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.
From gsantee:
Using Ohm's Law equations:
Power (P) is equal to Resistance (R) times the Current (I) Squared.
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 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.
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?
0.707
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?
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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