or
Subelement B
Electrical Math
Section 13
Power Relationships
What does the power factor equal in an R-L circuit having a 60 degree phase angle between the voltage and the current?
• 0.414
• 0.866
0.5
• 1.73

What does the power factor equal in an R-L circuit having a 60 degree phase angle between the voltage and the current?

0.5

From jordanm_78: \begin{align} Power Factor &= cos( ∢ {\hspace{2pt}\text{phase angle}}) \end{align} In this case, \begin{align} Power Factor &= cos(60°) \\ Power Factor &= 0.5 \end{align}

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If a resistance to which a constant voltage is applied is halved, what power dissipation will result?
Double.
• Halved.
• Remain the same.

If a resistance to which a constant voltage is applied is halved, what power dissipation will result?

Double.

From gsantee:

Using Ohm's Law equations:

Power (P) is equal to Resistance (R) times the Current (I) Squared.

• P= R x I²

Anytime R is halved, P will be halved as well.

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746 watts, corresponding to the lifting of 550 pounds at the rate of one-foot-per-second, is the equivalent of how much horsepower?
• One-quarter horsepower.
• One-half horsepower.
• Three-quarters horsepower.
One horsepower.

746 watts, corresponding to the lifting of 550 pounds at the rate of one-foot-per-second, is the equivalent of how much horsepower?

One horsepower.

The Imperial Horsepower of 746 watts is equivalent to 550 (550 ft-lbs/second), pounds at the rate of one-foot-per-second work.

1 Watt is 1 Newton-meter/second.

Converting from foot to meters, we get 1 foot of approximately 0.3048 meters.

1 pound = 0.454 kg = 0.454 ∗ 9.8= 4.492 N

So, we can determine that
550 ft/lb/s ∗ 0.3048 m/f ∗ 4.54 kg ∗ 9.8 Nm/s = 745.86 W
This shows that 550 ft/lb/s is equivalent to 746 Watts

By definition 1 HP is equal to 550 ft/lb/s

.

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In a circuit where the AC voltage and current are out of phase, how can the true power be determined?
By multiplying the apparent power times the power factor.
• By subtracting the apparent power from the power factor.
• By dividing the apparent power by the power factor.
• By multiplying the RMS voltage times the RMS current.

In a circuit where the AC voltage and current are out of phase, how can the true power be determined?

By multiplying the apparent power times the power factor.

From kd9fni:

Power Factor = True Power ÷ Apparent Power

True power = Apparent Power ∗ Power Factor

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What does the power factor equal in an R-L circuit having a 45 degree phase angle between the voltage and the current?
• 0.866
• 1.0
• 0.5
0.707

What does the power factor equal in an R-L circuit having a 45 degree phase angle between the voltage and the current?

0.707

From kd9fni:

Power factor = True Power ÷ Apparent Power.

If given an angle, calculate cos(45° degrees) = 0.707

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What does the power factor equal in an R-L circuit having a 30 degree phase angle between the voltage and the current?
• 1.73
0.866
• 0.5
• 0.577

What does the power factor equal in an R-L circuit having a 30 degree phase angle between the voltage and the current?

0.866

From kd9fni:

Power factor = True power ÷ Apparent Power

It can also be calculated by taking the phase difference between the voltage and current, so:

cos(30° degrees) = 0.866

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