Current vs Voltage in Inductance

We started the day comparing graph of voltage vs time and current vs time. When a graph of voltage vs time is sinusoidal starting from the origin, the graph of current vs time is 90* shifted to the left with amplitude of difference of c*omega. Instantaneous voltage can be calculated using the equation Vmax*sin(omega*t+flux)+initial velocity. We also learned from before that omega can be replaced by 2pi*frequency or 2pi/period. This equation is also related to instantaneous current, which is Imax*sin(omega*t+flux)-initial current. When graph of voltage vs time is squared, the graph changes in amplitude to be doubled.

We then learn about root mean square of current and voltage. It is the average of current or voltage in an AC circuit. We can calculate these by using max current divided by (sqrt)2. Same thing for average voltage. We are then given a question as below in pink. Calculate the max voltage when average voltage is 120. By multiplying 120 and (sqrt)2, we found max voltage to be 169.7 V.

Next, we are given a board with  4 different elements. We only need to use the capacitance for today's experiment. By using the logger pro sensor, we want to see the graph of voltage vs time and current vs time in logger pro.

The graph of current vs time is shown on the top in red color, and graph of potential(voltage) vs time is shown under it in green color. We can see the distinction between the two graphs in amplitudes.

The function generator is set at 10 Hz and 3V as below:

Using the equation of root mean square from earlier, we found the average current to be 0.313A by dividing the max current in the logger pro to (sqrt)2, and we found the average voltage to be 2.07V. 

Next, we are asked of the difference between a voltage vs time graph and current vs time graph. As we can see in the drawing below, the graph for current vs time is about 90* shifted to the left. The amplitude of these two graphs are different by capacitance*omega. We also learned the relationship between voltage and impedence is voltage equals to current*impedence, where resistor is in AC circuit. We are also given different euqations. First is the reactivity of capacitance, which is called khi capacitance. It is 1/capacitance*omega. We also knew from past lecture that omega is also 2pi*f; thus, we can substitute it in and find khi capacitance to be 1/2pi*fC. We also learned that khi Capacitance is average voltage per average capacitance. Lastly, we learned that instantaneous voltage is omega*Capacitance*max voltage.

We are then given an example with known as in red below. A capacitor with 0.02 microFarad, with a voltage of 50V giving a power to it. We also have frequency of 10 kHz. We are asked to find the current of it. First, we want to calculate the khi using 1/2pi*fC. After we find the khi, we can use it to calculate the current by dividing the voltage by the khi. 

Using the multimeter, we also found the root mean square of current and voltage to be 2.04 V, which is close to what we calculated from the logger pro (1.45% error), but the current is a little off with percent error as 36.4%.

Ja

Using the logger pro, we found the max voltage to be 1.9 and max current to be 0.13A. We can calculate the average voltage and current and find them to be 1.588 and 0.0757 respectively. By comparing it to the theoretical value, we can calculate the percent error and find it to be 0.199 and 17.63% respectively. We then learned finding Voltage from max voltage earlier. Now, we can derive it to find current, which is max voltage (-cos(omega*t+angle)) divided omega*inductors. Using these, we can find the khi capacitance from our data. First, we want to find khi capacitance. We know that we were using the frequency of 100 Hz and the capacitance was 100 microFarad. We can plug these known to the equation and found the khi capacitance to be 15.92. Then using this result, we can find the average current. We will multiply the average voltage by the khi capacitance and find the average current to be 22.1A We also learned that khi inductance is omega*inductors. Next, we learned about phase difference. It is change of time per period. Given a known of change of time as 0.24s and period as 0.01s, we are asked to find the phase difference. By using the formula, we found the phase difference to be 0.24. 

Using the inductor that we used from the past 2 lectures, we want to calculate the amount of the inductor using the equations that we just learned. Using the multimeter, we found the average current and voltage to be 0.052A and 1.192V respectively. Then, we calculate the khi capacity by dividing the current by voltage and we found it to be 23.4. Khi capacitance is equals to the khi inductance at max current, so we set this number equal to omega*inductance. We know the omega to be 2pi*f and the frequency is 100Hz. When we divide these, we found the inductance to be 0.0372H. 

