Help Solve Voltage Homework Questions for Engineering Student

[Luke0034]

Homework Statement

I am trying very hard to understand voltage. I am a second year engineering student in a physics II class. I am having trouble grasping the concept of voltage. Could you guys please help me out with a few questions.

1. Please correct me if I am wrong, but isn't voltage the difference of two separate electric potentials?
2. If you were to measure the electric potential at a certain point, you'd get a certain value "x" in J/C... but what is that measuring? What does that scalar represent? If you were to put a one coulomb charge at that point you'd have "x" J. But since you only measured one point in space, what is that in reference to? It takes "x" amount of Joules to do what? Is it just saying that at that point, that charge would have that much potential energy?
3. I am also confused about voltage drop across a resistor. So if you were to hook up one lead of a voltmeter on one side of the resistor in a circuit and the other lead to the other side, and you get a reading of "x" volts. That voltage to my knowledge represents the difference between the electric potential on one side and the electric potential on the other side... the drop. So my question is if you were to just have a wire going from the positive terminal of the battery to the negative terminal of the battery, and then you measured the voltage across that, wouldn't it be zero? If not, why wouldn't the reading be zero, there is nothing in the circuit to drop the voltage, so it should remain the same and x volts minus x volts is zero.
4. But here's where I'm confused, I've seen videos of people measuring the voltage of the battery by connecting the voltmeter to each terminal of the battery and getting the reading, but why did they get a value and not zero, since the difference should be zero?
5. Also, why have I heard of voltage being "the force" that pushing electrons through a circuit, when voltage is not a force. Force is measured in Newtons, and voltage is Joules/Coulomb. So I can't grasp how voltage could "push" electrons, given the units voltage has.
6. And the last thing I don't understand is why the voltage drop across all the resistors in a circuit must always be zero. The idea that the voltage drops to zero across a 10 ohm resistor in one circuit, and also drops to zero across 2 different 10 ohm resistors in a different circuit is mind boggling to me.

I know this is a lot of questions, and the questions may be confusing, but it is hard for me to word questions when I don't know what I'm talking about.

Homework Equations

V = IR

The Attempt at a Solution

My attempt has been thinking about this and searching the internet and getting conflicting answers for 3 hours.

[Thread moved to GP by mentor]

 

[TomHart]

I have heard people describe voltage, current, and resistance in terms of liquid flow. Voltage is analogous to pressure. Current is analogous to rate of water flow. Resistance is analogous to the smallness (if that's a word) of the hole that the water flows through. So think of a huge tank of water (a battery) that is two feet deep (representing say, 2 volts). If you drilled a 0.1 inch diameter hole (representing say, 10 ohms) at the bottom of that tank, you will get some amount of water flow (current) through that hole. Next, if you had another huge tank of water that is 10 feet deep (representing say, 10 volts) and had the same 0.1 inch diameter hole at the bottom of that tank, you will get a significantly greater rate of water flow (current) due to the greater pressure (voltage). Next, if you replaced that 0.1 inch diameter hole with a 1 inch diameter hole, that is equivalent to a large decrease in the resistance. The pressure (voltage) has not changed. But because the resistance has been significantly decreased, the current (water flow) will significantly increase. Next, if you drilled 9 more 1-inch holes at the bottom of the tank, that would be the equivalent of having 10 resistors in parallel. I hope that helps a little.

 

[Luke0034]

I have heard people describe voltage, current, and resistance in terms of liquid flow. Voltage is analogous to pressure. Current is analogous to rate of water flow. Resistance is analogous to the smallness (if that's a word) of the hole that the water flows through. So think of a huge tank of water (a battery) that is two feet deep (representing say, 2 volts). If you drilled a 0.1 inch diameter hole (representing say, 10 ohms) at the bottom of that tank, you will get some amount of water flow (current) through that hole. Next, if you had another huge tank of water that is 10 feet deep (representing say, 10 volts) and had the same 0.1 inch diameter hole at the bottom of that tank, you will get a significantly greater rate of water flow (current) due to the greater pressure (voltage). Next, if you replaced that 0.1 inch diameter hole with a 1 inch diameter hole, that is equivalent to a large decrease in the resistance. The pressure (voltage) has not changed. But because the resistance has been significantly decreased, the current (water flow) will significantly increase. Next, if you drilled 9 more 1-inch holes at the bottom of the tank, that would be the equivalent of having 10 resistors in parallel. I hope that helps a little.

