# Video: Physics Past Exam • 2017/2018 • Pack 1 • Question 38B

Physics Past Exam • 2017/2018 • Pack 1 • Question 38B

02:02

### Video Transcript

The figure below represents a p-n junction connected in series to a cell and a lamp. The second figure represents the same circuit, but the cell is connected with the opposite polarity. In which of the two circuits is the lamp lit up? Why?

Comparing figure one and figure two, we see they’re identical except, as the problem statement tells us, the polarity of the cell. In each circuit, we find a p-n junction, which is a way of controlling the direction that current can flow in a circuit.

A macroscopic view of each of these p-n junctions would show a net positive charge on the left-hand side and a net negative charge on the right-hand side. This means that, for each of our two figures, the junction tends to lead positive charges to the right and negative charges are drawn to the left.

Given that, lets’ consider how current flows in both figure one and figure two due to the power cell. In figure one, the direction of conventional current is for positive charges to effectively move counterclockwise. However, we see that this movement of positive charges is resisted by the p-n junction. The junction wants positive charges to move in the opposite direction, clockwise. Therefore, the junction resists the flow of current in figure one, and the lamp is not lit up.

In figure two, on the other hand, with its opposite cell polarity, positive charge wants to move in the clockwise direction. And looking at our p-n junction, we see that this direction is favored by the junction as well. The conventional current is therefore able to move through the junction and pass on through the light and light up the lamp.

When we consider these two p-n junctions with respect to the directions of the power cells, we say that junction in figure one is reverse-biased, that is, it opposes the flow of current, while the junction in figure two is forward-biased and therefore it works with the flow of current as generated by the power cell. As our answer then, we say that figure two is the one with the lamp lit. And the reason is that the circuit in figure two is forward-biased.