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S1-Q8.)       Concerning a three phase load, if KW is 170 and voltage is 480 at 240 amps, what is the
                          power factor?

                          a.) 0.95 lagging
                          PF = Watts/(Volts*Amps*1.732), 170,000/(480V*240A*1.732) not =.95.
                          b.) 0.95 leading
                          PF = Watts/(Volts*Amps*1.732), 170,000/(480V*240A*1.732) not =.95, and it is not leading.
                          c.) 0.85 leading
                          PF = Watts/(Volts*Amps*1.732), 170,000/(480V*240A*1.732) =.95, but is not leading.
                          d.) *0.85 lagging
                          Correct.  PF = Watts/(Volts*Amps*1.732), 170,000/(480V*240A*1.732) =.85 and is
                          lagging.


            S1-Q9.)       Rectification involves

                          a.) *Preventing Alternating Current from Alternating above and below 0 volts.
                          Correct.  Normally the negative going part of the sine-wave is reproduced above 0V
                          each time the sine-wave approaches 0.  The result is positive, pulsating, DC.
                          b.) Making Direct Current Alternate above and below 0 volts
                          This would then be alternating current and not DC.  While this is achievable with power inverters, it
                          is not rectification.
                          c.) Changing watts into Horse Power
                          Watts is the same as horsepower already.  746 watts = 1 Horsepower.
                          d.) Correcting Power Factor
                          This is achieved with Capacitors or Synchronous Condensers and has nothing to do with rectification.

            S1-Q10.)      If you think of a diode like a fluid system, its function is

                          a.) Allow current flow in two directions like a hydraulic motor
                          Diodes block in one direction.
                          b.) *Prevent current flow in one direction similar to a check valve
                          Correct.  Diodes block in one direction just like a check valve would stop flow from
                          going backwards in a fluid system, diodes prevent current from flowing the wrong
                          direction possibly damaging equipment.
                          c.) Monitor current flow like a flow meter
                          Diodes are used for blocking, not monitoring.
                          d.) Create flow like a hydraulic pump
                          Diodes do not create anything.

            S1-Q11.)      Which of the following will NOT affect the resistance of a conductor:


                          a.) Size
                          Size will directly affect the resistance of a conductor.  Cross sectional area of wire determines how
                          much current it can carry.  That determination is based on the amount of resistance the current will
                          see.
                          b.) Length
                          The resistance of the cross sectional area is multiplied by the number of feet of conductor.
                          c.) *Insulation
                          Correct.  Insulation determines the level of voltage the conductor can carry before it will fail.
                          The insulation has nothing to do with resistance.
                          d.) Temperature
                          Increased temperature directly affects resistance.  The higher the temperature, the higher the






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