voltage and current - Page 1
Forum section: electronics / beginners

author: parsa
author_rank: Newbie
timestamp: September 21, 2021, 07:31:52 am
content: doesn't current flow in the circuit with the power of voltage? so why the current goes up when the voltage goes down?!!!

author: ataradov
author_rank: Super Contributor
timestamp: September 21, 2021, 07:42:24 am
content: You probably need to provide more information on the specific case you are looking at. In the simplest case I = U / R. For the same R, if U goes down, then the I goes down as well.

author: parsa
author_rank: Newbie
timestamp: September 21, 2021, 07:48:04 am
content: believe me I know that,but some people say when the voltage is high,the current is low!!why??is that because of the resistance which when the voltage goes high,it goes high as well and doesn't let most of the current to flow?

author: Circlotron
author_rank: Super Contributor
timestamp: September 21, 2021, 07:56:15 am
content: For a given amount of watts, the higher the voltage the lower the current will be.120V x 10A = 1200W240V x 5A = 1200WBut that (generally) doesnotmean that if you raise the voltage it will automatically reduce the current. For a resistive load like a room heater for example, if you raise the voltage by 10% the current will also go up by 10%. And the wattage will go up by 1.1 x 1.1 = 1.21 = 21% increase.----------But just to be annoying, a switchmode power supply, if you raise the mains voltage the current it draws will decrease. And if you lower the supply voltage the input current will increase. The volts x amps stays approximately the same.

author: Zero999
author_rank: Super Contributor
timestamp: September 21, 2021, 07:59:16 am
content: In a resistive load, current is proportional to the voltage. I = V/R, so the current goes down, when the voltage goes down.I think you're referring to a switched mode power supply, which is a constant power load. If the load is kept constant, then yes the current does go up, when the voltage goes down. The power supply alters its input resistance, in opposition to Ohm's law, i.e. the input current goes up, as the voltage falls.A switched mode power supply is an unusual case. Most loads are resistive, or costant current. Going back to a resistive load, because the current increases, with the voltage, the power dissipation has a square relationship with the voltage.P = VII = V/RIf V is doubled, then I is also doubled, so the power is quadrupled.V = 6VR = 3RI = V/R = 6/3 = 2AP = VI = 6*2 = 12WDouble VV = 6*2 = 12VI = V/R = 12/3 = 4AP = 12*4 = 48WA quicker way to calculate the power is shown below by combining the formulae P = VI and I = V/RP = V*V/R = V2/R

author: parsa
author_rank: Newbie
timestamp: September 21, 2021, 02:49:39 pm
content: thank you so much

author: BeBuLamar
author_rank: Super Contributor
timestamp: September 21, 2021, 03:40:05 pm
content: Lots of people make the same kind of thinking. OK for example 2 heaters both are 1000W. The one that is designed to work on 120V would draw twice as much current as the one designed to work on 240V. But that because the 120V version has half the resistance.If you use the same heater and you increase the voltage then the current would go up.

author: fordem
author_rank: Regular Contributor
timestamp: September 23, 2021, 03:35:53 pm
content: Quote from: parsa on September 21, 2021, 07:31:52 amAs other respondents have said, you need to provide more details about the particular circuit or application that you are asking about.With motorized loads, as the voltage falls, more current is drawn from the circuit in order to maintain the required power.https://www.ecmweb.com/design/article/20901278/the-highs-and-lows-of-motor-voltage

author: timenutgoblin
author_rank: Regular Contributor
timestamp: September 23, 2021, 04:12:53 pm
content: Quote from: parsa on September 21, 2021, 07:31:52 amEEVBlog 1401https://youtu.be/fHoHYEZnVIgEEVBlog 1397https://youtu.be/AQK7RyecVW0Read up on MPPT (Maximum Power Point Tracking) and its applications using PV (Photo Voltaic) solar panels.

author: antenna
author_rank: Frequent Contributor
timestamp: September 23, 2021, 05:13:37 pm
content: Every power source has an internal resistance. Imagine an ideal supply set at 10v... if the source impedance or resistance is 0, infinite current can flow through your source/supply, and no matter what load you put on it, the voltage will stay at 10. This, however, is never the case. Every power source, battery, power supply, ham radio etc has a source impedance or resistance that can be modeled as some Ω or R±jX value in series with an ideal voltage source. Typically, this is a low resistance, and for sake of discussion, lets call it 1ohm. This 1Ω (inside the battery/source) will determine the current that can flow out of (and back to) said battery at whatever voltage the the internal chemistry/circuitry provides. Now, lets say you have a 10ohm load. That 10ohm load is now in series with that 1ohm source resistance and a voltage divider is set up. Now, for the same current, the voltage on the load is 10/11 of the ideal source voltage (even when measured at the battery terminals). When the load and source have the same resistance, you will see half the voltage on the load (which can make a battery very hot or destroy a power supply). And just like solar panels, peltier modules, etc, for a multitude of unique reasons, there is often a ratio of voltage to current that is most efficient for a given device which sometimes leads to the voltage and current not changing proportionally with a change in load.Edit: Do not put an ohm meter across a battery or supply to determine the source resistance, you will destroy it. The way you would do that for circuits designed to have matched impedances (signal generators, audio equipment, ham radio etc) is to vary the load while monitoring the voltage and seeing what resistance makes the voltage half of the unloaded value. This shouldn't be an issue for someone working with DC electronics though, and if you run into this issue, simply get a battery/supply rated for a bit more current than you intend to use at that voltage.

author: Circlotron
author_rank: Super Contributor
timestamp: September 23, 2021, 11:06:32 pm
content: Quote from: fordem on September 23, 2021, 03:35:53 pmThat article is talking about AC induction motors. With those, a reduction in supply voltage when the motor is fully loaded causes the motor to slow down. This causes the induced slip frequency in the rotor to increase beyond the optimum value so that the rotor reactance exceeds the resistance of the aluminium end rings. This means a lower power factor figure in the rotor induced current and consequently a lower power factor in the stator windings connected to the mains. As a result the motor pulls more amps as the voltage is reduced and the motor slows. So the current vs voltage situation can be a bit more complex that expected!

author: Rod
author_rank: Contributor
timestamp: September 24, 2021, 06:03:18 am
content: Quote from: BeBuLamar on September 21, 2021, 03:40:05 pmIt has 1/4 the resistance, not 1/2.

author: parsa
author_rank: Newbie
timestamp: September 24, 2021, 06:25:53 am
content: thanks man,that was exactly what I was looking for because I doubted the resistance.

author: tautech
author_rank: Super Contributor
timestamp: September 24, 2021, 09:50:26 am
content: An old favorite

author: Terry Bites
author_rank: Super Contributor
timestamp: September 24, 2021, 04:11:05 pm
content: It all depends entirely on your circuit.