Just a few questions relating to graphs

[ianb]

I'm not following the format because this isn't really a homework question -- I'm just studying for my finals and encountered a few graphs in the electricity unit, and want a few things cleared up.

So in a current-voltage graph, the slope is 1/R, and power (according to my teacher) is the area under the line. I only have a problem with the latter claim-- say we have a point on a linear line whose coordinate is (20,4). Should P = IV = 20 x 4 = 80 rather than P = Area = 20 x 4 x 0.5 = 40?

 

[Dick]

Yes, power should be area of the rectangle rather than area of the triangle.

 

[m2003]

Hello,

Thank you for your question. It's great to see that you are studying for your finals and taking the time to understand graphs in the electricity unit.

To start, let's clarify the meaning of slope and power in a current-voltage graph. The slope of a current-voltage graph is indeed equal to 1/R, where R is the resistance in the circuit. This is because Ohm's law states that V = IR, where V is voltage, I is current, and R is resistance. Therefore, the slope of the graph, which is the change in voltage over the change in current, is equal to R.

Now, let's address the issue with power. Your teacher is correct in saying that power is equal to the area under the line on a current-voltage graph. However, this only applies to non-linear lines. In the case of a linear line, such as the one you mentioned with coordinates (20,4), the power can be calculated using the formula P = IV, where I is the current and V is the voltage at that point. So in this case, P = 20 x 4 = 80, as you correctly stated.

However, if the line is non-linear, the power cannot be calculated using the simple formula P = IV. This is because the current and voltage values are constantly changing along the line, and the formula only applies to a single point on the graph. In this case, we use the area under the line to calculate the power, as it takes into account all the changing values of current and voltage along the line. This is why your teacher mentioned that power is equal to the area under the line in a current-voltage graph.

I hope this clears up any confusion you had about current-voltage graphs and power calculations. Good luck on your finals!

Best,