Why voltage increases when you increase resistance? - Page 1
Forum section: electronics / beginners

author: DuncanSteel
author_rank: Contributor
timestamp: August 04, 2016, 04:00:31 pm
content: HiSay you have 6.5V battery in series with a Potensiometer. You start turning the pot from 5 Ohms and go up to 10k Ohms.How comes that if you measure voltage across the pot, the voltage starts from 1V at 5 Ohms & goes upo to 6.5V at 10k Ohms ?I made a vid where you can see that the voltage is indeed changing.(Try chrome if it wont load in firefox)

author: rstofer
author_rank: Super Contributor
timestamp: August 04, 2016, 04:08:41 pm
content: The battery pack has an internal resistance. You can find out what it is by adjusting the potentiometer until the voltage reads exactly half of the open circuit voltage. At half voltage, the internal and external resistances are equal.Thevenin's Theorem comes to mind:https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

author: electr_peter
author_rank: Unknown
timestamp: August 04, 2016, 04:22:22 pm
content: I cannot see in the video how exactly circuit is connected, but my guess is that potentiometer is connected in series with battery as a rheostat (i.e. variable 2 terminal resistor). By rotating potentiometer knob you effectively create a varying resistive load to the 9V battery.9V battery is likely almost completely discharged (shows only 6.5V at max) and cannot handle the load.I would suggest to experiment with new battery as well - you will see less of a voltage change then. Also, try connecting battery to outer pins of potentiometer - this will give constant ~10kOhm resistive load to the battery irrespective of rotation. Then connect DMM across inner and outer potentiometer terminals - you should see voltages from zero to max battery voltage this way.

author: bson
author_rank: Unknown
timestamp: August 04, 2016, 04:31:58 pm
content: You're probably measuring between the wiper and ground.

author: madires
author_rank: Super Contributor
timestamp: August 04, 2016, 04:38:12 pm
content: It's wired as variable load. Luckily the current is limited by the battery's internal resistance, otherwise he would have learned that potentiometers got a power rating too:I = 6.5V / 5 Ohms = 1.3AP = 6.5V * 1.3A = 8.45W

author: rstofer
author_rank: Super Contributor
timestamp: August 04, 2016, 05:21:55 pm
content: Quote from: madires on August 04, 2016, 04:38:12 pmWhich is probably about 17 times the rated capacity of 1/2 Watt.Some batteries don't respond well to being short circuited. One of mine caught on fire! I think it was a rechargeable AA...

author: danadak
author_rank: Super Contributor
timestamp: August 04, 2016, 09:19:06 pm
content: There are a number of issues -1) At 5 ohms you are drawing 1A from batteries that do not have a lot of capabilityat that voltage. The earlier comment on internal resistance of the batteries. Sobattery chemistry getting heated up significantly.2) The pot at 5 ohms, if 1A is flowing out of the batteries, is experiencing 5W of heat,that can melt resistance elements.3) If you are spending a lot of time at 5 ohms plan on using a lot of batteries. Theywill not last very long.Regards, Dana.

author: Seekonk
author_rank: Super Contributor
timestamp: August 04, 2016, 10:09:52 pm
content: I am surprised the pot still works. That is a quick way to destroy it with that much current.

author: DuncanSteel
author_rank: Contributor
timestamp: August 04, 2016, 10:36:46 pm
content: Quote from: Seekonk on August 04, 2016, 10:09:52 pmI have burned several pots. Now im a little smarter when connecting them to a circuit.The reason it works is because I measured the short circuit current of this battery before connecting a pot.Turns out to be 35mA (yes suprizing, did the same to a AA battery & it gave out 1,2A) (so Iassumedthat a 35mA wont burn the pot)Ether way, so the voltage changes with resistance because of the Thévenin's theorem ?

author: Brumby
author_rank: Unknown
timestamp: August 05, 2016, 03:26:38 am
content: You don't need any theorems to explain this.It is a classic example of internal resistance of the battery. This is a factor in any electrical component and has the formal title ofEquivalent Series Resistance- orESR. Understanding this is extremely important in understanding how components actually work in the real world......(There are also capacitance and inductance parameters you should get to know - but you won't need to worry about those forthisexercise.)What you have created is a simple voltage divider that uses the internal resistance of the battery (which, from the conditions mentioned, appears to be somewhere around 28ohm{This is indicative only. Failing battery chemistry, variable usage and rest cycles will result in this varying significantly during experiments}) and the potentiometer, which you have wired as a simple variable resistor (only using the wiper and one end terminal).This rough sketch shows the circuit you have constructed, in two different layouts. In both cases, the dotted line represents the physical case of the battery, which is absolutely irrelevant in assessing the electrical circuit.The layout on the left is how it is commonly viewed, but the layout on the right makes it much more clear for seeing the voltage divider at work.When the potRis turned to minimum resistance (ie zero) there will be no voltage across it and the full voltage drop will occur across the internal resistancerof the battery. The resulting power loss will heat up the battery (how much will depend on how fresh the battery is).When the pot is turned to maximum (10K), then the pot will have a far greater resistance than the internal resistance of the battery - and will, therefore, have the greatest voltage drop.At points along the potentimeter's rotation, there will be a smooth but non-linear voltage change. This is what you have observed.On the point of burning the pots - let me guess.... At 10K, they survive. At zero, they survive. But when you move them just off from zero, the magic smoke appears...?

author: DuncanSteel
author_rank: Contributor
timestamp: August 05, 2016, 02:11:50 pm
content: I measured the inner resistance by turning the pot until I got half of the voltage 6,8V/2 = 3.4V, then measured the pot resistance 55 Ohm.Then increased pot resistance to 1,5k & calculated V.out= 1500/(55+1500) = 6,55VFinally measured the actual voltage, got 6.42VThank you Brumby, that was very good explanation. I now understand.

author: Rob Sims
author_rank: Regular Contributor
timestamp: August 05, 2016, 03:10:42 pm
content: Simple, The source voltage always stays the same. The voltage on the clamps changes with the load. . If the load resistance increase, the voltage on the clamps across the load will increase. Current draw is lower. If the load is infinite high (open circuit) the clamp voltage is the same as the source voltage. Current draw will be almost zero.If the load resistance decreases, the voltage on the clamps, across the load will decrease. Current draw is higher. If the load resistance is lowered to zero (short circuit), the current draw is at max and the voltage will be almost zero.Inner resistance of a power source is: Max voltage at open circuit divided by max current at short circuit.

author: ali6x944
author_rank: Frequent Contributor
timestamp: August 05, 2016, 03:59:34 pm
content: Is it like when a voltage regulator loses regulation when it is overloaded? I didn't quite understand...

author: Rob Sims
author_rank: Regular Contributor
timestamp: August 05, 2016, 05:42:09 pm
content: In case of a voltage regulator, the voltage never goes above a preset value. If the resistance of the load increases, the voltage on the clamps increases. At a preset voltage value, the voltage doesn't increase anymore on the clamps if the load resistance increases. It stays the same if the resistance of the load is still increasing. The current however will decrease. This is how a voltage regulator works.

author: ali6x944
author_rank: Frequent Contributor
timestamp: August 12, 2016, 09:52:33 pm
content: Thanks for clearing that