Is the textbook wrong or am I? (circuit with resistors, switches and a battery)

In summary: It appears that your textbook has a mistake. In summary, the given circuit has a voltage source with an internal resistance of (20/9)Ω and a voltage of 12.6V between the parts of S2 when S1 is shut off. Shutting off both S1 and S2 will result in a smaller combined resistance and a higher current, causing the voltage to be smaller than the previous value. The textbook may have a mistake in assuming that the new voltage is 10% lower than the old, rather than the old being 10% higher than the new.
  • #1
Eitan Levy
259
11

Homework Statement


This circuit is given:[/B]
upload_2017-12-16_18-10-55.png

Each resistor has a resistance of 30Ω, and the electrive force is 28V.
Now we shut S1, and the voltmeter displays a value lower by 10% from the previous one.
What is the internal resistance of the voltage source?
What is the voltage between the parts of S2?
Now we shut S2 (S1 stays shut), will the voltage displayed be higher of lower than the voltage displayed when only S1 closed?

Homework Equations


V=IR

The Attempt at a Solution


My answers are:
r=(20/9)Ω (Textbook says 2Ω)
V=12.6V (Again, a different answer according).
The voltage will be lower (Textbook says higher)
The book is brand new so it has some mistakes, that's why I am asking.
Thanks in advance!

*How I reached those values:
When both switches are open no current is flowing. V=28V.
When we shut S1 the new voltage become 25.2V, that means that 25.2=28-Ir (r for internal resistance).
Now the total resistance would be (60*30)/(60+30)+r=20+r. From here we can see: I=28/(20+r)
Two equations with two parameters.
I=1.26A, r=(20/9)Ω
The current on the top branch would be I/3=0.42A
The current on the branch below would be 2I/3=0.84A
The voltage: -0.42*30+0.84*30=12.6V.
Now, when we shut both switches no current would flow through the resistor right to S2.
The combined resistance becomes (30*30)/(30+30)+r, this gives a number smaller than the combined resistance before. That means the I will be higher this time, and because V=ε-Ir, V will be smaller.
 

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  • #2
Please describe how you reached those values. If not, it is impossible for us to know if and when you go wrong in your argumentation.
 
  • #3
Orodruin said:
Please describe how you reached those values. If not, it is impossible for us to know if and when you go wrong in your argumentation.
When both switches are open no current is flowing. V=28V.
When we shut S1 the new voltage become 25.2V, that means that 25.2=28-Ir (r for internal resistance).
Now the total resistance would be (60*30)/(60+30)+r=20+r. From here we can see: I=28/(20+r)
Two equations with two parameters.
I=1.26A, r=(20/9)Ω
The current on the top branch would be I/3=0.42A
The current on the branch below would be 2I/3=0.84A
The voltage: -0.42*30+0.84*30=12.6V.
Now, when we shut both switches no current would flow through the resistor right to S2.
The combined resistance becomes (30*30)/(30+30)+r, this gives a number smaller than the combined resistance before. That means the I will be higher this time, and because V=ε-Ir, V will be smaller.
Is everything correct?
 
  • #4
Eitan Levy said:
Now we shut S1, and the voltmeter displays a value lower by 10% from the previous one.
I suspect your textbook assumes that ##1.1 V_1 = V_0##, i.e., the old voltage is 10% higher than the new, which is not the same as saying that the new voltage is 10% lower than the old. At least, using ##V_1 = 28/1.1## V gives the internal resistance 2 ohm.

Eitan Levy said:
The combined resistance becomes (30*30)/(30+30)+r, this gives a number smaller than the combined resistance before. That means the I will be higher this time, and because V=ε-Ir, V will be smaller.
Sounds reasonable.
 
  • #5
Orodruin said:
I suspect your textbook assumes that ##1.1 V_1 = V_0##, i.e., the old voltage is 10% higher than the new, which is not the same as saying that the new voltage is 10% lower than the old. At least, using ##V_1 = 28/1.1## V gives the internal resistance 2 ohm.Sounds reasonable.
The exact words are "The new voltage equals to 90% from the previous one".
 
  • #6
Eitan Levy said:
The exact words are "The new voltage equals to 90% from the previous one".
Then it does seem wrong.
 
  • #7
Orodruin said:
Then it does seem wrong.
Alright, thanks.
 

1. Is it possible for the textbook to be wrong about a circuit?

Yes, it is possible for the textbook to have incorrect information about a circuit. Textbooks are not infallible and can contain errors or outdated information.

2. How do I know if the textbook is wrong or if I made a mistake?

The best way to determine if the textbook is wrong or if you made a mistake is to double check your work and compare it to other reliable sources. If you still have doubts, consult with a teacher or another expert in the field.

3. Can I trust my own analysis of a circuit over the textbook's explanation?

It is important to have confidence in your own analysis and understanding, but it is always a good idea to verify your conclusions with other sources. This will help you to catch any mistakes and deepen your understanding of the circuit.

4. What should I do if I find a mistake in my textbook?

If you believe you have found a mistake in your textbook, it is important to bring it to the attention of your teacher or the textbook publisher. This will help to ensure that future editions of the textbook will have accurate information.

5. How can I improve my understanding of circuits to avoid confusion between the textbook and my own analysis?

You can improve your understanding of circuits by practicing and experimenting with different circuits, seeking help from experts, and consulting multiple sources for information. It is also important to have a solid foundation in the fundamentals of circuit analysis.

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