Physics Solution Manual for 1100 and 2101

The current i in the circuit is equal to the voltage

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Unformatted text preview: 0.18), C23 = C2 + C3 (1) When the equivalent capacitance C23 is connected in series with C1, the resulting equivalent capacitance CS is, according to Equation 20.19 and Equation (1), −1 −1 −1 −1 1 1 1 1 1 1 1 CS = = + = + = + = 36 µ F C C C C C +C 67 µ F 45 µ F + 33 µ F 23 2 3 S 1 1 ______________________________________________________________________________ 99. REASONING a. When capacitors are wired in parallel, the total charge q supplied to them is the sum of the charges supplied to the individual capacitors, or q = q1 + q2. The individual charges can be obtained from q1 = C1V and q2 = C2V, since the capacitances, C1 and C2, and the voltage V are known. b. When capacitors are wired in series, the voltage V across them is equal to the sum of the voltages across the individual capacitors, or V = V1 + V2. However, the charge q on each capacitor is the same. The individual voltages can be obtained from V1 = q/C1 and V2 = q/C2. Chapter 20 Problems 1119 SOLUTION a. Substituting q1 = C1V and q2 = C2V (Equation 19.8) into q = q1 + q2, we have q = q1 + q2 = C1V + C2V = ( C1 + C2 ) V = ( 2.00 ×10−6 F + 4.00 ×10−6 F ) ( 60.0 V ) = 3.60 ×10−4 C b. Substituting V1 = q/C1 and V2 = q/C2 (Equation 19.8) into V = V1 + V2 gives V = V1 + V2 = q q + C1 C2 Solving this relation for q, we have V 60.0 V = 8.00 × 10−5 C 1 1 1 1 + + C 1 C2 2.00 ×10−6 F 4.00 ×10−6 F ______________________________________________________________________________ q= = 100. REASONING AND SOLUTION The 7.00 and 3.00-µF capacitors are in parallel. According to Equation 20.18, the equivalent capacitance of the two is 7.00 µF + 3.00 µF = 10.0 µF. This 10.0-µF capacitance is in series with the 5.00-µF capacitance. According to Equation 20.19, the equivalent capacitance of the complete arrangement can be obtained as follows: 1 1 1 –1 = + = 0.300 ( µ F ) C 10.0 µ F 5.00 µ F C= or 1 0.300 ( µ F ) –1 = 3.33 µ F The battery separates an amount of charge Q = CV = (3.33 × 10–6 F)(30.0 V) = 99.9 × 10–6 C This amount of charge resides on the 5.00 µ F capacitor, so its voltage is V5 = (99.9 × 10 –6 C)/(5.00 × 10 –6 F) = 20.0 V The loop rule gives the voltage across the 3.00 µ F capacitor to be V3 = 30.0 V – 20.0 V = 10.0 V This is also the voltage across the 7.00 µ F capacitor, since it is in parallel, so V7 = 10.0 V . ______________________________________________________________________________ 1120 ELECTRIC CIRCUITS 101. SSM REASONING The foil effectively converts the capacitor into two capacitors in series. Equation 19.10 gives the expression for the capacitance of a capacitor of plate area A and plate separation d (no dielectric): C0 = ε 0 A / d . We can use this expression to determine the capacitance of the individual capacitors created by the presence of the foil. Then using the fact that the "two capacitors" are in series, we can use Equation 20.19 to find the equivalent capacitance of the system. SOLUTION Since the foil is placed one-third of the way from one plate of the original capacitor to the other, we have d1 = (2 / 3)d , and d 2 = (1/ 3)d . Then C1 = ε0 A (2 / 3)d and C2 = ε0 A (1/ 3)d = = 3ε 0 A 2d 3ε 0 A d Since these two capacitors are effectively in series, it follows that 1 1 1 1 1 3d d = + = + = = Cs C1 C2 3ε 0 A / ( 2d ) 3ε 0 A / d 3ε 0 A ε 0 A But C0 = ε 0 A / d , so that d / ( ε 0 A ) = 1/ C0 , and we have 1 1 = or Cs = C0 Cs C0 ______________________________________________________________________________ 102. REASONING AND SOLUTION Charge is conserved during the re-equilibrium. Therefore, using q0 and qf to denote the initial and final charges, respectively, we have q10 + q20 = 18.0 µ C = q1f + q2f (1) After equilibrium has been established the capacitors will have equal voltages across them, since they are connected in parallel. Thus, Vf = q1f/C1 = q2f/C2, which leads to q1f = q2f(C1/C2) = q2f (2.00 µ F)/(8.00 µ F) = 0.250 q2f Substituting this result into Equation (1) gives 18.0 µ C = 0.250 q2f + q2f or q2f = 14.4 µ C Chapter 20 Problems 1121 Hence, –6 –6 Vf = q2f/C2 = (14.4 × 10 C)/(8.00 × 10 F) = 1.80 V ______________________________________________________________________________ 103. SSM REASONING The charge q on a discharging capacitor in a RC circuit is given by Equation 20.22: q = q0 e – t / RC , where q0 is the original charge at time t = 0 s. Once t (time for one pulse) and the ratio q / q0 are known, this expression can be solved for C. SOLUTION Since the pacemaker delivers 81 pulses per minute, the time for one pulse is 1 min 60.0 s × = 0.74 s/pulse 81 pulses 1.00 min) Since one pulse is delivered every time the fully-charged capacitor loses 63.2% of its original charge, the charge remaining is 36.8% of the original charge. Thus, we have q = (0.368)q0 , or q / q0 = 0.368 . From Equation 20.22, we have q = e – t / RC q0 Taking the natural logarithm of both sides, we have, q t ln = − q RC 0 Solving for C, we find −t −(0.74 s) = = 4.1 × 10−7...
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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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