Physics Solution Manual for 1100 and 2101

The resistor r1 is in series with r2345 and the

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Unformatted text preview: ________________________________ 93. REASONING A capacitor with a capacitance C stores a charge q when connected between the terminals of a battery of voltage V, according to q = CV (Equation 19.8). This equation can be used to calculate the voltage if q and C are known. In this problem we know that the two capacitors together store a total charge of 5.4 × 10−5 C. This is also the charge stored by the equivalent capacitor, which has the equivalent capacitance CP of the parallel combination, or CP = C1 + C2 (Equation 20.18). Thus, we can determine the battery voltage by using Equation 19.8 with the given total charge and the equivalent capacitance obtained from Equation 20.18. SOLUTION With q = qTotal and C = CP = C1 + C2 Equation 19.8 becomes qTotal = CPV = ( C1 + C2 ) V or V= qTotal C1 + C2 = 5.4 ×10−5 C = 9.0 V 2.0 ×10−6 F + 4.0 ×10−6 F 1116 ELECTRIC CIRCUITS 94. REASONING The equivalent capacitance CS of a set of three capacitors connected in 1 1 1 1 series is given by (Equation 20.19). In this case, we know that the = + + CS C1 C2 C3 equivalent capacitance is CS = 3.00 µF, and the capacitances of two of the individual capacitors in this series combination are C1 = 6.00 µF and C2 = 9.00 µF. We will use Equation 20.19 to determine the remaining capacitance C3. SOLUTION Solving Equation 20.19 for C3, we obtain 1 1 1 1 = −− C3 CS C1 C2 C3 = or 1 1 1 1 −− CS C1 C2 Therefore, the third capacitance is 1 = 18 µ F 1 1 1 − − 3.00 µ F 6.00 µ F 9.00 µ F ______________________________________________________________________________ C3 = 95. SSM REASONING AND SOLUTION Let C0 be the capacitance of an empty capacitor. Then the capacitances are as follows, according to the discussion following Equation 19.10: C1 = 3.00C0 and C2 = 5.00C0 The series capacitance of the two is 1 1 1 = + Cs 2.50C0 4.00C0 or Cs = 1.54C0 Now κC0 = 1.54C0 or κ = 1.54 ______________________________________________________________________________ 96. REASONING When capacitors are connected in parallel, each receives the entire voltage V of the battery. Thus, the total energy stored in the two capacitors is 1 C1V 2 + 1 C2V 2 (see 2 2 Equation 19.11b). When the capacitors are connected in series, the sum of the voltages across each capacitor equals the battery voltage: V1 + V2 = V . Thus, the voltage across each Chapter 20 Problems capacitor is series is less than the battery voltage, so the total energy, less than when the capacitors are wired in parallel. 1 2 1117 C1V1 + 1 C2V2 , is 2 2 2 SOLUTION a. The voltage across each capacitor is the battery voltage, or 60.0 V. The energy stored in both capacitors is Total energy = 1 C1V 2 + 1 C2V 2 = 2 2 = 1 2 ( 2.00 × 10 −6 1 2 ( C1 + C2 )V 2 ) F + 4.00 × 10−6 F ( 60.0 V ) = 1.08 × 10−2 J 2 b. According to the discussion in Section 20.12, the total energy stored by capacitors in series is Total energy = 1 CSV 2 , where CS is the equivalent capacitance of the series 2 combination: 1 1 1 1 1 = + = + −6 CS C1 C2 2.00 × 10 F 4.00 × 10−6 F Solving this equation yields CS = 1.33 × 10 Total energy = 1 2 −6 (20.19) F. The total energy is (1.33 × 10−6 F) ( 60.0 V )2 = 2.39 × 10−3 J ______________________________________________________________________________ 97. REASONING Our approach to this problem is to deal with the arrangement in parts. We will combine separately those parts that involve a series connection and those that involve a parallel connection. SOLUTION The 24, 12, and 8.0-µF capacitors are in series. Using Equation 20.19, we can find the equivalent capacitance for the three capacitors: 1 1 1 1 = + + Cs 24 µ F 12 µ F 8.0 µ F or Cs = 4.0 µ F This 4.0-µF capacitance is in parallel with the 4.0-µF capacitance already shown in the text diagram. Using Equation 20.18, we find that the equivalent capacitance for the parallel group is Cp = 4.0 µ F + 4.0 µ F = 8.0 µ F This 8.0-µF capacitance is between the 5.0 and the 6.0-µF capacitances and in series with them. Equation 20.19 can be used, then, to determine the equivalent capacitance between A and B in the text diagram: 1118 ELECTRIC CIRCUITS 1 1 1 1 = + + or Cs = 2.0 µ F Cs 5.0 µ F 8.0 µ F 6.0 µ F ______________________________________________________________________________ 98. REASONING To find the equivalent capacitance of the three capacitors, we must first, following CP = C1 + C2 + C3 + L (Equation 20.18), add the capacitances of the two parallel capacitors together. We must then combine the result of Equation 20.18 with the remaining 1 1 1 1 capacitance in accordance with =+ + + L (Equation 20.19). As Equation CS C1 C2 C3 20.19 shows, combining capacitors in series decreases the overall capacitance, and the resulting equivalent capacitance CS is smaller than any of the capacitances being added. Therefore, the way to maximize the overall equivalent capacitance is to choose the largest capacitance (C1) to be connected in series with the parallel combination C23 of the smaller capacitances. SOLUTION When C2 and C3 are connected in parallel, their equivalent capacitance C23 is, from CP = C1 + C2 + C3 + L (Equation 2...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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