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Discharging a Capacitor

Discussion in 'MCAT Study Question Q&A' started by collegelife101, 05.13.14.

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  1. collegelife101

    collegelife101

    Joined:
    05.02.14
    Messages:
    34
    Status:
    Pre-Medical
    Can someone please help explain this problem:

    A resistor, a fully charged capacitor, and a switch are all connected in series, with the switch initially open. The switch is then closed, allowing the capacitor to discharge. During this discharging process, which of the following quantities DECREASES?


    I.

    The current through the resistor

    II.

    The energy stored in the capacitor

    III.

    The voltage of the battery

    As the capacitor discharges, the voltage across its plates decreases according to V = Q/C. Because the voltage drop across the resistor must equal the capacitor’s voltage, VC – VR = 0, the voltage across the resistor correspondingly decreases. Therefore, by V = IR, so does the current across the resistor, meaning I is true. The power dissipated by a resistor is given by P = I2R, so as the current decreases, so does the power dissipated across the resistor, meaning III is true. Moreover, as the charge of the capacitor decreases, so does the potential energy stored, according to PE = Q2/2C. Thus II is true.

    I'm a little confused on where the battery is in this situation. If the power dissipated by a resistor decreases, how does that tell us that the voltage of the battery is decreasing?

    Thanks!
     
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  3. Rhino1000

    Rhino1000 2+ Year Member

    Joined:
    12.30.12
    Messages:
    446
    Status:
    Pre-Medical
    There is no battery in the passage. Need I say more?
     

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