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Unformatted text preview: Physics 217 HW6 Solutions learned T E X for by Jonathan Newport March 25, 2005 Purcell 4.18 Let P be the power dissipated in the resistor R since P = I 2 R I = V R + R i → P = V 2 R ( R + R i ) 2 to maximize the power dissipated, we take the derivative with respect to R and set equal to zero 0 = dP dR = ( R i + R ) 2 V 2 2 V 2 R ( R i + R ) ( R i + R ) 4 → R = R i In addition, dP dR < 0 for R > R i and dP dR > 0 for R < R i . And so it goes for the maximum power... Purcell 4.25 Figure 1: The energy dissipated in a resistor is again E = I 2 R . From Eq 4.34 we have dQ dt = = V e t RC R 1 To find the total energy dissipated we need to integrate over all times the capacitor is discharging through the resistor, which is the instant we connect the capacitor (charged to V mind you) to the resistor out to infinity Z ∞ P dt = Z ∞ V R e t RC dt = V R RC 2 Z ∞ e x dx = 1 2 CV 2 which is nary more than the equation for the capacitor’s stored energy....
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 Spring '06
 LECLAIR, A
 Power, Electric charge, Ri, Purcell

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