The Physics
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Opus in profectus

Electric Power

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Practice

practice problem 1

Read the following description of the execution of Kenneth Stewart in Virginia on 23 September 1998.

The electric cycle, 1825 volts at approximately 7.5 amps for 30 seconds, then 240 volts at approximately 1.5 amps for 60 seconds… a 5 second pause intervenes, and the cycle is repeated, was designed to render the condemned brain dead within the first few moments. The function of the remainder of the cycle is to stop the body's organs so that a physician can certify that death has occurred….

Amnesty International, 1998

Determine…

  1. the total electric energy delivered to Mr. Stewart on the day of his execution.
  2. his change in temperature assuming assuming a standard 70 kg mass and a body made entirely of water.

solution

  1. Fairly straightforward. Energy is power times time. Electric power is voltage times current. Energy is a scalar, so just add up the parts of the cycle and double each to get the total.

    E = ∑Pt = ∑VIt
    E = 2[(1825 V)(7.5 A)(30 s) + (240 V)(1.5 A)(60 s)]
    E = 864,000 J

  2. The electric energy applied to Stewart's body was transformed into heat. Use the sensible heat equation and assume that the human body is basically a bag of water.

    E = Q = mc∆T  
     
    ∆T =  Q
    mc
    ∆T =  (864,000 J)
    (70 kg)(4190 J/kg°C)
    ∆T ≈ 3 °C  
     

practice problem 2

Humboldt State University in northern California has a unique laboratory devoted to appropriate technology called the Campus Center for Appropriate Technology (CCAT). One of the many innovations pioneered at CCAT is a project known as "pedal power" — a set of stationary bicycles connected to electric generators that provide current for household use.

Most of the energy used to fuel the CCAT house comes from the Sun, but a battery bank is also charged with 24 V permanent magnet DC generators run by exercise bikes in the basement. The flywheel on each bike is furnished with a groove. Through this groove is a belt that is attached to the generator's central shaft. The average person pedaling 70 rpm spins the shaft creating an electrical current of 4 A…. This allows someone to get a comfortable workout while at the same time charging the batteries.

Auto-Free Times, 2001

Now answer these questions…
  1. Determine the power output of person pedaling a stationary bicycle at the CCAT house.
  2. How much work does the person do in one minute?
  3. Over what total distance have the pedals moved in one minute? (Assume a crank arm length of 18 cm.)
  4. How fast are the person's feet moving while generating power at the CCAT house?
  5. With what average force does a typical person in the CCAT house push down on the pedals?

solution

  1. Pick the appropriate electric power equation.

    P = VI
    P = (24 V)(4 A)
    P = 96 W

  2. Work is power times time.

    W = Pt
    W = (96 W)(60 s)
    W = 5,760 J

  3. This is a geometry problem. In one minute, the pedaler's foot would crank through 70 circumferences.

    ∆s = 70 × 2πr
    ∆s = 140π(0.18 m)
    ∆s = 79 m

  4. This is a basic kinematics problem.

      v = 
    ∆s
    ∆t
     
      v =  79 m
      60 s
      v = 1.3 m/s
     

  5. Use the force-velocity form of the power equation (P = Fv cos θ). Solve it for force. (We need to assume that the force applied by the foot to the crank arm is always parallel to the pedal's velocity.) Insert numbers. Compute solution.

      F = 
    P
    v
     
      F =  96 W
      1.3 m/s
      F = 73 N
     

practice problem 3

Two similar looking AA batteries
Which of these apparently similar rechargeable batteries is "better"? Answer this by computing the…
  1. charge in…
    1. milliamp hours [mAh]
    2. coulombs [C]
  2. potential energy in…
    1. milliwatt hours [mWh]
    2. joues [J]

Compile your findings in a table like this one.

Two apparently similar rechargeable batteries
type voltage charge energy
(V) (mAh) (C) (mWh) (J)
NiZn 1.6     2500  
NiMH 1.2 2500      

solution

There is no correct sequence for completing this table. Here's one I just thought of. Start with the definition of potential difference or voltage…

V =  W
q

Solve it two ways.

q =  W
V
 ⇐ 
V =  W
q
W = qV  ⇐ 
V =  W
q

Compute the charge of the NiZn battery first.

  qNiZn =  W
  V
  qNiZn =  2500 mWh
  1.6 V
  qNiZn = 1563 mAh
 

Compute the energy of the NiMH battery (or the work it can do) using similar logic.

WNiMH = qV
WNiMH = (2500 mAh)(1.2 V)
WNiMH = 3000 mWh

Convert the charges to coulombs using the definition of current.

q = I∆t  ⇐ 
I =  ∆q
∆t

Shift the decimal 3 places when converting from milliamps to amps. Replace an hour with 3600 s.

qNiZn = I∆t
qNiZn = (1.563 A)(3600 s)
qNiZn = 5625 C
qNiMH = I∆t
qNiMH = (2.500 A)(3600 s)
qNiMH = 9000 C

Convert the energies to joules using the defintion of power.

W = P∆t  ⇐ 
P =  ∆W
∆t

Shift the decimal 3 places when converting from milliwatts to watts. Replace an hour with 3600 s.

WNiZn = P∆t
WNiZn = (2.500 W)(3600 s)
WNiZn = 9000 J
WNiMH = P∆t
WNiMH = (3.000 W)(3600 s)
WNiMH = 10,800 J

Compile the results in a table.

Two apparently similar rechargeable batteries
type voltage charge energy
(V) (mAh) (C) (mWh) (J)
NiZn 1.6 1563 5625 2500 09,000
NiMH 1.2 2500 9000 3000 10,800

The nickel metal hydride battery can hold more energy than the nickel zinc battery, which translates to longer periods between rechargings. That makes the NiMH the "better" battery. Higher voltage might be useful for some applications, but I don't know of any. AA, AAA, C, and D cells all should be around 1.5 V. Electrical engineers who design battery powered devices count on this.

practice problem 4

Write something completely different.

solution

Answer it.