Day 16 RC Circuit

Quantitative Measurement on RC System: 

For this experiment, we were asked to create RC circuit with a resistor with resistor, power supply, and capacitor. We used logger pro to measure the voltage, and before we also make some prediction that the potential voltage and time should be inversely related. As the result below, we were right.

On discharge(pink) one, we know that B in the natural exponential equation should be vary close to zero. However, the Amp should be close to 4.5 V (original V from power supply). For conclusion, C as our charge time should be the same for both discharge and charge.

To be able to find out the C value which is 0.003620 sec by theoretically method. We first used multimeter to determine the resistance of our resistor. (R = 2.15 k"omega") 

We next used their relationship that Time =1/Resistance*Capacitance to solve the value of charge time. 

As the result, t=0.00465 sec, it mean there were about 22% error. This error might come from the capacitors value straight off the labeling. Below is our calculation. 

Day 15 Capacitors

Self Made Capacitor:

For experiment we place to aluminum foil between physic book's pages. The foils were connected to the multimeter to be able to determine the capacitance. As area remain the same the thickness of pages act like distance, which cause the change of capacitance. We have tested 1, 10, and 20 pages and notice smaller distance cause bigger capacitance.(Graph below)

Capacitors in Series & Parallel:

For this experiment we given two resistor, we measured each capacitance which come up two solutions. 1st the parallel resistor come up the solution equal to their sum. But for series the capacitance is equal to Csum = 1/C1 +1/C2.

Day 14 Parallel and Series Circuits 04/16/2014

Voltage:

In this lab we were ask to find the voltage and current by ammeters. As the result, the initial voltage and final voltage are the same, which means the voltage won't loss during the circuits. However, the current stay constant at any position on the circuit. 

Parallel Circuits:

Find the voltage and current under the parallel circuits condition. As result, the voltage remain constant, and the initial current is equal to the final current. 

Decoding and Measuring Resistors:

Decoding the resistors by the key provided, and calculate the percent discrepancy by subtract the difference between coded and measure then divide by the coded one. As the result, the gold one is much accurate which is not suppose to be.

The Equivalent Resistance for a Network

Key:

                     Resistance of Series = sum of all resistance

                     Resistance of Parallel = 1/R1+1/R2+1/R3+............. then flip it up side down 

Loop Rule

Follow the step below:

• Draw Circuit
• Assign current to each element
• Apply loop rule to each loop
• Mathematics

About the loop rule, when current direction is same as loop direction then it is negative.

About the voltage, from (-) -> (+) is positive, but from (+) -> (-) is negative.

 

Day 13 Electric Potential 04/14/2014

Electric Potential Lab:

This is a lab mainly focus on the potential difference between two points, and as we know the changing of voltage will give us work as long as we know the q. On this lab, power supply is 15 V, and the potential difference between two aluminum paints are 15.54 V. However, random two points on same aluminum paint is always zero, so it means there is no work require to move on this aluminum conductor. 

These is the potential difference on each displacement, and also their increment ratio;however, we increase distance 1 cm each time.

Next

Calculated the work to move a given charge:

a) x = 0cm to x = 3cm

b) x = 4cm to x = 6cm

c) x = 5cm to x = 2cm

Result

First one q = 1 so it's differential potential energy are there work , then rest can be found by the picture above. 

This is potential vs position graph, and it shows when distance increase the potential energy increase. I think it is pretty similar to work, which longer distance given bigger energy. 

Day 12 Electric Potential 04/09/2014

Quiz:

100% Correct ~ LOL

Change Water Temperature by Electricity

In this lab, we heat up the water by coiled wired heater, which is about 42 cm. It attached to a 4.5 V battery, and place in the water for 10 minutes. In this lab, we were asked to find the temperature change with given wired heater length, time, and voltage.

First we calculate the resistance (R) of wired heater ( Power*Length / Area )then use the resistance to find the current (I ) (voltage / Resistance). Also Put in the uncertainty by differentiate two different power.

After solving current (I), we solver power by (P = I * V) With solved power we can solve delta T by Q = mc(delta T; however, power * t = Q. Use the value of uncertainty to plug in exact same formula that we were use will gives us the final uncertainty.

========================================================================

With double voltage I = 2V/R => P = 2V/R * 2V = 4V^2/R (which is aquatically)
changing T = (p*t) / (m*c) = 4V^2t / (Rmc).
========================================================================

This picture shows the temperature change from logger pro. The blue one is the 9 V and red is  4.5V

According to logger pro for the initial experiment the temperature started at about 23°C and concluded at about 25°C for a 2°C temperature change. The doubled voltage experiment started at 24.6°C and ended at 32.1°C for a 7.5°C temperature change.

Day 11 Current & Resistance & Voltage 04/07/2014

Light Bulbs & Batteries Experiment:

One Battery

Two Batteries

This is the experiment asked us to find a way to light up the bulb. The first one bulb light up with one battery.

But we want it to be twice lighter, so we add another battery to make it twice lighter. According to the picture above we see 2nd light bulb is twice lighter than first one.

Current:

For this experiment we were ask to choose a resistor to connect to ammeter and power supply. Since we know the voltage by power supply we can measure the resistant inside the resistor which determined by power meter. For each trail we increase the voltage by power supply. Below is our data and other group's.

This data shows that #1's resistant is much larger than #2 . The Voltage on both group are almost similar, but the current on each group has lot difference. 

Here comes the graph

According to the graph above the voltage and current's relationship is linear and proportional. However, the slope of the line is the resistant of resistor by V = RI, so #1 resistor has bigger resistant than #2 for sure.

The reason why #1 has greater resistant is because #1 is a longer resistor which electron need more time to pass through.

Copper & Silver material resistant test:

This is Lab we has given couple coil wrapped around silver and copper with same diameter except one. However there are only one copper then rest are nickle&silver. The goal is to determine the resistant of each coil.

We using the power meter to determine the resistance, and we got 1.9 as our unaffected resistant. 

These data prove that when length get larger the resistance also get larger.

Here are the graph:

The graph again shows the proportional relationship between length and resistance which length increase the resistance also increase.

However the diameter is total different which diameter increase the resistant decrease due to the larger cross sectional area which is inverse relationship between resistance and diameter. 

03/26/2014

Gauss Law

 Here comes the solution 1~4:

#5

 Question 5: Electric Field Inside a Charged Solid Sphere

Show that the electric field INSIDE the solid uniformly charged sphere varies as:

E = [k Q/(R_{charged sphere})²](R/R_{charged sphere})

where R_{charged sphere} is the radius of the charged sphere (10 mm in the simulation), Q is the total charge on the sphere (adjustable with the Q slider), and R is the distance from the center of the sphere at the position where the electric field is being calculated.

E = kQ/r^2

   = k (charge density)V/r^2

    =k (Q/r^2) (charge density)/r^2

     =kQV/VR*r^2

     = kQ/r^2 * r^2/R^3

      =kQ/R^2*r/R