Thank you so much, and what would be the equivalent of resistors in series?

What would be a capacitor in this analogy?

 

[TomHart]

Hmmm. I guess resistors in series would be like different pipe sizes connected end-to-end. I guess at this point, it may be more helpful to think of the hole being cut into the side of the tank and have a horizontal pipe attached to the hole. So if you had one huge-diameter pipe attached to a hole of the same diameter in the tank, and then connect the output of that huge pipe to a pin-hole size pipe, you should be able to see that the pin-hole size pipe would dominate the amount of water that flows. It would be like placing a 1 MΩ resistor in series with a 1 Ω resistor. The 1 Ω resistor is negligible.

I don't really have a good understanding of fluid flow, so I can't take this analogy very far. So as far as a capacitor, I'm not sure.

 

[John Park]

Yes, the capacitor's tricky in that model. It might be a large vessel with an inlet at one side and an outlet at the other and water in the vessel, but a flexible membrane dividing the vessel down the middle. If one side of the tank fills up more, the membrane will bulge a bit but no water will actually get through. On the other hand if the water level on one side oscillates, the membrane will bulge in and out and transmit the pressure changes to the other side. I'm not sure that's accurate in detail, but it gives some idea of what a capacitor does.

 

[cnh1995]

If you were to measure the electric potential at a certain point

In a circuit, you should first pick a point and call it the 'reference node' or 'ground'. All the potentials in the circuit are with respect to the ground. So naturally, potential of the ground will be 0V. Circuits work on potential 'difference'. So what you measure is the potential 'difference' between two points.

Also, why have I heard of voltage being "the force" that pushing electrons through a circuit, when voltage is not a force. Force is measured in Newtons, and voltage is Joules/Coulomb. So I can't grasp how voltage could "push" electrons, given the units voltage has.

Voltage is not a force. Voltage (EMF precisely) of the battery creates electric fields of different magnitudes in various circuit components and wires. Electric field is the gradient of potential. But the field inside the wires is negligible compared to that in the resistors, which means there is a negligible potential gradient. Hence, potential of a wire is assumed to be constant and hence, voltage "across" the wire is zero.

See if this thread helps..
https://www.physicsforums.com/posts/5687446/

 

[Luke0034]

In a circuit, you should first pick a point and call it the 'reference node' or 'ground'. All the potentials in the circuit are with respect to the ground. So naturally, potential of the ground will be 0V. Circuits work on potential 'difference'. So what you measure is the potential 'difference' between two points.

Voltage is not a force. Voltage (EMF precisely) of the battery creates electric fields of different magnitudes in various circuit components and wires. Electric field is the gradient of potential. But the field inside the wires is negligible compared to that in the resistors, which means there is a negligible potential gradient. Hence, potential of a wire is assumed to be constant and hence, voltage "across" the wire is zero.

See if this thread helps..
https://www.physicsforums.com/posts/5687446/

Thanks Sheldon, I freaking love you!

 

[anorlunda]

Usually, I don't like the water analogy, but I must admit that it helps sometimes.

Credit this picture to PF member @Jony130 .
It includes resistors, capacitors, inductors, a switch, even a diode. It can work with AC or DC.
Very clever.

 

[David Lewis]

Also, why have I heard of voltage being "the force" that pushing electrons through a circuit, when voltage is not a force. Force is measured in Newtons, and voltage is joules/coulomb. So I can't grasp how voltage could "push" electrons, given the units voltage has.

The force that pushes electrons is electrostatic force. Electromotive force is a bit of a misnomer. EMF is an energy concentration, not a force.