Electricity, current, and resistivity (12th day).

Spring 2015

Professor Mason

April 9th class

Light Bulb and Energy

First thing in class, we needed to make a sketch of the set up between battery, light bulb, and wire. We first needed to sketch two ways of how the light bulb would light by using the battery and wire. Then, we also needed to make two sketches of how the bulb would not light up and also the reasons. It would light up because the electricity is going through positive and negative charges through the wire. It would not light up because there is no flow going through the positive and negative charges. We could also double the light by putting two batteries together. We also did an experiment with electroscope machine, and we were asked to predict what would happen if we put two batteries on the tip of that machine, and the result was nothing happened.

We learned that the fundamental definition of energy is the capacity to do work using energy. For example, the battery is giving energy and the bulb is using it up.

Another experiment with battery, wire, and light bulb is making a circuit, so the bulb would light up. We need the second wire for this experiment to go back from the bulb to have positive and negative charges working together in order for the force to go in one direction. We need it to have equilibrium of the charges. 

Voltage, Current, and Power

In class, we had an imaginary experiment dealing with waterfall and hydro electric generator. We need the flow of water in order to run the generator; however, there are two things that we should know about the waterfall, and those are the potential energy and the flow rate. We need a faster and heavier waterfall, which we could get from rising the height of the water fall; then, we need to have a bigger rate of flow of the waterfall in order to run the generator. We come up with an equation of voltage, flow rate (current) and power [V = J/C] or energy per unit charge, [I = C/s] or Ampere = Charges per second, and [P = J/s] or energy per unit time. The relationship between Voltage, Flow Rate, and Power is [P = I V] or Voltage times Current, which will become J/s in the end. Then we need to do a calculation to find the P, which I and V are already given. For example, the I is 12 mA (mili Ampere) and V is 1.5V (Volt), so that P is [0.12 * 1.5 = 0.18 J/s]. 

Next, in class, we were given a question, which asked us to pick the correct one from model A to model D. We chose model D as in the lab manual showed us because the direction of the current will be shown and it will be the same in both wires, which means the current going in and out are the same. 

Current in the wire [I = dq/dt] (C/s) = Ampere. 

We did an experiment in class, which required us to deal with ampere meter. The picture on the right is how the tools are set up. We set up the wire from the negative charge of the battery to the positive charge of the ampere meter; then, we put two wires from the battery (+ and -) and attach them to the light bulb circuit. After that, we attach one of the circuit to the ampere meter.  

Drift Velocity and Current and Ohm's Law
Today, we found out the equation of Current with the function of drift velocity is [I = RHOn q Vd A], which RHOn is the density (charges per unit volume), q is the charge of electron, Vd is the drift velocity, and A is the cross-sectional area perpendicular to the flow (the thicker the wire, the easier the flow is). In class, we also needed to draw graph  I vs V and V vs I; it looks like the picture attached. The equation of the slope is [m = dV/dI] or the derivative of V/I.

Resistance

We also learned resistance in class, which we come up with an equation of [V = I R], which R is the resistivity. The longer the wire, the bigger the resistance is. It also depends on the Area of the wire. For example, wire 1's length is 200cm and has a diameter of 0.25cm; wire 2 has a length of 200cm as well and has a diameter of 0.32cm. The smaller the area is, the bigger the resistance is; therefore, in this case, the wire with the diameter of 0.25cm has a bigger resistance, whilst R is proportional to Length. 

Resistance is proportional to the material as well (copper is good conductor, so the bigger the conductivity is, the smaller the resistance is). It also depends on the length and the area, which leads to an equation of [R = Rho L/A], which Rho is the resistivity, L is the length, and A is the area of the wire. 

Resistance as a function of temperature has a formula of {R = Ro[1+alpha(T-To)]}, which alpha is the temperature coefficient, and it depends on the type of metal. The colder the temperature is, the easier for the electron to go through; as the temperature increases, the resistance increases because there is more collision happening in the wire, thus resistivity is proportional to temperature.