Circuit problem: Figuring out Voltage Drops

[joej24]

Homework Statement

Homework Equations

V = IR

The Attempt at a Solution

Why does the voltmeter measure only the voltage drop in V₃?

Is it because the Voltmeter is specifically placed between points a and b?
I originally thought that the Voltmeter would measure the voltage drop up to where it is located.

 

[gneill]

What do you mean by "up to where it is located"?

The voltmeter has two leads (connection points); it can only tell you the potential difference that exists between those two leads. Anything else would be magic

 

[joej24]

Starting from the positive end of the battery and ending where the voltmeter is located. What are leads and where are they located? Why does the voltmeter only show the pot. diff. between these points too?

 

[gneill]

Starting from the positive end of the battery and ending where the voltmeter is located. What are leads and where are they located? Why does the voltmeter only show the pot. diff. between these points too?

Have you ever seen a voltmeter in real life? It has two flexible wires (usually one is red and the other black) that end in rigid "test probes". The user touches the ends of the probes to the circuit at the points, between which, he wishes to determine the potential difference.

In your circuit the voltmeter leads are connected to points a and b, so it will measure the potential difference between points a and b.

 

[joej24]

Oh okay. And no, I am not too familiar with voltmeters.

So the voltmeter isn't really part of the circuit? The current first passes through r (50 Ohms) from the positive terminal of the battery and then splits at J. It meets back at the point right next to R3. It then passes through R3, past b, and goes back to the negative terminal of the battery.

The current that passes through b is 6 A and that current also passes through R3. So V = IR = 6*150 = 900 V.

 

[joej24]

How does the location of point a affect the potential difference measured by the voltmeter?
If a was located right after r (the battery's internal resistance) would the potential difference be changed in any way?

Is this correct?

V = IR
Total Resistance that is between a and b is 150 + (1/600 + 1/300) ^ -1 = 350 Ohms
V = 6 * 350 = 2100, which is the energy lost between a charge from point a to point b?

 

[gneill]

Oh okay. And no, I am not too familiar with voltmeters.

So the voltmeter isn't really part of the circuit? The current first passes through r (50 Ohms) from the positive terminal of the battery and then splits at J. It meets back at the point right next to R3. It then passes through R3, past b, and goes back to the negative terminal of the battery.

The current that passes through b is 6 A and that current also passes through R3. So V = IR = 6*150 = 900 V.

Looks good.

Voltmeters are usually designed to present a very high resistance to the circuit under test (they "look like" a very high resistance value) so that they don't disturb the operation of the circuit.

 

[gneill]

How does the location of point a affect the potential difference measured by the voltmeter?
If a was located right after r (the battery's internal resistance) would the potential difference be changed in any way?

Yes, of course it would change; you would be measuring the potential from point b to a location different from a in the circuit.

Is this correct?

V = IR
Total Resistance that is between a and b is 150 + (1/600 + 1/300) ^ -1 = 350 Ohms
V = 6 * 350 = 2100, which is the energy lost between a charge from point a to point b?

Yes, that's good.