1. Problems With Assistance Module 2 – Problem 7
Go straight to the First Step
Filename: PWA_Mod02_Prob07.ppt
You can see a brief introduction starting on the next slide, or go right to the problem.
Go straight to the Problem Statement
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2. Overview of this Problem
In this problem, we will use the following concepts:
Voltage Divider Rule
Voltmeters
Go straight to the First Step
Go straight to the Problem Statement
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3. Textbook Coverage
The material for this problem is covered in your textbook in the following chapters:
Circuits by Carlson: Chapter 3
Electric Circuits 6th Ed. by Nilsson and Riedel: Chapter 3
Basic Engineering Circuit Analysis 6th Ed. by Irwin and Wu: Chapter 2
Fundamentals of Electric Circuits by Alexander and Sadiku: Chapter 2
Introduction to Electric Circuits 2nd Ed. by Dorf: Chapter 1
This is the material in your circuit texts that you might consult to get more help on this problem.
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4. Coverage in this Module
The material for this problem is covered in this module in the following presentations:
DPKC_Mod02_Part02 and DPKC_Mod02_Part03.
This is the material in this computer module that you might consult for more explanation. These are presentations of key concepts that you should find in this problem.
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5. Problem Statement
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
This is the basic problem. We will take it step by step.
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6. Solution – First Step – Where to Start?
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
How should we start this problem? What is the first step?
Try to decide on the first step before going to the next slide.
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7. Problem Solution – First Step
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
How should we start this problem? What is the best first step?
Attach an arbitrary voltage source to this circuit at terminals A and B.
Write a voltage-divider rule equation.
Replace the meter with its equivalent resistance.
Divide the voltage (2[V]) by the resistance (10[kW]) to get the current.
Divide the voltage (50[V]) by the resistance (10[kW]) to get the current.
Click on the step that you think should be next.

8. Your Choice for First Step – Attach an Arbitrary Voltage Source to This Circuit at Terminals A and B.
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
This is not the best choice for the first step.
We could indeed attach a voltage source to terminals A and B. After doing this we could then solve for currents and voltages that would result from that. If we were to do this, we would want to apply a 50[V] source, since this would give us the full scale reading we desire. Thus, this approach would work, but it adds a step we may not need. So, even though this is a valid approach, we would recommend that you go back and try again.

9. Your Choice for First Step Was – Write a voltage-divider rule equation
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
This is a good choice for the second step, not the first step. We will be writing a voltage divider rule relationship, but to do this, we need to know what to do with the voltmeter in the diagram. We are not yet ready for this step. So, even though this is a valid approach, we would recommend that you go back and try again.

10. Your Choice for First Step Was – Replace the meter with its equivalent resistance
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
This is the best possible choice for the first step.
The key to solving this problem is to convert it to something we can solve with our circuits tools. A voltmeter is, from a circuits standpoint, just a resistor. So, if we replace it with a resistor, we can solve. Let’s make the replacement, removing the voltmeter and putting a 10[kW] resistor in its place.

11. Your Choice for First Step Was – Divide the Voltage (2[V]) by the Resistance (10[kW]) to Get the Current
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
This is not the best choice for the first step.
If the meter is reading full scale, then the current through the meter will be 2[V] divided by the resistance of the meter, 10[kW]. However, we don’t have any particular use for this current. Actually, it could be used to solve this problem, but not easily. We recommend that you go back and try again.

12. Your Choice for First Step Was – Divide the Voltage (50[V]) by the Resistance (10[kW]) to Get the Current
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
This is not the best choice for the first step.
This does not even make any sense, since the voltage 50[V] is not across the 10[kW] resistance of the voltmeter. It is often tempting to divide a voltage by a resistance to get a current, but it only makes sense to do so when the voltage is across that resistance. We recommend that you go back and try again.

13. Replace the Voltmeter With Its Equivalent Resistance
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
We have replaced the voltmeter with its equivalent resistance, which is 10[kW]. We have have called this resistor RM.
Now, it is probably clear to you that RV and RM are in series, and that we can use the voltage divider rule to write an equation relating vT and vM. We do this below.
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14. Values for the Voltage Divider Rule?
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
What values can be inserted in the voltage divider rule below? The resistor value RM is known, so we have already inserted that. However, this still leaves us with three unknowns. RV is the one we are looking for, so we expect that we might be able to find appropriate values for vT and vM.
What can we use?
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15. Full Scale Values
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
We asked in the last slide which values for vT and vM we can use? We can use any pair of values that are valid at the same time. We could use the lowest reading values, although using zero for both does not help us find RV. The values we generally use are the full-scale values, since they occur at the same time, and are often known. We have
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16. Solving for RV
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
Now, we can just solve for RV. We get
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17. Looking at Part b)
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
Having solved for RV, we now have the circuit shown above. Now, we have been told that the voltage at terminals A and B is 18[V], which means that this is the value for vT for this part. If we substitute in this value, we can find vM.
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18. Solution for Part b)
A voltmeter is used in series with a resistor to obtain a new voltmeter with the ability to read a higher voltage. The combination is given in the circuit here. The voltmeter has a meter resistance of 10[kW], and reads up to 2[V] full-scale.
The goal is to have a new voltmeter that can read up to 50[V]. What value of resistor RV should be used to make this happen?
When 18[V] is applied to this series combination at terminals A and B, with RV as found in part a), what will the voltmeter read on the 2[V] scale?
We have solved for the voltage vM, for the case of this applied voltage. What will the voltmeter with the full scale reading of 2[V] read? This voltmeter is represented here by the resistor RM. It reads the voltage vM! This is the voltage across it. So, we have the solution, which is
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19. What is the deal with this Full-Scale Reading?
The full-scale reading is a very useful tool in solving many of these meter problems. Many meters are characterized by their full-scale readings. But, the most important part is that this is one value that we can associate with other full-scale readings in a multi-range meter. The key point is this: If we use one meter but several connections to get multiple ranges, each range reads full scale at the same time. Thus, all these values are valid together, and we can solve circuits using them.
Go back to Overview slide.