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quiz6 - battery V = ε R(r R = 12.6x5/5.08 = 12.4V When the...

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October 6, 2009 PHY 2054/3808 Quiz VI (1) What is the potential difference between the plates of a 3.3-F capacitor which stores sufficient energy to operate a 75 W light bulb for one minute? How much energy do we need to light up a 75 W bulb for 60 s? E = 75x60 = 4500 J The energy and potential difference of a capacitor are related by E = ½CV 2 . Solving for V, we get V = √[2x4500/3.3] = 52.22V (2) Problem 18.49 The potential difference across the headlight bulbs is given by V = IR, where I is the current flowing through the series combination of the headlight bulb and the internal resistance of the
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Unformatted text preview: battery. V = ε R/(r+R) = 12.6x5/5.08 = 12.4V When the starter motor is engaged, There is extra current being drawn. In the battery, ε = V + Ir. Therefore V = ε - r[35 + V/R]. Solving for V = [ε - 35r]/[1+r/R] = 9.65 V. The message is that the voltage output of a battery can drop significantly at high currents. That is how the headlights dim when the battery is old (higher internal resistance) and the engine is started....
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