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Unformatted text preview: Physics 204 morning summer 2008 Homework 4 solutions Chapter 20 Problems 4,6,7,25,75 4. REASONING a. According to Ohm’s law, the current is equal to the voltage between the cell walls divided by the resistance. +
b. The number of Na ions that flows through the cell wall is the total charge that flows divided by the charge of each ion. The total charge is equal to the current multiplied by the time. SOLUTION a. The current is I= V 75 × 10−3 V
=
= 1.5 × 10−11 A (20.2) 9
R 5.0 × 10 Ω + b. The number of Na ions is the total charge q that flows divided by the charge +e on each ion, or Δq / e . The charge is the product of the current I and the time t, according to Equation 20.1, so that Δq I Δt (1.5 × 10−11 A ) ( 0.50 s )
+
=
=
= 4.7 × 107 Number of Na ions = −19
e
e
1.60 × 10
C 6. REASONING According to Ohm’s law, the resistance is the voltage of the battery divided by the current that the battery delivers. The current is the charge divided by the time during which it flows, as stated in Equation 20.1. We know the time, but are not given the charge directly. However, we can determine the charge from the energy delivered to the resistor, because this energy comes from the battery, and the potential difference between the battery terminals is the difference in electric potential energy per unit charge, according to Equation 19.4. Thus, a 9.0‐V battery supplies 9.0 J of energy to each coulomb of charge passing through it. To calculate the charge, then, we need only divide the energy from the battery by the 9.0 V potential difference. SOLUTION Ohm’s law indicates that the resistance R is the voltage V of the battery divided by the current I, or R = V/I. According to Equation 20.1, the current I is the amount of charge Δq divided by the time Δt, or I = Δq/Δt. Using these two equations, we have R= V
V
V Δt
=
= I Δq / Δt Δq According to Equation 19.4, the potential difference ΔV is the difference Δ(EPE) in the electric potential energy Δ ( EPE )
. However, it is customary to denote the potential difference across a Δq
Δ ( EPE )
Δ ( EPE )
battery by V, rather than ΔV, so V =
. Solving this expression for the charge gives Δq =
. Using Δq
V
divided by the charge Δq, or ΔV = this result in the expression for the resistance, we find that ( 9.0 V ) ( 6 × 3600 s ) = 16 Ω V Δt
V Δt
V 2 Δt
R=
=
=
=
Δq Δ ( EPE ) /V Δ ( EPE )
1.1× 105 J
2 7. REASONING As discussed in Section 20.1, the voltage gives the energy per unit charge. Thus, we can determine the energy delivered to the toaster by multiplying the voltage V by the charge Δq that flows during a time Δt of one minute. The charge can be obtained by solving Equation 20.1, I = (Δq)/(Δt), since the current I can be obtained from Ohm’s law. SOLUTION Remembering that voltage is energy per unit charge, we have Energy = V Δq Solving Equation 20.1 for Δq gives Δq = I Δt, which can be substituted in the previous result to give Energy = V Δq = VI Δt According to Ohm’s law (Equation 20.2), the current is I = V/R, which can be substituted in the energy expression to show that 2
V 2 Δt (120 V ) ( 60 s )
⎛ V⎞
Energy = VI Δt = V ⎜ ⎟ Δt =
=
= 6.2 × 10 4 J ⎝ R⎠
14 Ω
R 25. SSM REASONING According to Equation 6.10b, the energy used is Energy = Pt, where P is the power and t is the time. According to Equation 20.6a, the power is P = IV, where I is the current and V is the voltage. Thus, Energy = IVt, and we apply this result first to the dryer and then to the computer. SOLUTION The energy used by the dryer is ⎛ 60 s ⎞
7
Energy = Pt = IVt = (16 A)(240 V)(45 min) ⎝
⎠ = 1.04 × 10 J
1.00 min Converts minutes
to seconds For the computer, we have Energy = 1.04 × 10 7 J = IVt = (2.7 A )(120 V) t Solving for t we find t= 1.04 × 107 J
⎛ 1.00 h ⎞
= 3.21 × 104 s = 3.21 × 104 s ⎜
⎟ = 8.9 h
( 2.7 A )(120 V )
⎝ 3600 s ⎠ ( ) 75. SSM REASONING The current I can be found by using Kirchhoff's loop rule. Once the current is known, the voltage between points A and B can be determined. SOLUTION a. We assume that the current is directed clockwise around the circuit. Starting at the upper‐left corner and going clockwise around the circuit, we set the potential drops equal to the potential rises: (5.0 Ω) I + (27 Ω)I + 10.0 V + (12 Ω ) I + (8.0 Ω )I = Potential drops Solving for the current gives I = 0.38 A . 30.0 V
Potential rises b. The voltage between points A and B is c. V AB = 30.0 V – (0.38 A)(27 Ω) = 2.0 × 101 V Point B is at the higher potential. ...
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This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.
 Spring '09
 rollino
 Physics

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