Lastly we learned about the rest of the understanding of inductance. First, we concluded that on phase difference, inductance circuit will shove the graph backwards, where as the capacitance circuit will show it forward. Lastly, we learned about impedance. Impedance works the same as resistance. When there is resistance, the impedance is equal to impedance. The voltage and current are the same, so there is zero phase difference. Then in inductance, current is perpendicular to voltage. The phase difference is 90*, and the impedance is ewual to the khi inductance. In capacitance, the voltage and current is facing the opposite way, causin the phase difference to be 180*. The impedance is equal to the khi capacitance.

Relationship of Inductance, Resistance, impedance, and khis

We started the day deeper understanding about inductance. Recall that khi capacitance is 1/omega*capacitance, where as khi inductance is omega*inductance. We can calculate the .
Given the known as below, 10 ohm resistor, 100 microFarad capacitor, 2 frequencies at 10 Hz and 1000Hz, calculate impedance and Khi Capacitance for each frequency. Using the formula that we knew from past lecture, we first find the omega, which is 2pi*f. For 10 Hz frequency, we find the omega to be 62.83. Then we calculate the Khi Capacitance, using the formula earlier. After finding the khi, we find the impedance using the formula that we just learned with the Khi inductor as zero because inductor does not take place in this equation minus the khi capacitance that we just found. We find the answer of the impedance to be 159.15 ohm for 10 Hz frequency and 10.10 ohm for 1000 Hz frequency using the same method.
Next, we want to find the max voltage and current using the logger pro. We read the max voltage to be 1.009 V and max current to be 0.0127 ohm for 10 Hz frequency and 0.999 V, 0.0182 ohm respectively for 1000 Hz. With these values, we can calculate the root mean square. For the 10 Hz, we found the average voltage to be 0.7135 V and the average current to be 0.00898 ohm and for the 1000 Hz, we found the average voltage to be 0.7064 V and the average current to be 0.01287 ohm. These are the theoretical value. To find the experimental value, we used the multimeter. For the 10 Hz, we found the average voltage to be 0.726 V and the average current to be 0.00561 ohm and for the 1000 Hz, we found the average voltage to be 0.689 V and the average current to be 0.0062 ohm. We then calculated the impedance from these values by dividing the average voltage by the average current. For 10 Hz, we found impedance to be 129, and for 1000 Hz, we found impedance to be 110.

The set up using the 10 Hz frequency is as below. The inductors, resistors, power, and current are all in series and connected to the sensor.

When we run the logger pro, the graph comes out as below. The sinusoidal graph in red is current vs time, and in blue is potential (voltage) vs time. The amplitudes between the two graph are different, but they start at about the same origin.

For 1000 Hz, the graph of current vs time is as below.

The graph of potential(voltage) vs time is as below.

We then calculate the phase difference for 10 Hz and 100 Hz. Using the formula that we learned from the past lecture, which is change of time/period. Period is also 1/f; thus, we can use change in time*frequency.

We learn that when capacitor increases, the resistance decreases and the relationships between these two elements make the angle between them. And we also learn that resistance is at max when the capacitance is at zero.

To find the angle, first, we want to find te khi capacitor. We know that khi capacitor is 1/omega*c, but ee know that omega is just 2pi*f and the cspacitor is 100 microFarad. From this, we found the 10 Hz khi Capacitor to be 159.15 -!; 15.9 for 100 Hz. Then, tan of an angle is khi inductors-khi capacitors/resistance. Thus, we can use the khi that we found and 10 ohm resistor. And we found the angle for 10 Hz and 100 Hz to be 86.4* and 57.87* respectively.

We then went through some equations and understanding about resonance frequency. We first learned that current is at maximum when khi capacitance is equal to khi inductance. The peak graph would be as below in red. Then we learn about resonance frequency, which is 1/2pi(sqrt)LC.
After learning these, we are then given a practice question. A resistor, inductor, capacitor, and battery is set in series in a closed circuit. We are firstly asked to draw the diagram as below in green. Then, we are given the value for each of them as written on the diagram. We are asked to find the resonance frequency of the circuit. First, we convert the values to SI. For the inductance, we found it to be 2*10^-4 Henry, and for the capacitance, we found it to be 5*10^-6 Farad. Then, we can plug in the values that we found to find the frequency to be 5.03 kHz. Next, we were asked to find current when frequency is 3 kHz. We first calculate the khi capacitance using the equation 1/omega*capacitance, but omega is also 2pi*f. Using the frequency, we found the khi capacitance to be 10.6. Then calculate the khi inductance, using multiplication of omega and inductance and again, we use frequency instead of omega and found the khi to be 3.77. Then we can find the impedence. With resistance as 10 ohm, and using the khi inductance and capacitance that we just found, we can subtract them, and found the impedence to be 12.1. Lastly, we can find the current using average voltage divided by the impedence amd we found the current to be at 2.06A. Last question, we were asked to find the disappated power at 3kHz frequency. Since we know that the resistor is the one that dissipates power. Therefore, we can calculate using the resistance. Power is current(squared) multiplied by resistance. Using the current that we just found, we can find the power dissipated to be 42.4 W.

