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16. We determine each capacitance from the slope of the appropriate line in the graph.
Thus,
C
1
= (12
µ
C)/(2.0 V) = 6.0
µ
F.
Similarly,
C
2
= 4.0
µ
F and
C
3
= 2.0
µ
F.
The total
equivalent capacitance is
C
123
=
(
(
C
1
)
−
1
+ (
C
3
+
C
2
)
−
1
)
−
1
= 3.0
µ
F.
This implies that the charge on capacitor 1 is (3.0
µ
F)(6.0 V) = 18
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Unformatted text preview: µ C. The voltage across capacitor 1 is therefore (18 µ C)/( 6.0 µ F) = 3.0 V. From the discussion in section 254, we conclude that the voltage across capacitor 2 must be 6.0 V – 3.0 V = 3.0 V. Consequently, the charge on capacitor 2 is (4.0 µ F)( 3.0 V) = 12 µ C....
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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
 Capacitance

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