# Conventional current problem

In my revision notes, there's this problem whereby a high potential difference is applied between the electrodes of a hydrogen discharge tube such that the gas is ionised. Electrons in the gas then move towards the positive electrode, while the positive ions move towards the negative electrode.

The solution given is then to calculate the current due to the movement of the positive ions and the current due to the movement of electrons and add them together. This follows the definition of current given in my notes:

Current is defined as the rate of flow of charge, Q or the amount of charge that is moving past a chosen point (along a wire) per second as measured by an ammeter. However, as a side-note, it also states that:

Based on convention, the current due to the movement of electrons can be treated to be in the same direction as the current due to the movement of the positive ions as the conventional current flows from a position of higher potential to a lower potential.

But, how can this be true when the two charged particles are moving in opposite directions? If the movement of the positive ions were to be treated to be the conventional direction and the movement of the negative ions were to be treated in the opposite, would I have to minus the two currents instead of adding them together. I'm kind of confused at this part.

Say the direction the positive ions are going is the positive direction. Then the positive ions contribute $\frac{N_+ q_+}{T}$ to the current if you see $N_+$ of them pass in the positive direction over the time $T$. The electrons are going the other way so they contribute $- \frac{N_- q_-}{T}$. So yes you do subtract when putting them together, but $q_- < 0$ and $q_+ >0$ so you subtract a negative quantity. That makes sure that even if $\mid {N_+ q_+} \mid = \mid{N_- q_-}\mid$, they don't cancel out.