HW05_Solutions - HOMEWORK 5: SOLUTIONS Q26.19: ANSWER:...

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25-1 HOMEWORK 5: SOLUTIONS Q26.19: ANSWER: think of relations to the known SI units. For example, energy stored in a capacitor is U=CV 2 /2 [J], while the dissipated power in a resistor is P=I 2 R [J/s]. Divide U/P= CV 2 /( 2I 2 R). From the Ohm’s law: V=RI, therefore U/P= CR/2 [J/(J/s)=s], so RC has units of seconds. 26.41. IDENTIFY: The capacitors, which are in parallel, will discharge exponentially through the resistors. SET UP: Since V is proportional to Q , V must obey the same exponential equation as Q , V = V 0 e t / RC . The current is I = (V 0 / R) e t / RC . EXECUTE: (a) Solve for time when the potential across each capacitor is 10.0 V: t = RC ln( V / V 0 ) = –(80.0 Ω)(35.0 μ F) ln(10/45) = 4210 μ s = 4.21 ms (b) I = (V 0 / R) e t / RC . Using the above values, with V 0 = 45.0 V, gives I = 0.125 A. EVALUATE: Since the current and the potential both obey the same exponential equation, they are both reduced by the same factor (0.222) in 4.21 ms. 26.48. IDENTIFY: When the capacitor is fully charged the voltage V across the capacitor equals the battery emf and Q CV = . For a charging capacitor, ( ) / 1 t RC qQ e = . SET UP: ln x ex = EXECUTE: (a) 64 (5.90 10 F)(28.0 V) 1.65 10 C. Q CV −− = = × (b) / (1 ) t RC = , so / 1 t RC q e Q = and .
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This note was uploaded on 02/11/2012 for the course PHYSICS 222 taught by Professor Ogilvie during the Fall '05 term at Iowa State.

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HW05_Solutions - HOMEWORK 5: SOLUTIONS Q26.19: ANSWER:...

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