Multiple-Concept Example 9 reviews the concepts that are important to this problem. A light bulb is wired in series with a 134-0 resistor, and they are connected across a 120.0-V source. The power delivered to the light bulb is 21.1 W. What is the resistance of the light bulb? There are two possible answers. Give the larger of the two. Number i Units Light bulb M P₁ V R₂
Multiple-Concept Example 9 reviews the concepts that are important to this problem. A light bulb is wired in series with a 134-0 resistor, and they are connected across a 120.0-V source. The power delivered to the light bulb is 21.1 W. What is the resistance of the light bulb? There are two possible answers. Give the larger of the two. Number i Units Light bulb M P₁ V R₂
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter29: Direct Current (dc) Circuits
Section: Chapter Questions
Problem 19PQ
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Question
![Multiple-Concept Example 9 reviews the concepts that are important to this problem. A light bulb is wired in series with a 134-0
resistor, and they are connected across a 120.0-V source. The power delivered to the light bulb is 21.1 W. What is the resistance of the
light bulb? There are two possible answers. Give the larger of the two.
Number i
Units
Light bulb
M
P₁
V
R₂](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff82a0189-a001-438e-8e10-429fbfe68738%2F100c487c-df80-4a9a-a6a0-ac7d2f3fcac3%2Fpetv4xi_processed.png&w=3840&q=75)
Transcribed Image Text:Multiple-Concept Example 9 reviews the concepts that are important to this problem. A light bulb is wired in series with a 134-0
resistor, and they are connected across a 120.0-V source. The power delivered to the light bulb is 21.1 W. What is the resistance of the
light bulb? There are two possible answers. Give the larger of the two.
Number i
Units
Light bulb
M
P₁
V
R₂
![Modeling the Problem
STEP 1 Power The power P₁ delivered to the 6.00- resistor is given by P₁ = 1²R₁ (Equation 20.6b), where I is the current through the resistor and R₁ is the resistance.
In this expression R₁ is a known quantity, and the current I will be determined in Step 2.
STEP 2 Ohm's Law The current I in the circuit depends on the voltage V of the battery and the equivalent resistance Rs of the two resistors in series (see Figure
20.15). This dependence is given by Ohm's law (Equation 20.2) as
V
1 =
Rs
This result for the current can be substituted into Equation 20.6b, as indicated at the right. Note from the data table that the voltage is given. In Step 3 we will evaluate
the equivalent resistance from the individual resistances R₁ and R₂.
STEP 3 Equivalent Resistance Since the two resistors are wired in series, the equivalent resistance Rs is the sum of the two resistances (Equation 20.16):
Rs = R₁ + R₂
The resistances R₁ and R₂ are known. We substitute this expression for Rs into Equation 1, as shown in the right column.
Solution
P₁ = P²R₁
?
P₁ = P²R₁
I=
R$
I =
?
P₁ = I²R₁
V
Rs
Rs = R₁ + R₂
(20.6b)
(20.6b)
(1)
(20.6b)
(1)
(20.16)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff82a0189-a001-438e-8e10-429fbfe68738%2F100c487c-df80-4a9a-a6a0-ac7d2f3fcac3%2Fz5ba6g_processed.png&w=3840&q=75)
Transcribed Image Text:Modeling the Problem
STEP 1 Power The power P₁ delivered to the 6.00- resistor is given by P₁ = 1²R₁ (Equation 20.6b), where I is the current through the resistor and R₁ is the resistance.
In this expression R₁ is a known quantity, and the current I will be determined in Step 2.
STEP 2 Ohm's Law The current I in the circuit depends on the voltage V of the battery and the equivalent resistance Rs of the two resistors in series (see Figure
20.15). This dependence is given by Ohm's law (Equation 20.2) as
V
1 =
Rs
This result for the current can be substituted into Equation 20.6b, as indicated at the right. Note from the data table that the voltage is given. In Step 3 we will evaluate
the equivalent resistance from the individual resistances R₁ and R₂.
STEP 3 Equivalent Resistance Since the two resistors are wired in series, the equivalent resistance Rs is the sum of the two resistances (Equation 20.16):
Rs = R₁ + R₂
The resistances R₁ and R₂ are known. We substitute this expression for Rs into Equation 1, as shown in the right column.
Solution
P₁ = P²R₁
?
P₁ = P²R₁
I=
R$
I =
?
P₁ = I²R₁
V
Rs
Rs = R₁ + R₂
(20.6b)
(20.6b)
(1)
(20.6b)
(1)
(20.16)
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