716_PartUniversity Physics Solution

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

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24-6 Chapter 24 24.18. IDENTIFY: 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. / CQV = . SET UP: 1 C and 2 C are in parallel and 3 C is in series with the parallel combination of 1 C and 2 C . EXECUTE: (a) 12 and CC are in parallel and so have the same potential across them: 6 2 6 2 40.0 10 C 13.33 V 3.00 10 F Q VV C × == = = × . Therefore, 66 11 1 (13.33 V)(3.00 10 F) 80.0 10 C QV C == × . Since 3 C is in series with the parallel combination of and , its charge must be equal to their combined charge: 6 3 40.0 10 C 80.0 10 C 120.0 10 C C −− = × . (b) The total capacitance is found from tot 12 3 1 1 1 9.00 10 F 5.00 10 F CCC =+ = + ×× 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 × = × . EVALUATE: 6 3 3 6 3 120.0 10 C 24.0 V Q V C × = × . 13 ab VVV = + . 24.19. IDENTIFY and SET UP: 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. EXECUTE: () ( ) 1 / 150 C / 3.00 F 50 V VQC = and are in parallel, so 2 50 V V = 31 120 V 70 V =− = EVALUATE: Now that we know the voltages, we could also calculate Q for the other two capacitors. 24.20. IDENTIFY and SET UP: 0 A C d = P . For two capacitors in series, eq 1 2 111 C . EXECUTE: 2 1 1 1 0 2 eq 0 1 0 d A d C A dd 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 + . EVALUATE: eq C is smaller than either 1 C or 2 C . 24.21. IDENTIFY and SET UP: 0 A C d = P . For two capacitors in parallel, eq 1 2 C = + . EXECUTE: 01 02 0 1 2 eq 1 2 A AA A C d + =+= + = PP P . So the combined capacitance for two capacitors in parallel is that of a single capacitor of their combined area A A + and common plate separation d . EVALUATE: eq C is larger than either 1 C or 2 C . 24.22. IDENTIFY: Simplify the network by replacing series and parallel combinations of capacitors by their equivalents. SET UP: For capacitors in series the voltages add and the charges are the same; eq 1 2 C =++ " For capacitors in parallel the voltages are the same and the charges add; eq 1 2 C = ++ " Q C V = . EXECUTE: (a) The equivalent capacitance of the 5.0 F and 8.0 F capacitors in parallel is 13.0 F. 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. (b) tot tot (3.47 F)(50.0 V) 174 C QC V = (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. EVALUATE: 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 . 10 13 9 17.4 V 13.4 V 19.3 V 50.1 V ++= + + = . The sum of the voltages equals the applied voltage, apart from a small difference due to rounding.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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

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