36 b. Ib 2.95 c. P 733.22 Wand Qs 159.66 d. ?s,I = 750.40 VA @ a [ ? lagging. O unity or O leading] PF of 97.71 2 3.00 ?1527 VAR Submit these values

Explanation / Answer

IN THE QUESTION VOLTAGE 250 AT ANGLE ZERO IS TAKEN AS REFERENCE .

SO YOU HAVE OBTAINED SOURCE VOLTAGE AS 254.36 AT ANGLE 3 DEGREE.

IN QUESTION IT IS MENTIONED POWER FACTOR OF VOLTAGE SOURCE .

SO IT IS P.F. = COS (3 degree )=0.9986 = 99.86 % with respect to given voltage .

and as the angle is positive it is a leading type source.

the answer you have written 97.7 % is the power factor of source current with respect to source voltage that is

COS(15.27-3 )=0.977 =97.7 % AND THAT IS LEADING TYPE AS ANGLE IS POSITIVE ,NOT THE SOURCE VOLTAGE .

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