- #1

PhysicsInNJ

- 44

- 1

I'm having trouble understanding how voltage drops.

My understanding of voltage is a difference in potential, with potential being how much work is necessary to move a test charge to a specific orientation.

So, if we have a circuit with say, a 12V battery and a 1 ohm resistor, and I had a +1 coulomb charge sitting at the negative terminal I would have to apply 12 joules of work to move it to the positive terminal.

My confusion begins here: why, If I do this in real life, no matter where I take check the potential difference with a voltmeter I always will get 12V whether from the negative terminal or, after the resistor.

Is it because the wire/resistor have charge carriers and essentially become "terminals" wherever I decide to place my voltmeter?

Then, when considering a simple voltage divider with two resistors of equal value, what happens to the other half of the potential? Shouldn't it require the same amount of work to bring that test charge to the positive terminal?

My understanding of voltage is a difference in potential, with potential being how much work is necessary to move a test charge to a specific orientation.

So, if we have a circuit with say, a 12V battery and a 1 ohm resistor, and I had a +1 coulomb charge sitting at the negative terminal I would have to apply 12 joules of work to move it to the positive terminal.

My confusion begins here: why, If I do this in real life, no matter where I take check the potential difference with a voltmeter I always will get 12V whether from the negative terminal or, after the resistor.

Is it because the wire/resistor have charge carriers and essentially become "terminals" wherever I decide to place my voltmeter?

Then, when considering a simple voltage divider with two resistors of equal value, what happens to the other half of the potential? Shouldn't it require the same amount of work to bring that test charge to the positive terminal?

Last edited by a moderator: