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47 questions By Tony R. Kuphaldt

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  • Question 10 of 47

    Calculate the current in this circuit, and also the amount of mechanical power (in units of “horsepower”) required to turn this alternator (assume 100% efficiency):




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  • Question 11 of 47

    A student is pondering the behavior of a simple series RC circuit:





    It is clear by now that the 4 kΩ capacitive reactance does not directly add to the 3 kΩ resistance to make 7 kΩ total. Instead, the addition of impedances is vectorial:



    XC2 + R2
    = Ztotal




    ZC + ZR = Ztotal




    (4kΩ ∠ - 90o) + (3kΩ ∠ 0o) = (5kΩ ∠ - 53.13o)



    It is also clear to this student that the component voltage drops form a vectorial sum as well, so that 4 volts dropped across the capacitor in series with 3 volts dropped across the resistor really does add up to 5 volts total source voltage:


    VC + VR = Vtotal




    (4V ∠ - 90o) + (3V ∠ 0o) = (5V ∠ - 53.13o)



    What surprises the student, though, is power. In calculating power for each component, the student arrives at 4 mW for the capacitor (4 volts times 1 milliamp) and 3 mW for the resistor (3 volts times 1 milliamp), but only 5 mW for the total circuit power (5 volts times 1 milliamp). In DC circuits, component power dissipations always added, no matter how strangely their voltages and currents might be related. The student honestly expected the total power to be 7 mW, but that doesn’t make sense with 5 volts total voltage and 1 mA total current.

    Then it occurs to the student that power might add vectorially just like impedances and voltage drops. In fact, this seems to be the only way the numbers make any sense:





    However, after plotting this triangle the student is once again beset with doubt. According to the Law of Energy Conservation, total power in must equal total power out. If the source is inputting 5 mW of power total to this circuit, there should be no possible way that the resistor is dissipating 3 mW and the capacitor is dissipating 4 mW. That would constitute more energy leaving the circuit than what is entering!

    What is wrong with this student’s power triangle diagram? How may we make sense of the figures obtained by multiplying voltage by current for each component, and for the total circuit?

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  • Question 12 of 47

    Calculate the current in this circuit, and also the amount of mechanical power (in units of “horsepower”) required to turn this alternator (assume 100% efficiency):




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