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716_PartUniversity Physics Solution

716_PartUniversity Physics Solution - 24-6 Chapter 24 24.18...

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24-6 Chapter 24 24.18. I DENTIFY : For capacitors in parallel the voltages are the same and the charges add. For capacitors in series, the charges are the same and the voltages add. / C Q V = . S ET U P : 1 C and 2 C are in parallel and 3 C is in series with the parallel combination of 1 C and 2 C . E XECUTE : (a) 1 2 and C C are in parallel and so have the same potential across them: 6 2 1 2 6 2 40.0 10 C 13.33 V 3.00 10 F Q V V C × = = = = × . Therefore, 6 6 1 1 1 (13.33 V)(3.00 10 F) 80.0 10 C Q VC = = × = × . Since 3 C is in series with the parallel combination of 1 2 and C C , its charge must be equal to their combined charge: 6 6 6 3 40.0 10 C 80.0 10 C 120.0 10 C C = × + × = × . (b) The total capacitance is found from 6 6 tot 12 3 1 1 1 1 1 9.00 10 F 5.00 10 F C C C = + = + × × and tot 3.21 F C μ = . 6 tot 6 tot 120.0 10 C 37.4 V 3.21 10 F ab Q V C × = = = × . E VALUATE : 6 3 3 6 3 120.0 10 C 24.0 V 5.00 10 F Q V C × = = = × . 1 3 ab V V V = + . 24.19. I DENTIFY and S ET U P : Use the rules for V for capacitors in series and parallel: for capacitors in parallel the voltages are the same and for capacitors in series the voltages add. E XECUTE : ( ) ( ) 1 1 1 / 150 C / 3.00 F 50 V V Q C μ μ = = = 1 2 and C C are in parallel, so 2 50 V V = 3 1 120 V 70 V V V = = E VALUATE : Now that we know the voltages, we could also calculate Q for the other two capacitors. 24.20. I DENTIFY and S ET U P : 0 A C d = P . For two capacitors in series, eq 1 2 1 1 1 C C C = + . E XECUTE : 2 1 1 1 0 2 eq 1 2 0 1 0 1 1 d A d C C C A d d A = + = + = + P P P . This shows that the combined capacitance for two capacitors in series is the same as that for a capacitor of area A and separation 1 2 ( ) d d + . E VALUATE : eq C is smaller than either 1 C or 2 C . 24.21. I DENTIFY and S ET U P : 0 A C d = P . For two capacitors in parallel, eq 1 2 C C C = + . E XECUTE : 0 1 0 2 0 1 2 eq 1 2 ( ) A A A A C C C d d d + = + = + = P P P . So the combined capacitance for two capacitors in parallel is that of a single capacitor of their combined area 1 2 ( ) A A + and common plate separation d . E VALUATE : eq C is larger than either 1 C or 2 C . 24.22. I DENTIFY : Simplify the network by replacing series and parallel combinations of capacitors by their equivalents. S ET U P : For capacitors in series the voltages add and the charges are the same; eq 1 2 1 1 1 C C C = + + " For capacitors in parallel the voltages are the same and the charges add; eq 1 2 C C C = + + " Q C V = . E XECUTE : (a) The equivalent capacitance of the 5.0 F μ and 8.0 F μ capacitors in parallel is 13.0 F. μ When these two capacitors are replaced by their equivalent we get the network sketched in Figure 24.22. The equivalent capacitance of these three capacitors in series is 3.47 F. μ (b) tot tot (3.47 F)(50.0 V) 174 C Q C V μ μ = = = (c) Q tot is the same as Q for each of the capacitors in the series combination shown in Figure 24.22, so Q for each of the capacitors is 174 C. μ E VALUATE : The voltages across each capacitor in Figure 24.22 are tot 10 10 17.4 V Q V C = = , tot 13 13 13.4 V Q V C = = and tot 9 9 19.3 V Q V C = = .
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