positive plate of the fully charged capacitor – making its leftmost plate (the one closest to the negative terminal of the battery) the negative plate, as it should be. (e) As stated in (b), the electrons travel up through point b. (f) As stated in (c), the electrons travel up through point c. 25. We note that the total equivalent capacitance is C123 = [(C3)−1+ (C1 + C2)−1]−1= 6 µF. (a) Thus, the charge that passed point ais C123 Vbatt= (6 µF)(12 V) = 72 µC. Dividing this by the value e= 1.60×10−19 C gives the number of electrons: 4.5×1014, which travel to the left – towards the positive terminal of the battery. (b) The equivalent capacitance of the parallel pair is C12 = C1 + C2= 12 µF. Thus, the voltage across the pair (which is the same as the voltage across
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This note was uploaded on 01/25/2011 for the course PHYS 244 taught by Professor Ac during the Spring '10 term at The Petroleum Institute.