Resistance and Voltage, switch diagram

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vysero
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Homework Statement


Text problem statement: a DC circuit with switches, all resistorless when closed and all resistors identical, with an ideal voltmeter; below it are eight switch configurations labelled for the circuit.
[/B]
Circuit schematic with a battery, a voltmeter V, three resistors and switches S1, S2 and S3, followed by a table of eight configurations A to H giving each switch as open or closed.


Homework Equations


V=IR

The Attempt at a Solution


I said: A>D=F>B=C>G>H=E

I assumed higher resistance would lead to higher readings. However here are the answers:

H=E>C=B>G>D>F>A

Can someone explain to me why this is? As you can see H=E which combined gives a total resistance of (2/3)ohms is, according to the answers, going to give the largest volt meter reading. Why is this true?
 
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vysero said:

Homework Statement


View attachment 80648
Circuit schematic with a battery, a voltmeter, three resistors and three switches labeled S1, S2 and S3, with a table listing open and closed states of the switches for configurations A through H.

Homework Equations


V=IR[/B]

The Attempt at a Solution


I said: A>D=F>B=C>G>H=E

I assumed higher resistance would lead to higher readings. However here are the answers:[/B]
Why assume that? The volt-meter only measures the voltage drop across that one resistor. Its resistance is fixed, so the voltage is proportional to the current through that resistor. Any current supplied by the battery must pass through that resistor, some how is that current related to the overall resistance of the circuit?

H=E>C=B>G>D>F>A

Can someone explain to me why this is? As you can see H=E which combined gives a total resistance of (2/3)ohms is, according to the answers, going to give the largest volt meter reading. Why is this true?
How do you get any particular resistance value?

No resistance value is given for any of the resistors. Are we to assume that they all have the same resistance?
 
SammyS said:
Why assume that? The volt-meter only measures the voltage drop across that one resistor. Its resistance is fixed, so the voltage is proportional to the current through that resistor. Any current supplied by the battery must pass through that resistor, some how is that current related to the overall resistance of the circuit?

How do you get any particular resistance value?

No resistance value is given for any of the resistors. Are we to assume that they all have the same resistance?

OH geesh sorry I feel like a dork I forgot a part of the question here is the explanation:

Text block describing a DC circuit with several resistanceless switches, identical resistors, and an ideal voltmeter, with all switches open, followed by eight switch configurations.
 
vysero said:
OH geesh sorry I feel like a dork I forgot a part of the question here is the explanation:

View attachment 80652
Printed problem text describing a DC circuit with several resistanceless switches, identical resistors and an ideal voltmeter, with all switches open, followed by eight switch configurations.

You could type some of this out.

Part of the explanation has been given. We need a response from you. You have over 100 posts, so you should know how things work here.

How did you arrive at your order? (in addition to having voltage behavior reversed.)
 
I am not sure how I had voltage behavior reversed. V = IR so and increase in resistance will increase voltage. If I = 1 and R = 4 then V = 4 if I = 1 and R = 10 then V = 10. 1<10

However, I did not take into account that the measurement of V was across only one resistor. I am currently trying to work out how to get the V for that one resistor in each configuration.