# Quick query regarding voltage sources

1. Aug 8, 2009

If you had a DC voltage source that supplied a voltage V volts and you connected a resistor of R1 ohms to the + and - terminals of this source then shorted the resistor with a wire of little but measurable resistance - R2 ohms would the following happen:

*would the wire draw high current from the voltage source. ie current drawn = V / R2 and if this didn't cause the wire to burn would this current be limited by the ability of the circuitry controlling the voltage source and hence if it couldn't reach the current needed to meet the resistance of the wire ( R = V / I ) would it increase the resistance of this wire? So could we have a voltage controlled variable resistor?

*is the above physically possible? And if so how does this work?

Cheers!

2. Aug 8, 2009

Throw away resistor 1 since it's shorted. You are asking what happens when we short a power supply?
The voltage of the power supply drops until it reaches it's maximum power output P_max=U^2/R_wire

3. Aug 8, 2009

### TurtleMeister

If the source could supply it the current draw would be V / ((R1 * R2) / (R1 + R2))
Yes.
No. The voltage and current would drop (or a fuse would blow). See previous question and answer.
Yes, but not the way you described.
Field Effect Transistors (FET) are sometimes used as voltage controlled resistors. You can find info on the web about FETs.

4. Aug 8, 2009

### Carl Pugh

Yes you can have a voltage controlled variable resistor. (Though not as you described)

An Incandescent light bulb is an example of a voltage controlled resistor.

In the past the feedback in a Wein bridge oscillators was frequently controlled by the variable resistance of an incadescent light bulb.

5. Aug 8, 2009

### negitron

No, that would be a current-controlled resistor.

6. Aug 8, 2009

Thankyou to everyone for the feedback :)

7. Aug 9, 2009

### vk6kro

Have you come across the concept of internal resistance?

All batteries and most simple power supplies have an effective internal resistance, which doesn't really exist, but gives a way of predicting the behaviour of the power source.

This resistance appears in series with the open circuit voltage of the device. So, a 9 volt battery might have an internal resistance of 10 ohms. If you put a low resistance wire across the battery, it would not deliver more than (9volts / 10 ohms) or 0.9 amps into the wire.

If you put a 10 ohm resistor across the battery, half the 9 volts would be across the internal 10 ohm resistor and half across the external 10 ohm resistor. So, the external resistor would have 4.5 volt across it and 0.45 Amps flowing in it.

This internal resistance increases with age in batteries and, eventually, the battery cannot deliver enough current to operate whatever it is operating and then it has to be replaced with one that has less internal resistance.