Find the resistance to produce a specified potential drop

[Vladi]

Homework Statement

In Fig. 26-5, how large must R be if the potential drop from A to B is 12 V?
(The figure has been pasted into the attached file).
The answer to this problem is 3.0 ohms. I keep trying to solve this problem but I get a wrong answer. The answer I get from my calculations does not make sense. How can resistance be negative? Any help is appreciated.

Homework Equations

-R=V/I-->V=(R)(I)

The Attempt at a Solution

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My attempt has been attached to this post. The signs that are used within my calculations were determined by the bullet points.

 

[Nidum]

The diagram shows that the current flowing into this series connected line of components is 3 A and you are also told that the total potential drop from A to B is 12 V .

Start by working out the potential drop across the 2 Ohm resistance . What do you think that is ?

 

[Vladi]

As I stated with the bullet points, the potential drop across a resistor is always negative. Thus, the potential drop across the 2 Ohm resistor must be -6 volts. This must also mean that the potential drop across the resistor R is -R*3. The wider side of a battery is positive. The shorter side of a battery is negative. If I move from negative to positive across a battery, the potential drop is positive. If I move from positive to negative, the potential drop is negative. The current determines which direction I should be calculating. In this case, it is left to right.
Using the logic stated above, I came up with the following:
12V=-6.0V-(2.0 Ohms)(3.0 A)+9.0V-(R)(3.0A)

 

[haruspex]

the potential drop across the 2 Ohm resistor must be -6 volts

Not sure if you mean that. If the "drop" is -6V that would mean the potential goes up by 6V as the current passes through it.
The potential drops 6V across the first battery, another 6 through the first resistor, then rises 9V across the second battery.

 

[Vladi]

Not sure if you mean that. If the "drop" is -6V that would mean the potential goes up by 6V as the current passes through it.
The potential drops 6V across the first battery, another 6 through the first resistor, then rises 9V across the second battery.

You're right. If you are saying drop, that means you are subtracting voltage. If you are rising, that means you are adding. Saying there is a potential drop of -6 volts is redundant and confusing.

 

[Vladi]

You're right. If you are saying there is a potential drop, that means you are subtracting voltage. If you are saying there is a potential rise, that means you are adding voltage. Saying there is a potential drop of -6 volts is redundant and confusing. There is always a potential drop across a resistor. If you are going from a short end of a battery to a long end, there is a POTENTIAL rise. If you are going from a long end of a battery to a short end, there is a POTENTIAL drop. I'm still adjusting myself to the terms. Thanks for the clarification. Are my calculations incorrect? If so, what is it? Thank you for your time.

 

[haruspex]

You're right. If you are saying drop, that means you are subtracting voltage. If you are rising, that means you are adding. Saying there is a potential drop of -6 volts is redundant and confusing.

Do you see that this confusion has led to sign errors in your calculation?
Look at the second sentence in my post #4. Can you continue in that style through the remaining components? Can you then turn that into a voltage equation with the right signs?

 

[Vladi]

Do you see that this confusion has led to sign errors in your calculation?
Look at the second sentence in my post #4. Can you continue in that style through the remaining components? Can you then turn that into a voltage equation with the right signs?

The voltage equation that I posted before has one mistake. Because the question asks how large R must be for the potential drop to be 12 volts from a to b, this implies that the 12 volts is negative within the equation!
-12V=-6.0V-(2.0 Ohms)(3.0 A)+9.0V-(R)(3.0A)
-->R=3 ohms
Thank you for you help! This problem has helped me solidify some of the new terminology.