There are a lot of questions asking what happens if you change the current, but keeping the same voltage. If more current is flowing to me it means you can damage the device. But what if it is the other way around? What happens if you increase the voltage, but have the same current?

Let's say I power on a device with an adapter that outputs 5 volts @ 2 amperes. If I would like to connect that same device to an adapter that outputs 20 volts @ 2 amperes, will I burn it? I think you burn a device by passing too much current, no? Could increasing the voltage also burn it if you are passing the same current?

  • 5
    \$\begingroup\$ i think that you do not fully understand the relationship between voltage and current. ... there is resistance involved .... google ohm's law \$\endgroup\$
    – jsotola
    Mar 4, 2018 at 7:31
  • \$\begingroup\$ the only way that you can keep the same voltage and also increase the current flow, is to reduce the device resistance ... if the device does not change, then you cannot force more current without raising the voltage \$\endgroup\$
    – jsotola
    Mar 4, 2018 at 7:33
  • \$\begingroup\$ I see... so the device is acting as a resistor? That is why if I increase the voltage the current will increase? \$\endgroup\$
    – Tono Nam
    Mar 4, 2018 at 7:36
  • \$\begingroup\$ if a device draws 2A at 5V and you connect it to a power adapter that outputs 20V, and that adapter is unable to supply more than 2A current, then the power adapter will most likely overload and its output will drop to 5V or so \$\endgroup\$
    – jsotola
    Mar 4, 2018 at 7:37
  • 2
    \$\begingroup\$ "There are a lot of questions asking what happens if you change the amps keeping the same voltage.". No, there are not a lot of questions asking that. \$\endgroup\$ Mar 4, 2018 at 7:49

3 Answers 3


The load determines the ratio of current to voltage at its terminals.

If you control the voltage you supply, it determines the current.

If you control the current you supply, it determines the voltage.

It's possible to break a load by supplying, or letting it define, too much of either.

Let's compare a 5 V, 2 A adapter, and a 20 V, 2 A adapter, driving different loads.

We'll drive a 100 ohm load, which has a thermal limit of 1 watt. With 5 V, it will draw 50 mA and dissipate 250 mW, and run happily. With 20 V, it will draw 200 mA and dissipate 4 watts, and eventually overheat.

Now let's drive a 1 ohm load, which has a thermal limit of 10 watts. With a 2 A supply, it will collapse the input voltage to 2 V, and dissipate 4 watts. It doesn't matter whether it's the 5 V supply, or the 20 V supply driving it; if they both put out a constant current of 2 A, then the load will drop their voltage to 2 V. Different adapters may behave differently to the current limit however; some will deliver a constant current, some will shutdown for a moment and try to restart, and repeat that cycle continuously, and some will deliver a lower current (so-called foldback limiting) to protect themselves.

Now let's connect the gate-source junction of a FET, that has a voltage limit of 15 V. On 5 V, it will draw essentially no current and survive. On 20 V, it will draw essentially no current, punch through and be destroyed.

  • \$\begingroup\$ But if both adapters output the same current why will the device get damage? I am still lost. I am just learning about ohm's law :/ \$\endgroup\$
    – Tono Nam
    Mar 4, 2018 at 7:51
  • \$\begingroup\$ added update to my answer \$\endgroup\$
    – Neil_UK
    Mar 4, 2018 at 12:33

The thing to keep in mind is not only voltage and current in understanding this; you must also calculate power.

Consider this:


simulate this circuit – Schematic created using CircuitLab

Notice all the circuit values are 1 in the circuit - (1 volt, 1 ohm, 1 amp, and 1 watt). There is no need for a calculator on this circuit since if you apply the value of 1 to any two of the variables in any of those Ohm's law formulas, the mathematical result will always be 1 again.

The power supply is supplying 1 volt @ 1 ampere and is therefore producing 1 watt of power. If the power supply is producing power, then that power mathematically must be being dissipated (in the form of heat) somewhere else in the circuit.

Since current-reading meters, or ammeters, have near-zero resistance, the ammeter is not consuming or dissipating any meaningful amount of power. How do we know this? Let's say the ammeter resistance inside of it is 0.01 ohms (which is reasonable). If the ammeter is passing/showing 1 ampere of current, then the power dissipation (P = I^2*R) = 1 (ampere) squared times 0.01 (ohms) = 0.01 watts. This is a minuscule amount of power dissipation and can safely be ignored in this case.

