Why is the voltage across this parallel circuit positive?

[s3a]

Homework Statement

The problem along with its solution are attached in TheProblemAsWellAsSolution.jpeg file. For this post, I'm only referring to part (a), since part (b) is the same thing with another number.

Homework Equations

V = IR
Parallel circuit: $I = I_1 + I_2 + . . . + I_n$

The Attempt at a Solution

I understand how to get i = 4$i_x$ - 4.0, and I understand that multiplying both sides of that equation by R = 10 Ω gives iR = 40##i_x − 40.0, but when looking at the circuit, I'm confused as to why ν_R = 40.0 V instead of −40.0 V. I thought that when (positive) current goes from the positive to negative end of a resistor, that there is a voltage drop, as the following website implies.: http://faculty.wwu.edu/vawter/PhysicsNet/Topics/DC-Current/DropGain.html

Could someone please clarify this for me?

Any input would be GREATLY appreciated!

 

[davidbenari]

The +40V is correct. Do proper mesh analysis and you'll see that in fact positive current is going through the positive terminal.

 

[davidbenari]

$4*2A$ goes downwards and $4A$ goes upwards. This gives $4A$ downwards (which means it enters the positive terminal). Finally $4*10=40V$

 

[s3a]

So a voltage drop (i.e., a negative potential difference) is when positive current goes from the negative end of a resistor to the positive end?

 

[davidbenari]

No, positive current goes through the positive terminal for there to be a voltage drop.

 

[davidbenari]

By which I mean it has to cross the positive terminal first.

 

[davidbenari]

In my opinion its easier to figure out where the positive current is going and then think about the way terminals should be oriented such that $V_{+}-V_{-}>0$ and work from there to answer other things.

 

[phinds]

So a voltage drop (i.e., a negative potential difference) is when positive current goes from the negative end of a resistor to the positive end?

Resistor don't HAVE a positive end or a negative end, they are passive elements. When current flows into a resistor at one end and out the other end then the end that the current flows into has a higher potential than the end the current flows out of. If the current has a negative sign, that just means it's flowing opposite to how the arrow is drawn, but the end of the resistor that the current actually flows into is the end that has the higher potential.

 

[s3a]

Since the current goes into the resistor, I see why the + is "north" and the - "south", but what's going on with the value of the potential difference being positive in the solution? In class, my teacher said that a voltage drop is positive, but then what's going on in this link ( http://faculty.wwu.edu/vawter/PhysicsNet/Topics/DC-Current/DropGain.html )?

 

[davidbenari]

If they say there is $10V$ across a resistor and have drawn a (+) sign and a (-) sign next to it then what they're telling you is $V_{+}-V_{-}$. I think that link is going to confuse you although what they're saying is true.

If you want an explanation for what they say in the link here it is:

Traditionally when people say change in a quantity they refer to $Final-Initial$. The "change in voltage" as they specify it in your link is $V_b-V_a$ since they traverse the resistor in the direction a->b

But when we say voltage drop across a resistor we really mean $V_a-V_b$, and this is what they mean by "voltage drop is the negative of the change in voltage".

"Voltage drop" is more useful and intuitive in my opinion. If you imagine having a voltage as being on a hill then we can imagine that traversing a resistor is the equivalent to falling off that hill. The way you know how much you fell is $V_{a}-V_{b}$ as this would be a positive quantity if positive current was indeed traversing a->b

Does that answer your question?