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Unformatted text preview: V 1 = 2.0 V. The above expression consequently implies C 1 = 4 C 2 . Next we note that the graph shows that, at C 3 = 6.0 µ F, the voltage across C 1 is exactly half of the battery voltage. Thus, 1 2 = C 2 + 6.0 µ F C 1 + C 2 + 6.0 µ F = C 2 + 6.0 µ F 4 C 2 + C 2 + 6.0 µ F which leads to C 2 = 2.0 µ F. We conclude, too, that C 1 = 8.0 µ F....
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.
 Spring '10
 Rebello,NobelS

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