Day 14 Current Voltage and Resistance

At the beginning of the class, we first watched a video about lighting a light bulb

Then, we did an activity called arrangement that cause light and made 3 arrangements with light bulb, battery, and a wire that cause light, and 2 that do not cause light.

We also used two batteries to make the bulb lit twice brighter.

In this picture, the professor is using a rubber bar to rub by animal skin and put the bar on the ball, we discovered that two pieces of sheet-metal were separate.

In this picture, we used the battery to touch the ball, but nothing happened.

 We made a simple electric circuit and defined that battery give energy, energy is getting used up, and we need a wire because we want the energy to go back from the bulb to battery. And the reason why the bulb would light is because there is a circuit between the battery and the bulb that have the equilibrium of charge.

We also discussed the concept of electric potential difference of voltage. 

In this picture, we used the equation Power = V*I to calculate power to be 0.18 J/S.

In the pictures, it is the model C for current flow which there will be less current in the return wire than the coming wire.

We did an activity called measuring current and we measured that the value of the current in amps to be 132 mA in positive direction and 130 mA in the opposite direction

In the picture, we did a problem and calculated vd (drift velocity and current) by the equation

 I = ρnqvdA. 

Next, Ohm's law was introduced to us and we drew a I vs. V graph for V = RI where R is ohm. And this equation is only true when we have a good wire.

In these pictures, the professor used logger pro to graph I vs. V graph.

This is the I vs. V graph. It is a linear line.

In this picture, we found that unit of R which is J*S/C^2 which is also called ohm (Ω).

Then, we did an example problem about resistivity. And we found that the resistivity is proportional to the length and the less resistance the more electric conductive.

Finally, in the picture, we found that R is inversely proportional to cross section area of the wire, and we found that R = P*L/A 

Conclusion

In today's class, we first discussed about how to light a light bulb in various ways with batteries and wires. Then we talked about the relationship between voltage, current, and power. And finally we were introduced the ohm's law and have some applications and problems on ohm's law.