# Given Ohm's law, how can current increase if voltage increases, given fixed resistance?

According to Ohm's law, $V=IR$ (voltage equals current times resistance).

So if the voltage increases, then the current increases provided that the resistance remains constant.

I know that Voltage or potential difference means work done per unit positive charge in bringing that charge from one point to another.

So according to Ohm's law, if the work done per unit charge increases then current will increase. How can this be true? Point out my mistakes.

## 3 Answers

I think you've answered that yourself. If you are putting more work into moving unit of charge, then that unit of charge is going to move faster (all else being constant). Current is the flow electric charge across a surface at specific rate (1 ampere = 1 coulomb per second) and hence - more voltage, more work, faster flow (rate), higher current.

how can this be true?

If we connect 1.5V cell to a 10 ohm resistor, the current is, by Ohm's law, 0.15A and the power delivered to the resistor by the cell is 0.225W.

Now, connect a 9V battery in place of the 1.5V cell. The current is now, by Ohm's law, 0.9A and the power delivered to the resistor is now 8.1W.

The 9V battery must deliver far more power to the resistor than the 1.5V cell does.

There's nothing mysterious here. Why do you think the current should not increase if the voltage increases?

It is better to think of Ohm's Law as I=V/R.

What it is telling you is that if you apply a voltage (V) to a resistive material (characterised by R), then that voltage is capable of driving a current I. The material could be anything, a piece of copper, or the plasma in a star.

The voltage is constantly supplying energy to the electrons in the material, but the resistivity is constantly taking that energy back out (converting it to thermal energy). The higher the voltage, the more energy you can give to the electrons and hence the higher the current. On the other hand, the higher the resistance, the more energy is taken away from the electron flow and hence the lower the current.

## protected by AccidentalFourierTransformJul 29 '18 at 17:40

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?