Simple Question: potential difference between two points on circuit

[mtu620]

Homework Statement

See attached image. I'm having trouble applying the concept of potential difference in a circuit.
Vb-Va=?
Va-Vb=?

Homework Equations

Vab=Vb-Va

The Attempt at a Solution

The overall voltage is 5V-2V=3V. The direction of the current is defined by the higher voltage batter (i.e. ccw). Because current is defined as direction of positive charge flow,
Vb-Va=-3V
Va-Vb=+3V
Thus A is at a higher potential.

Is my reasoning correct?

 

[BruceW]

There will also be voltage drops in the circuit, so the potential at A and B will depend on where the voltage drops happen. You haven't drawn the voltage drops (i.e. resistors or whatever). But they do need to be in there somewhere, so that Kirchhoff's laws are obeyed.

About the concept of potential difference: the keyword is difference. So if you have a battery for example, there will be some potential on one side and a different potential on the other side. I like to imagine it as the altitude of a person walking around a hilly countryside. So when the person makes a round trip, he knows he will be back at the same altitude he started at. (This explains why the emf's will equal the voltage drops in any given loop of a circuit).

 

[mtu620]

I have attached a redrawn circuit.

Thanks. That does help. However, I am still confused on the signs for potential difference.


Req=3 ohms and V=3V, thus I=1A. Also, V1 + V2 =Vbatt=3V

What I'm confused about is if
Va-Vb=+3V or
Vb-Va=+3V or
neither?

and why?

Thanks so much again!

 

[BruceW]

Yes. Getting the sign right can be difficult. There's 2 things to remember: The larger terminal of a battery is at a higher potential than the smaller terminal. (They are also called positive and negative, respectively). Also, the potential either side of a resistor decreases in the direction of the current through it. (This is because the resistor dissipates energy). In other words, if you follow the current through the resistor, you'll find the potential decreases.

So maybe now you have the problem: 'what direction is the current going?' Well, using the rules just above, you can find the direction which current must go so that Kirchhoff's laws are obeyed in the circuit.

Once you know which way the current is going (and what its value is), then you can find the potential difference between any two points on the circuit. (Using the same rules above).

 

[mtu620]

Thanks, Bruce.
From what you've said, I've gathered that if the resistor is traveled in the direction of the current (ccw, 1A in this example) then the potential difference across the resistor is -IR.
Thus, traveling from + to - of battery (in the direction of the current), the potential difference is -Vbatt.

Thus, Va-Vb=-3V and
Vb-Va=+3V.

 

[BruceW]

yes, the current is going ccw. And yes, if we 'travel ccw', the potential difference across the resistor in this direction will be -IR. Also, we are going ccw, so we are going from -ve to +ve terminals of the battery. (since the -ve terminal is on the left and the +ve is on the right). So the potential difference is +Vbatt.

 

[love.singh93]

I have attached a redrawn circuit.

Thanks. That does help. However, I am still confused on the signs for potential difference.


Req=3 ohms and V=3V, thus I=1A. Also, V1 + V2 =Vbatt=3V

What I'm confused about is if
Va-Vb=+3V or
Vb-Va=+3V or
neither?

and why?

Thanks so much again!

buddy ...
you take 2 points a and b
there is no matter of that a is near the +ve end of battery and b is near the -ve end of battery
we need to find pot. diff between 2 batteries
V at a will be (+5) - (+2)=+3 volt
V at b will be (-5) - (-2)= -3 volt
thus Va-Vb=(3)-(-3)= 6 volt
and Vb-VA=(-3)-(3)= -6volt