Laboratory 1 Sae Yoo 904135449 A After setting the voltage source to 5 0 volts we measured it using the voltmeter and a DVM Both instruments gave higher precision readings than what the voltage source gave us Dials only gave readings up to tenths place while voltmeter and DVM gave readings up to ten thousandths and thousandths place respectively B 1 We cannot hook a voltage source to an ammeter directly because the ammeter has low resistance and we know from V IR that voltage over low resistance creates a high current which can cause irreversible damage to the instrument 2 If there is a power source during the reading the reading would be inaccurate because the ohm meter creates its own power with that power source C The lines show the connectivity where connecting on the same line will connect in a series D This section is to verify Ohm s Law using a voltage source and resistors with 20k and 10k We set up a circuit with voltmeter and the oscilloscope as an ammeter We can verify by setting different V values and recording the current on the ammeter For 20k we measured For 10k we measured And got the following linear graph Current I v Voltage V 0 0 f x 0x 0 0 0 0 0 0 0 0 0 2 3 4 5 6 7 8 9 Graph 1 Graph of 10 k resistor Current I v Voltage V 0 0 f x 0x 0 0 0 0 0 0 0 5 Graph of 20 k resistor 10 15 20 25 From this we know that the placement of the voltmeter can affect the measurement readings due to non ideal conditions of the equipment Ideally we would expect a resistor to have infinite resistance so we can completely measure all the voltage that runs through the voltmeter however that is not the case Also the ammeter should have no resistance so that the current is not affected in any way however there is a small amount of internal resistance which we will calculate in the next section From taking calculations using Ohm s law we can see that there is a certain amount of error but that gets smaller and smaller as the resistor value increases E To find the internal resistance of the ammeter we use Ohm s law but replace the R with R Rinternal and solve for Rinternal after measuring current I and voltage using a setup like this This shows how non ideal real life instruments are and how it can deviate from theoretical values F We used the multimeter set to measure ohms to measure the resistance of a 10M First time we measured with the resistor in the bread board and then second time with us holding it with our fingers Then we did the same with a 10K resistor The values we got are in the table below We can only expect this to happen because the voltage would not be completely going through the voltmeter but some of it is going through the fingers This is well due to the fact that the human body is over 65 water This bypasses the measuring system thus recording a lower value of resistance G We now take a Watt resistor and hook it up to our power source using 5V and not exceeding 1 A output We want to choose a resistor that can dissipate a very high amount of watts Using the equation of power P IV V IR P V 2 R we find 2 0 25 R V 25 R R 25 100 0 25 We see that 100 is the highest resistance at which the resistor will not burn out the threshold resistance We will choose a low resistance 5 When we plugged in the 5 resistor into the voltage output of 5V we saw that it became a bit yellow and it glowed red for an instant before smoking a little and burning out We then throw away the dead resistor H For this we set up a 47 incandescent light bulb and put it in a circuit on the breadboard Then without exceeding 6 5V we measured varying voltages and currents as shown below Then graphed it to get a linear representation Current I vs Voltage V 0 16 f x 0 02x 0 04 0 14 0 12 0 1 0 08 0 06 0 04 0 02 0 0 1 2 3 4 5 6 I We used an ammeter voltmeter 1k resistor 100k varying resistor or potentiometer and diode on the breadboard in a circuit as follows First we took measurements with varying voltages from 350mV to 750mV Then we graphed it to see the relationship in a linear format Diode Voltage V vs Current I 4 5 Logarithmic 4 3 5 Logarithmic Logarithmic 3 2 5 2 Logarithmic 1 5 1 0 5 0 0 1 2 3 4 5 6 7 J When we put the full 5 0V across in the circuit nothing changed It would have been different if we put the voltage source directly across it without any kind of resistors Probably would have rendered the diode useless K Using two 10k resistors an ammeter and a power source we set up the following circuit We will apply a V 15Volts We measure an output voltage and we will also find the current of the short circuit I ss We will use V and I ss to calculate the Thevenin resistance RTh We will repeat this step with a load 10k Without load V 15V V out V Th 7 54 V 8 I ss 1 54 mA RTh V Th 7 54 V 4 896 k I ss 1 54 mA With load V 15V V out V Th 7 54 V I ss 0 53 mA RTh V Th 7 54 V 14 226 k I ss 0 53 mA L We set up the function generator and hooked it up with a speaker and drove a sine wave We heard a steady noise associated with a 1kHz sine wave and turned up the frequency until we could not hear it anymore The range was around 17 4kHz M We set up our oscilloscope and function generator and plugged in the function generator into Channel 1 of the oscilloscope We then ran the function generator on the sine wave on 1 000 Hertz 1kHz We got a clearer view of the sine wave as we adjust the vertical gain horizontal sweep and trigger controls We then changed the shape of our function generator to the square wave We zoomed into the portion of the wave where it shoots up We measured the rise time time to pass from 10 to 90 of its full amplitude As we zoomed in we saw that the vertical line that shoots up is actually a sloping line We find that the rise time is about 10 nanoseconds We then plugged our cords into the function generator s SYNC OUT or TTP connector We saw that this port always outputs a square function We then found the period …