Unformatted text preview: bulb = 74. REASONING AND SOLUTION Ohm’s law gives V 1.5 V
=
= 0.054 Ω
I
28 A
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r= 75. REASONING When the terminal voltage of the battery (emf = 9.00 V) is 8.90 V, the
voltage drop across the internal resistance r is 9.00 V − 8.90 V = 0.10 V. According to
Ohm’s law, this voltage drop is the current I times the internal resistance. Thus, Ohm’s law
will allow us to calculate the current.
SOLUTION Using Ohm’s law we find that I= V
0.10 V
=
= 8.3 A
r 0.012 Ω 76. REASONING The following drawing shows the battery (emf = ξ ), its internal resistance r,
and the light bulb (represented as a resistor). The voltage between the terminals of the
battery is the voltage VAB between the points A and B in the drawing. This voltage is not
equal to the emf of the battery, because part of the emf is needed to make the current I go
through the internal resistance. Ohm’s law states that this part of the emf is I r. The current
can be determined from the relation P = I VAB, since the power P delivered to the light bulb
and the voltage VAB across it are known. Chapter 20 Problems 1103 Light bulb I
A − + + −
r = 0.10 Ω ξ B SOLUTION The terminal voltage VAB is equal to the emf ξ of the battery minus the voltage across the internal resistance r, which is I r (Equation 20.2): VAB = ξ − I r. Solving
this equation for the emf gives ξ = VAB + I r (1) The current also goes through the light bulb, and it is related to the power P delivered to the
bulb and the voltage VAB according to I = P/VAB (Equation 20.6a). Substituting this
expression for the current into Equation (1) gives
P VAB ξ = VAB + I r = VAB + r 24 W (
= 11.8 V + 0.10 Ω ) = 12.0 V 11.8 V ______________________________________________________________________________ 77. REASONING The voltage V across the terminals of a battery is equal to the emf of the
battery minus the voltage Vr across the internal resistance of the battery: V = Emf − Vr.
Therefore, we have that
Emf = V + Vr (1) The power P being dissipated by the internal resistance is equal to the product of the voltage
Vr across the internal resistance and the current I: P = IVr (Equation 20.6a). Therefore, we
can express the voltage Vr across the internal resistance as
Vr = P
I (2) 1104 ELECTRIC CIRCUITS SOLUTION Substituting Equation (2) into Equation (1), we obtain
Emf = V + Vr = V + P
34.0 W
= 23.4 V +
= 24.0 V
I
55.0 A 78. REASONING The power P delivered to a resistor such as the light bulb is found from
P = I 2R (20.6b) where I is the current in the resistor and R is the resistance. When either battery is connected
to the bulb, the circuit can be modeled as a singleloop circuit with the battery emf in series
with two resistors, namely, the internal resistance r and the resistance R of the bulb. As the
two resistors are in series, their equivalent resistance RS is given by RS = R1 + R2 + R3 + L
(Equation 20.16):
(1)
RS = r + R
The current I supplied by a battery connected to a resistor RS is, according to Ohm’s law, I= V
RS (20.2) where V is equal to the emf of the battery. Thus, V is the same for both circuits, since both
batteries have the same emf, even though their internal resistances (rwet, rdry) are different.
SOLUTION Applying Equation 20.6b separately to each circuit, we obtain
2
Pwet = I wet R 2
Pdry = I dry R and (2) Taking the ratio of Equations (2) eliminates the bulb resistance R, yielding Pwet
Pdry = 2
I wet R
2
I dry R = 2
I wet (3) 2
I dry Substituting Equation (20.2) for both Iwet and Idry into Equation (3), we obtain Pwet
Pdry = 2
I wet
2
I dry V RS,wet
= 2 = RS,dry 2 RS,wet
V RS,dry 2 (4) Chapter 20 Problems 1105 Replacing the equivalent series resistances in Equation (4) by Equation (1), we find that
2 2 RS,dry rdry + R 0.33 Ω + 1.50 Ω 2
= = = = 1.39 Pdry RS,wet rwet + R 0.050 Ω + 1.50 Ω ______________________________________________________________________________
Pwet 79. 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
upperleft 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 =
144444444 2444444444
4
3
Potential drops 30.0 V
13
2
Potential rises Solving for the current gives I = 0.38 A .
b. The voltage between points A and B is
V AB = 30.0 V – (0.38 A)(27 Ω ) = 2.0 × 101 V c. Point B is at the higher potential.
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80. REASONING In part a of the drawing in the text, the current goes from lefttoright
through the resistor. Since the current always goes from a higher to a lower potential, the
left end of the resistor is + and the right end is –. In part b, the current goes from righttoleft through the resistor. The...
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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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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