(b) and (c) At C3= 0, the graph indicates V1= 2.0 V. The above expression consequently implies C1= 4C2. Next we note that the graph shows that, at C3= 6.0 µF, the voltage across C1is exactly half of the battery voltage. Thus, 12= C2+ 6.0 µFC1+ C2+ 6.0 µF= C2+ 6.0 µF4C2+ C2+ 6.0 µFwhich leads to C2= 2.0 µF. We conclude, too, that C1 = 8.0 µF. 28. Initially the capacitors C1, C2, and C3form a combination equivalent to a single capacitor which we denote
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