Electro question: Drop Voltage with 1 ohm resistors

[atrain77a]

This might be a silly one, but I'm a bit stumped:

Suppose you have a 6V battery and an unlimited supply of 1 ohm resistors.

How do you drop the potential from 6V to 4V using the resistors?

Off the top of my head I think you just put two of the resistors into a series attached to the positive side (cathode??) of the battery and you're at 4V.

However, since current isn't mentioned anywhere (think V=IR) I'm a bit confused. Is it possible to simply lower potential by stringing resistors together in series without worrying about current drawn or did I miss something simple? Thanks a bunch. -Stumped T. Day

 

[mewmew]

Current is only created in the presence of a potential difference and resistance, so they didn't mention it because without the resistors you have no current. With resistors in series the current is the same, I = I1=I2=... also, so it will be constant. If they are in parallel than I=I1+I2... and the voltage stays the same.

You should be able to use that to construct a circuit that has a final potential difference between two points of 4V. The potential difference between your battery though will always be 6V.

Also just to make sure you know, to get current or resistance in a circuit you must have a potential difference, so if you have two resistors hooked up to the + battery terminal nothing will happen.

 

[Claude Bile]

A simple voltage divider will do. Connect 3 1 Ohm resistors in series, the voltage drop across two 1 ohm resistors will be 4 Volts (you can verify this using Ohm's Law). You can use this potential as your new 'battery' (Although in practice, I would recommend this configuration since 2 Amps are flowing through the circuit).

Claude.