Lastly, we are doing an experiment to see the experimental and theoretical value for the resonance frequency when we connect the inductor that we used the day before, largest capacitance on the board, which is the 470 microFarad, and a 10 ohm resistor that is also on the board in series as below.

Using the multimeter, we can watch the current. We then tweak the frequency on function generator until we find the range of the current when it's at its peak. The value obtained is the frequency, which we found to be about 0.80Hz.
Then we went through a little bit more of lecture about Power factor. We learned that we can find average power using 1/2 impedence*max current(square) at a cosine angle. That cosine angle is the power factor. We also took notes that when inductance is equal to capacitance, current is at maximum and there is no phase difference between them.

Inductors on Oscilloscope

We started the day getting to know the physical form of inductor. There are two metal tips on the corner of the inductor to be connected to the power supply, so the current can flow through. We learned from past lecture that inductor stores the current, just like capacitor stores voltage. We also learned that inductor is flux per current. Today, we predict the inductor vs time graph. We know that voltage vs time of inductor is a 1/ln graph. So we found that inductor vs time is an exponential graph.

We first are given a question, with known as below. Number of coils is 110, length of 0.05m, radius of 5.12 * 10^-4m. First, we calculate the area of the 18 gauge copper coil using the area of circle formula. We know the density of copper to be 1.72 * 10^-8. By plugging in the numbers for density multiplied by length per area, we can find the resistance to be 0.3 ohm. We then calculate the inductance using formula of inductance from past lecture to get the number of coils, which is inductance = myu*N(square)*A/l. For this, we use are of the inductor, which is a square and we also use the magnetic permeability of free space, which we know to be 4pi*10^-7. Plugging in the knowns, we found the inductance to be 760*10^-6 H. We then can calculate the tao by inductance divided by resistance, which we found to be 5.07*10^-6.

Lastly, we can find the period by multiplying the tao by 5.

We are then given a 880 coils inductor. By connecting it to the oscilloscope and the power generator using the alligator cables, we firstly set the function generator to a square wave form. Then we adjust position, and vertical and horizontal scales. Lastly, we adjust frequency of the function generator. We find the voltage vs time graph to be as below:

From the graph, we obtained the half- time to be 0.4 microsecond. We then calculate the tao by half-time divided by ln 2. Then we can also calculate the inductance by multiplying the resistance and tao. Lastly, we can use the formula of inductance from past lecture to get the number of coils, which is inductance = myu*N(square)*A/l. There is an error in the inductor's calibration.

Using the last color of the resistor, which is brown. We can find the uncertainty for the resistor in the table as we learned before, which is 5% +/- 2*10^-7

We then given a circuit with an inductor, resistors, and voltage as in orange diagram below. First, we want to calculate Tao by dividing the inductors by the resistor, both in line 1. After finding that, we can calculate the half time, which is 5* the Tao that we found. Now we want to find current of each line. For first current, we will use the equation Max current multiplied by 1-e^-t/Tao. We found the Tao earlier and we can use the time as 170 ms. Max current is the voltage in the circuit divided by the resistor in first line. We found the answer to be 0.165 A. For the second current, we can just calculate the current using the voltage divided by the resistor in the second line and we found the answer to be 0.0616 A.

Next, we want to find the potential drop across the resistor. We know that the voltage in the second line is just the 45V. On the first line, we can calculate the voltage by multiplying the current that we found by the resistor on the first line and we found the voltage to be 19.8. When we subtract this we found the potential drop to be 25 V.

We can also calculate the time for inductor's voltage to drop to 11V. By subtracting the initial voltage to the final voltage. Then divide them by the first line's resistor to find the current on that line. Then we can find the time by using the current equation earlier. By using current max and Tao on line one, we found the time to be 409ms.