So, if the ammeter is not dissipating any power, who's left to dissipate the 1 watt of power that the power supply is producing? It must be the resistor. Since the resistor is dissipating that 1 watt of power, and since power is always dissipated in the form of heat, the resistor temperature increases in unison (linearly) with the power that it is having to dissipate.

Now, what happens if we change the Voltage (E) to 2 volts instead of 1 volt? The 1 ohm resistor will now have 2 volts across its leads. (It will be dropping 2 volts.)

Let's do the Ohm's law math now.


  • Circuit Voltage = 2 V
  • Circuit Resistance = 1 ohm (again, ignoring the small ammeter resistance)
  • Circuit Current (I) = E/R = 2 V divided by 1 ohm = 2 amperes

Calculations based on Ohm's law:

  • Power supply is producing: P = I*E = 2 volts * 2 amperes = 4 watts
  • Resistor is dissipating: P = E^2/R = 2 V squared divided by 1 ohm = 4 watts

So as can be seen, if the load (device) resistance remains constant, then an increase in input voltage will cause the circuit power to increase quite a bit. For every doubling of input voltage, the circuit power increases by a factor of four. And remember, the circuit power produced by the power supply mathematically must be being dissipated by the load or device connected to that power supply. (They are equal at all times.)

In your question, you asked what if a 5 V, 2 A adapter powering a device were replaced with a 20 V, 2 A adapter.

Let's assume that the device consumes all of the power given to it from the initial adapter (5ampereV, 2ampereA):

  • The resistance of the device then must be: R = E/I = 5 V/2 A = 2.5 ohms
  • The power dissipated by the device must be: P = I*E = 5 V*2 A = 10 watts

Now you replace the first 5 V, 2 A adapter with a 20 V, 2 A adapter:

  • Assume the resistance of the device remains the same (2.5 ohms) since no changes were made to it.
  • The power supply voltage now changes from 5 V to 20 V, which means that the device must now dissipate 20 V squared divided by 2.5 ohms = 400/2.5 = 160 watts!

Fortunately, your new adapter can only supply 20 V * 2 A = 40 W of power.

The voltage on the 20V adapter will likely drop until it meets its maximum power output while still trying to maintain 2 A of output current - it's still going to try to deliver 40 W of power which means that one way or the other (either by over-voltage or over-current or both), you're still damaging your poor device which is only designed to handle 10 W.

Power is the meaningful calculation in many cases such as this one. Whether you're dealing with a 20 V, 2 A or a 2 V, 20 A power supply, either way the math says that the maximum power dissipation will be 40 W. That's why they are called power supplies, since any combination of output voltage and current can never exceed the P = I*E law.

Note: All of the above assumes that your device (load) is constant, like a resistor (or resistive load) would be.

Things change when applying too much or too little input voltage to electronic devices, as many times they do not represent a resistive load. They are nonetheless susceptible to damage should the input voltage rise high enough to damage the internal semi-conductors (transistors, etc.) as well as passive components (capacitors, etc.)

  • 3
    \$\begingroup\$ Why do you use E as symbol for Voltage? U or V are common. \$\endgroup\$
    – Ariser
    Mar 4, 2018 at 11:13

For a resistor-like load increasing the voltage will increase the current. This is what Ohm discovered and is neatly given by \$ V = IR \$. For a fixed R, V and I are proportional.

When you connect a device to an adequate power-supply the current will be determined by the device, not by the supply. The Irish national power grid has a peak capacity of 5,000,000 kW (5 GW). If I switch on a 30 W lamp it will only draw that much power from the grid, not the whole 5 GW.

Not all devices are resistive loads, however. Many will vary depending on what the device is doing. e.g., Laptop on standby, phone display off, on, watching video, making a call, etc. The current drawn changes.

Lets say I power on a device with an adapter that outputs 5 volts @ 2 amps. If I would to connect that same device to an adapter that outputs 20 volts @ 2 amps will I burn it?

Electronic devices usually have a voltage tolerance. Exceeding these usually destroys the devices.

I think you burn a device by passing to much current no?

That can do it but the high voltage on its own, without much current, can do it too.

Increasing the voltage could also burn it if you are passing the same current?

In general increasing the voltage will increase the current. Some devices have built-in regulators - e.g., your mobile phone's internal battery charger - and will try to control the current. If you exceed the maximum rated voltage though you will destroy the controller.


Not the answer you're looking for? Browse other questions tagged or ask your own question.