{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics Solution Manual for 1100 and 2101

# An application of this rule shows that the proton

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: F 6 R ln(q/q0 ) (1.8 ×10 Ω) ln(0.368) ______________________________________________________________________________ C= 104. REASONING The time constant of an RC circuit is given by Equation 20.21 as τ = RC, where R is the resistance and C is the capacitance in the circuit. The two resistors are wired in parallel, so we can obtain the equivalent resistance by using Equation 20.17. The two capacitors are also wired in parallel, and their equivalent capacitance is given by Equation 20.18. The time constant is the product of the equivalent resistance and equivalent capacitance. 1122 ELECTRIC CIRCUITS SOLUTION The equivalent resistance of the two resistors in parallel is 1 1 1 3 = + = RP 2.0 kΩ 4.0 kΩ 4.0 kΩ R P = 1.3 kΩ or (20.17) The equivalent capacitance is CP = 3.0 µ F + 6.0 µ F = 9.0 µ F (20.18) The time constant for the charge to build up is ( )( ) τ = RP CP = 1.3 × 103 Ω 9.0 × 10−6 F = 1.2 × 10−2 s ______________________________________________________________________________ 105. REASONING AND SOLUTION According to Equation 20.21, we find that τ 3.0 s = 2.0 × 10 4 Ω C 150 × 10 – 6 F ______________________________________________________________________________ R= = 106. REASONING The charging of a capacitor is described by Equation 20.20, which provides a direct solution to this problem. SOLUTION According to Equation 20.20, in a series RC circuit the charge q on the capacitor at a time t is given by ( q = q0 1 − e−t / τ ) where q0 is the equilibrium charge that has accumulated on the capacitor after a very long time and τ is the time constant. For q = 0.800q0 this equation becomes ( q = 0.800q0 = q0 1 − e−t / τ ) or 0.200 = e−t / τ Taking the natural logarithm of both sides of this result gives ( ln ( 0.200 ) = ln e −t / τ ) or ln ( 0.200 ) = − t τ Therefore, the number of time constants needed for the capacitor to be charged to 80.0% of its equilibrium charge is t τ = −ln ( 0.200 ) = − ( −1.61) = 1.61 Chapter 20 Problems 1123 107. REASONING In either part of the drawing the time constant τ of the circuit is τ = RCeq , according to Equation 20.21, where R is the resistance and Ceq is the equivalent capacitance of the capacitor combination. We will apply this equation to both circuits. To obtain the equivalent capacitance, we will analyze the capacitor combination in parts. For the parallel capacitors CP = C1 + C2 + C3 + ... applies (Equation 20.18), while for the series capacitors − − − CS 1 = C1−1 + C2 1 + C3 1 + ... applies (Equation 20.19). C C C R R C C C C C +− +− (a) (b) SOLUTION Using Equation 20.21, we write the time constant of each circuit as follows: τ a = RCeq, a and τ b = RCeq, b Dividing these two equations allows us to eliminate the unknown resistance algebraically: τ b RCeq, b = τ a RCeq, a or Ceq, b Ceq, a τb = τa (1) To obtain the equivalent capacitance in part a of the drawing, we note that the two capacitors in series in each branch of the parallel combination have an equivalent capacitance CS that can be determined using Equation 20.19 1 11 =+ CS C C or CS = 1 C 2 (2) Using Equation 20.18, we find that the parallel combination in part a of the drawing has an equivalent capacitance of Ceq, a = 1 C + 1 C = C 2 2 (3) To obtain the equivalent capacitance in part b of the drawing, we note that the two capacitors in series have an equivalent capacitance of 1 C , according to Equation (2). The two 2 1124 ELECTRIC CIRCUITS capacitors in parallel have an equivalent capacitance of 2C, according to Equation 20.18. Finally, then, we have a series combination of 1 C and 2C, for which Equation 20.19 applies: 2 1 Ceq, b = 1 1 5 + = 1C 2C 2C 2 or 2 Ceq, b = 5 C (4) Using Equations (3) and (4) in Equation (1), we find that Ceq, b τb = τa Ceq, a 2C 5 = ( 0.72 s ) = 0.29 s C 108. REASONING To find the current, we can use the relation that the power P is the product of the current I and the voltage V, since the power and voltage are known. SOLUTION Solving P = I V (Equation 20.6a) for the current, we have P 0.11 W = = 0.024 A V 4.5 V ______________________________________________________________________________ I= 109. SSM REASONING AND SOLUTION The charges stored on capacitors in series are equal and equal to the charge separated by the battery. The total energy stored in the capacitors is Energy = Energy = Q2 Q2 + 2C1 2C2 1 Q2 1 C +C 2 1 2 According to Equation 20.19, the quantity in the parentheses is just the reciprocal of the equivalent capacitance C of the circuit, so Q2 2C ______________________________________________________________________________ Energy = Chapter 20 Problems 1125 110. REASONING AND SOLUTION First determine the total charge delivered to the battery using Equation 20.1: 3600 s 5 ∆q = I ∆t = (6.0 A)(5.0 h) = 1.1×10 C 1h To find the energy delivered to the battery, multiply this charge by the energy per unit charge (i.e., the voltage) to get Energy = (∆q)V = (1.1 × 105 C)(12 V) = 1.3 × 106 J ______________________________________________________________________________ 111. SSM REASONING AND...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online