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

  • Thread starter Eitan Levy
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  • #1
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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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Answers and Replies

  • #2
Orodruin
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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
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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
Orodruin
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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.

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
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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
Orodruin
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The exact words are "The new voltage equals to 90% from the previous one".
Then it does seem wrong.
 
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