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Thermal heat from a Capacitor

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A resistor, a capacitor, a battery and a switch are connected to a circuit. The capacitor will be charged from the empty state to the full state.

    During the period in which the capacitor is charged, how much thermal heat is emitted from the resistor?

    [tex]C = 2 \mu F[/tex]
    [tex]R = 2 \Omega[/tex]
    [tex]E = 6 V[/tex]

    2. Relevant equations

    [tex]U = RI [/tex]
    [tex]Q = CV[/tex]

    3. The attempt at a solution

    The amount of charge stored in the capacitor is: 6 V * 2 [tex]\mu F[/tex] = 12 [tex]\mu C [/tex].

    The amount of work done by the battery is 12 [tex]\mu C [/tex]* 6 V *1/2 = 36 [tex]\mu J [/tex], during charging.

    The amout of heat is (wrongly) 3/4 * 36 [tex] \mu J [/tex].
     
    Last edited: May 23, 2009
  2. jcsd
  3. May 23, 2009 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    What is the energy stored in the capacitor in terms of C and V?

    Assume the energy stored in the capacitor is transformed into heat in the resistor when the capacitor discharges. How does this relate to the reverse when it charges?

    AM
     
  4. May 24, 2009 #3
    The energy (=QV) is 12 [tex]\mu C [/tex] * 6 V = 66 [tex]\mu J [/tex]. But somethig must be wrong, as I have:
    Perhaps, it is 12 [tex]\mu C [/tex]* 6 V = 72 [tex]\mu J [/tex]. ??? Dunno why ???

    If we suppose the logic right, we get a result for the energy emitted:

    72 [tex]\mu J [/tex] - 66 [tex]\mu J [/tex] = 6 [tex]\mu J [/tex]

    The amount of the thermal heat emitted is the difference between the energy emitted during discharging and the work done by the battery.

    After-thought
    NB. If I am right, the unit check shows that my calculations are wrong. E != CQ, E = C*Q^2. Now, I must be mistaken.
     
    Last edited: May 24, 2009
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