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Let's say that the load is connected to a 12V battery consumes 3A current. When the 12V battery is replaced with 24V battery causes 6A current flow for the same load.

I'm just trying to understand if the load consumes only the required amps from the source battery regardless of the voltage of the battery or does the change in voltage has direct impact on current flowing through the load

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  • \$\begingroup\$ 'the required amps' is a bit of a misinterpretation; if the load is supplied with a voltage that is above the rated value then it will draw too much current. The resistance of the load and the supply voltage determine the current, and this is governed by Ohm's law: \$I=\frac{V}{R}\$, so doubling the voltage, doubles the current. \$\endgroup\$
    – Chu
    Commented Oct 22, 2017 at 10:48
  • \$\begingroup\$ @Chu Does that means the electronic devices are made in consideration with the grid volt system (120/240 V etc). and if the devices made for 240V system is connected to 120V system then it will not be able to draw much current? \$\endgroup\$
    – VVK
    Commented Oct 22, 2017 at 10:55
  • \$\begingroup\$ Electronic devices are not connected directly to an AC power source - the AC is changed to DC at the required voltage, and with consideration of the current demands of the device. The circuit that does the conversion from AC to DC is designed to work with a given supply voltage, so it will probably not work as desired at a lower or higher voltage source. \$\endgroup\$
    – Chu
    Commented Oct 22, 2017 at 11:01
  • \$\begingroup\$ @Chu That means the electronic devices made to operate on both (120V/240V) systems has a conversion ckt which can handle both incoming voltages and outputs to a constant DC voltage. right? \$\endgroup\$
    – VVK
    Commented Oct 22, 2017 at 11:19
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    \$\begingroup\$ @Chu not all electronic components are resistors; those that are not resistors don’t obey the Ohm’s law. Moreover, PSUs are electronic devices that are connected directly to an AC power source. \$\endgroup\$
    – user2233709
    Commented Oct 22, 2017 at 11:34

2 Answers 2

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It depends what the load is.

If the load is a resistor with good stability, and adequate power handling for the higher supply, then yes, at twice the voltage it will draw twice the current.

For other loads, the situation is more complicated.

If a resistor has a positive temperature coefficient, like a filament lamp, then it will take less than twice the current at twice the voltage, as it heats up and increases resistance.

If its power handling is only good for the lower supply, it may take twice the current briefly, then burn out and drop to zero.

More complicated loads, like a universal input voltage power brick, may take constant power when charging the laptop battery, so take twice the current at 120v that it takes at 240v.

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    \$\begingroup\$ take half the current at 120v that it takes at 240v? Do you mean takes double current at 120v that it takes at 240v \$\endgroup\$
    – VVK
    Commented Oct 22, 2017 at 12:31
  • \$\begingroup\$ @VVK doh! Good catch, I'm glad someone's awake. Fixed. \$\endgroup\$
    – Neil_UK
    Commented Oct 22, 2017 at 12:43
  • \$\begingroup\$ If the load is a resistor with good stability, and adequate power handling for the higher supply, then yes, at twice the voltage it will draw twice the current assuming the ideal supply. \$\endgroup\$
    – 0___________
    Commented Oct 22, 2017 at 12:47
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    \$\begingroup\$ @PeterJ_01 No. Such a resistor fed with twice the voltage will draw twice the current. The quality of any power supply involved is irrelevant. It's outside the system under consideration, which is the resistor, and the voltage on its terminals. It could be an AC supply, and we're talking about instantaneous voltage and current. If I'd wanted the supply to be part of the system, I would have specified it, which would have cluttered the answer with irrelevancies. \$\endgroup\$
    – Neil_UK
    Commented Oct 22, 2017 at 13:09
  • \$\begingroup\$ @Neil_UK if it is the theoretical situation, the question should be closed as it is the Ohm equation, and it is a primary school problem \$\endgroup\$
    – 0___________
    Commented Oct 22, 2017 at 16:10
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Well really all electrical devices are hungry for the watts as watts is the Working power regardless of voltage and as a general rule if you double the voltage the amperage required to create the watts the device requires is halted why would an device draw double the amps when the voltage is doubled when only half the amps are needed to create the same wattage which is all the device needs to operate so long as it can handle the voltage put through it.

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    \$\begingroup\$ What if the device is a simple power resistor? \$\endgroup\$
    – Velvet
    Commented Sep 4 at 7:23

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