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.
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.
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.
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.
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