We can also find the energy dissipated after switch is open for 170ms. By using the equation of emf energy from past lecture, we can calculate it by multiplying half of current(square) by inductance on the first line. We find it to be 4.76*10^-4 J.

Inductors

We started the day answering questions on the active physics website. Given a loop as shown below, we anwered the questions as below. Using the program, we can answer the questions as below

These questions help us understand of the relationship between flux to magnetic field, force, current

Recall about the right hand rule that we learned in the past lectures connecting force, magnetic field, and current. Given a setup as below, the current goes through the tube metal from the side thats closer to us, with magnetic field going from north to south, which is pointing upward. This means that the force is going away from the magnet. When we let the current run through, the tube metal rolls over away from the magnets and fell from the table, which proves that our prediction was correct. When we switched the positive and negative sides of the metal, which causes the current to go through the opposite way, the tube metal rolls closer to the magnet.

We then do some more questions. Given a loop as shown below, we again anwered the questions as below.

Understanding inductors:

We learned that inductor is flux per current. It works the same was as capacitors because they both store energy. When voltage increases, current decreases. When magnetic field increases, back emf is also increasing. The more number of coils also increases the resistance of it.
Inductor is also emf produced by the inductors divided by dI/dt. Using the algebra, we found dI/dt to be the emf/inductor. We know that emf is in voltage; thus, voltage is inductors multiplied by dI/dt.
Capacitor can also be related to the time, by replacing charge with multiplication of current and change of time, we find current to be I*change of time per voltage.

The equations below are some of the already simplified equations from the laws that we have learned before.

We knew from past lecture that flux is multiplication of number of loops, magnetic field, area vector, in an angle of cosine. We can replace magnetic field with myu, number of loops, divided by the length. When we integrate them, we find flux to be that multiplied by the time. When we derive the equation in terms of time, we found the equation to be in dI/dt. Recall that voltage is induction multiplied by dI/dt, we can replace voltage to this and get rid of the derivation. Finally, we found the induction to be myu*N(square)*A/l.

We are then given a question with known radius of 0.1 cm, number of loops 100, length is 4 cm. We know that magnetic permeability of copper is 1.26 * 10^-6. First, we convert all the units into SI units, so the radius and length would be in meters. Using the equation that we derived, we just plugged in all the known and found the answer to be 9.9*10^-7.
We then learn that in DC circuit, induction has no effect. It only has effect in AC circuit. Also, we predicted the graph of voltage vs. time when current vs. time graph is a flat horizontal line. We predicted it to be a vertically straight line because  based on the equation that we know, when current does not change, resistance increases or decreases, which causes the voltage to stay vertically at a particular time.
We also learned that emf is 1/2 LI(square).

We learned about time constant in the past. Time constant is RC. When time constant foes to zero, the
Lastly, we another set of questions from the active physics website as below.

Cathode Ray and Oscilloscope

Cathode Ray:

We started the day understanding about cathode ray. The way it works is

Next, we learn about Cathode Ray Tube. In it, there is a mechanism that emits electrons. First, the emitted electrons are concentrated into a tight beam by thr focusing plate. Right after, the electrons are subjected to different plates that alter their x, y axis positioning by using electric potential difference.

We learned a force in electron from past lecture that m*a in electric is q*E. These two understanding can be equated and we can find acceleration to be qE/m. But electric field we know to be V/d. The electric field can then be replaced, so the equation is qV/dm.

From 4A we learned that y is Vx*t. When we substitute this to the equation of acceleration we found earlier, we can find the time to be qVLD/mdV(square).

Oscilloscope:

We follow the instructions exactly as the lab manual says to familiarize ourselves to the use of oscilloscope.

Second, we switch over to sinusoidal wave and square wave outputs of the fuction generatur. Square wave comes out as below:

Sinusoidal output comes out as below:

The function generator is then hooked up to channel 2 of oscilloscope. When both function generators have same frequencies, it will create a circle as below.

When both function generator are given 180 Hz, it shows as below.

We are then given a box and we had to determine the type of current from the oscilloscope by using AC and DC.
When we used red and black AC, it comes out as below:

When we used red and black DC, it comes out as below:

When we used Black and Green and Black and Blue DC, it shows as below:

When we used Red and Blue and Red and Green DC. it shows as below:

When we used Red and Yellow (DC and AC), it shows as below:

When we used every other combination of colors (DC and AC), it shows as below: