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Audio Description – Course 1, Ohm’s Law and Kirchhoff’s Voltage Law (KVL)
– Clip 1 Introduction – Clip 2 Ohm’s Law – Clip 3 Course Description – Clip 4 Course Objectives – Clip 5 Course Parameters – Clip 6 Course Equipment Part 1 – Clip 7 Introduction Instrumentations – Clip 8 Voltage Test – Clip 9 2 wire resistance test – Clip 10 4 wire resistance test – Clip 11 Calculations 1 – Clip 12 Measurements for the load – Clip 13 Calculations 2 – Clip 14 Introduction 2 heating wire – Clip 15 4 wire resistance measurement – Clip 16 Power measurement and heat increase – Clip 17 Calculations 3 – Clip 18 Calculations 3 – Clip 19 All resistors in series – Clip 20 CAll resistors in detail – Clip 21 Total resistance measurement – Clip 22 Calculations 4 – Clip 23 Resistance measurement 10 Ohm – Clip 24 Resistance measurement 15 Ohm 1 – Clip 25 Resistance measurement 15 Ohm 2 – Clip 26 Resistance measurement 33 Ohm – Clip 27 Calculations 5 – Clip 28 Calculations 6 – Clip 29 Voltage delivery – Clip 30 Voltage Drop 10 Ohm – Clip 31 Voltage Drop 15 Ohm 1 – Clip 32 Voltage Drop 15 Ohm 2 – Clip 33 Voltage Drop 33 Ohm – Clip 34 Current measurement – Clip 35 Calculations 7 – Clip 36 Calculations 8
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Course 1 contains two physical laws in electricity. The first one is Ohm’s law and the second is Kirchhoff’s second law of voltage also called (KVL). The video is produced by Vortices Dynamics and lectured by TheOldScientist, aka Thomas Imlauer. This content is available for EDU, Bronze, Silver and Gold Members as well as a per pay per post option.
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Here the classical triangle of the Ohm’s law. You calculate by covering the parameter you want. For Voltage you multiply current = I times R = resistance. For current you divide R = resistance over V = voltage. For Resistance you divide I = current over V = voltage
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The course description does list the following scenarios. 1. Calculating all values for a closed circuit. 2. Measure all values to confirm calculated values as well as derive unknowns. 3. Learn how to predict the circuit behaviour. We look into some phenomenons which are not that much mentioned based on Kirchhoff’s law. The voltage drop.
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Course objectives:
1. You will be able to calculate all values for a circuit. 2. You can apply all measurements and confirm calculations 3. All power requirements can be correct defined and applied.
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Course Parameters
All calculations and measurements are on Direct Current.
1. Voltage is a battery with 12 Volt 2. A Bench power supply with 20 volt and 5A. 3. Resistor or load is a light bulb 4. Second resistor is a one foot long or 12 cm heating wire
Questions are based on
V = 5, 10, 12 Volt , R=? I=? P=?
We use these resistors as well to prove Kirchoff’s law. It is fundamental to understand the consequences of voltage drop when not caused by the load.
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Here in part one we focus on the 12 volt battery and the light bulb as load. We measure with the precision Keithley 2110 Bench Digital Multimeter 5.5.
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Hello and welcome to my course #1 electricity and the subject is Ohm’s law and (Kirchoff’s Law). What we are going to do in this course is to work through the formulas. How we can derive certain values from the formulas. We will solve questions we will raise and will replicate the result of the formulas here in the experimental setup based on our parameters of load, resistance and power supply. I will walk you through how to measure the values and before we start we want to raise the obvious questions. We have a power supply and we have a load. We want to know how much power this load requires to run. We also want to investigate if our power supply is powerful enough to provide the power or current for the load. We measure all these details here simultaneously. We are going to measure voltage and we are measuring current and power. And that over time. We take the measurements here with my precision Keithley 2110 Bench Digital Multimeter 5.5. We will also use measurements from the DMM on the left hand side, from which I can also add calculation into a spreadsheet, via a USB feed. The exercise is constructed in that way that it should be easy for anybody to follow up or replicate. After the course you should all be in the position, based on direct current, to calculate the values in your closed circuit and be 100% certain how much power your system requires and how much power you have to deliver and what load can consume this power.
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ON the formula I defined 12 Volt as our power supply, here as a small micro battery, a deep cycle battery and this does have 12.874 volt. We are going to monitor that over time. when we connecting the load. We will also go to a deeper more accurate measurement . We will apply the 4 wire measurement. This is important for resistance measurements to exclude lead resistance from the measurements.
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As load I will use a small 12 Volt light bulb. Very important to note. Let’s assume I know nothing about my load. In some of the cases that will the circumstances you have to deal with. We measure the load and your load is and has resistance. Resistance is in Ohm’s law a factor for the calculation of power. We see here 2.771 Ohm. We take a note of that for the calculation. We will see later on if this value does reflect the power rating of the light bulb. At the moment I have applied a two wire measurement. I will add for you now the 4 wire measurement as an example of difference in accuracy.lip 10
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You can see the value of 2.745 Ohm at the moment. It is set to a 4 wire measurement. If I toggle back to the standard value of two wire. You see two wire measures, 2.770 Ohm. Coming back to 4 wire it reads 2.741 Ohm. It is a small change because the leads are fairly thick. We have very low resistance in there. Even though this resistance in the lead wire can be excluded from the calculation. In high power factors circuits or especially when resistance is less than 100 Ohm for the load this measurement is crucial to conduct. A lead resistance of one Ohm can drop the voltage considerable and removes the current from getting delivered. Second, the resistance in the lead is causing the wire to heat up and dissipate energy we want to preserve for the load only. The only option to avoid that is to use large and short diameter leadsClip 11
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We integrate our measurements into the formula. The power source does measure 12.873 Volt. The resistance of the load did measure 2.740 Ohm. Ohm’s formula for current is I=V/R or 12.873/2.740 Ohm = 4.698 A. That seem like a very high value. If that correct. The calculation for power is I*V or 12.873 * 4.698 = 60.5 Watt . What we witness here is typical for light bulbs based on filaments. We measure the resistance in the cold state of the light bulb. When current flows through the filament it heats up to the maximum temperature which can be handled by voltage based on the increased resistance of the wire. The heated filament becomes high resistive to a point where the voltage can not deliver more current. We see a very bright light based on the Wolfram or Tungsten wire up to (3,422 ° “Celsius”, 6,192 ° “Fahrenheit”), the highest melting point for any metal. We see later how much current is really drawn .
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Here I have now the circuit connected. I have the battery and I have the load. I use the Keithley 2110 to measure the value of voltage and the value for current at the same time. I have a so called “Second Display” active. I display in the first row in bigger letters the current in Ampere to Direct Current which is required to drive the load. In the second row the voltage in smaller letters. At the moment we have 12.751 Volt on the battery. When we connect the load the voltage will drop. That has an influence on the current delivery we have to take that later into consideration. Lets have a look. We connect the load and we see a current draw and voltage drop. The first values are 0.3763 A and 12.554 Volt. I keep that running for a small time until changes are very small or it stabilizes for a longer period of time. In fact that drop will continue until the battery can not deliver current anymore and the voltage drops to a low point. We take 0.3725 A and for the voltage 12.433 Volt for our calculation and compare that with our first measurements to see if that reflects our predicted values for power.Clip 13
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We see now two triangles as briefly mentioned before. The power triangle of Ohm’s law. You treat that in the same way as the standard Ohm’s law by covering the field you want to solve for. Power or P is I*E. Yes, Voltage is here represented as “E” and not as “V”. Looking at our measure values we have for E = 12.433 Volt and for current, I = 0.3275 A. We solve now for power and have P = E*I or 12.433 volt * 0.3725 A = 4.631 Watt. That is more like it. The light bulb has a power rating of 5 Watt. We calculate now the resistance based on that current and get from R= E/I or 12.433 Volt / 0.3725 A = 33.38 Ohm. This is the resistance the filament presents for that voltage and power rating. Increasing the voltage would destroy the filament of the light bulb. That shows that is is not that simple to calculate a load when the load is dissipating heat. We come to a second example where heat plays a role as load as well.Clip 14
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In my second example we will use heating wire. The length is one foot or 27 cm. We measure the temperature change on the handheld Tenma DMM. We will do the same exercise as with the light bulb but you have noticed that we had a bit of a problem to calculate the resistance available under load. What we measure is the initial current draw. That is only a very small fraction of a second until the filament reaches it’s operating temperature which presents the final resistance for the current. We measure for that 33.38 Om which gave us the correct power rating value of around 5 Watt. We will see if we can apply the same to the heating wire.
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We take our first measurement and here we apply right away the 4 wire measurement for accuracy. We get a value of 4.793 Ohm. The temperature of the wire is close to 21 degree Celsius. In the following test we will use 5 Volt from a Bench power supply to drive the load.Clip 16
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We measure at 5 Volt the current which is drawn by the load or heating wire. We also measure the voltage drop which represents the energy which can be delivered. The power supply will display the delivered current and the temperature will rise under power. Let’s have a look and right after closing the circuit I get 0.72 A and final 0.74 A displayed on the power supply. The temperature is in an instant at 51 degree Celsius. My assumption is that the temperature measurement is not in real time based on the loose connection to the wire. The initial 5 Volt is dropped to 3.5585 volt. We have temperature now at 80 degree Celsius. The voltage drop does slightly decrease but the current draw is stable. At 95 degree Celsius we have a voltage drop display of 3.5660 volt. Voltage drop and current draw is stable now at 100 degree Celsius. We take that values now into our formulas for calculations.
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We integrate our measurements into the formula. E = 5 volt, I = 0.740 A, Voltage drop is to 3.567 Volt. The resistance measured of the room temperature heating wire was 4.795 Ohm. Power or P= I*E or 0.74 A * 5 Volt = 3.7 Watt. We can now verify our voltage drop by calculating E = R*I or 4.795 Ohm * 0.74 A = 3.548 Volt. Within tolerance the result is confirmed.Clip 18
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Kirchhoff’s second Law of voltage (KVL) will give us a closer view of how power is delivered. This law is like Ohm’s law a fundamental law and needs to be known by anybody working with electricity or in electrical engineering. Here by calculation of ac circuit are already large errors are made. Goals is it how to measure and how to spot those errors for your (Device UNder Test) (DUT)Clip 19
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Now I like to demonstrate to you Kirchhoff’s law, second Law of voltage which is about the voltage drop between each resistor or load. Voltage drop has to occur in order to deliver energy. However voltage drop should not occur in the leads wire to the load. If there is any resistance between the leads or within the load, which is not account for the energy dissipation which is required for the load. That has to be marked or defined as a loss or inefficiency. For every design you build you have to make sure that every component is qualified to deliver or transform 100% of the power required. We do that by measuring the complete circuitry for the lowest loss and for the highest delivery of power. What do we have here, we have 4 individual resistors or loads. We have a bench power supply, which will deliver 10 Volt and we measure the voltage drop across each individual resistor. Each resistor or load will drop the voltage accordingly. But what I don’t want to see here at the end or coming from the leads wire is a voltage drop. I want to see the full value of 10 Volt is delivered. That is the reason I use leads with a high wire diameter and stranded to avoid eddy current
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Kirchhoff’s second law of voltage states that each voltage drop across each resistor or load added together, minus the voltage delivers is equal zero. We have at the first resistor 10 Ohm, They all have a tolerance of 5%. We will measure that in detail. 15 Ohm on the second resistor. 15 Ohm on the third resistor and 33 OHm on the last resistor. We will conduct a precise measurement for resistor via 4 wire test and voltage drop for each resistor as well as the total voltage of the system and resistance.
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We perform now a benchmark measurement of the total resistance of the circuit, including the leads wire for our calculation of maximum delivered power. This is done via the 4 wire test. We measure a total of 73.04 Ohm. We take a note of that and go now to each individual resistor and measure the resistance.Clip 22
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We integrate our measurements into the formula. E = 10 Volt, R =73.039 Ohm.
Solve for current.
V/R = 10/73.039 = 0.1369 A.
Solve for power.
P = I*E = 0.1369 A* 10 Volt = 1.37 Watt
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The first resistor, measured with 10 Ohm, is very accurate 10 ohm including small leads wire resistance. Lets move on to the next tab.15 Ohm
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The next tab is a 15 Ohm resistor and as you can see it is almost correct. 15.068 Ohm. this is also very accurate. the 5% tolerance would provide a much larger value deriving from the ideal of 15 Ohm. Let go to the next resistor
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The next 15 OHm resistor show a little bit a higher value. 15.230 Ohm. Lets go to the last resistorClip 26
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This is a 33 Ohm resistor and as you can see a little bit less than 33 OHm. 32.660 Ohm. We take a note of that. We have now all the resistors measured. this allows us to calculate the power requirement for each load or resistor, based on 10 Volt we deliver. Lets go now to the next stage, voltage delivery and current consumption and voltage drop from each resistor and take a note of that.Clip 27
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We integrate our measurements into the formula.
V= 10 Volt,
R1, 10 Ohm, M = 10.004 Ohm
R2, 15 Ohm, M = 15.068 Ohm
R3, 15 Ohm, M = 15.230 Ohm.
R4, 33 Ohm, M = 32.668 Ohm
Solve for Current for each resistor
R1 = 10/10.004 – 1 A
R2 = 10/15.068 = 0.664 A
R3 = 10/15.230 = 0.657 A
R4 = 10/32.668 = 0.306 A
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We integrate our measurements into the formula.
V= 10 Volt,
R1, 10 Ohm, M = 10.004 Ohm
R2, 15 Ohm, M = 15.068 Ohm
R3, 15 Ohm, M = 15.230 Ohm.
R4, 33 Ohm, M = 32.668 Ohm
Solve for power for each resistor
R1 = 10 Volt*1 A = 10 Watt
R2 = 10 Volt*0.664 A = 6.64 Watt
R3 = 10 Volt*0.657 A = 6.57 Watt
R4 = 10 Volt*0.306 A = 3.06 WattClip 29
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The voltage drop across all resistor will now be measured. I expect almost no voltage drop occurs. We measure 9.9384 Volt is delivered to the load and we have a current requirement of 130 mA. We calculated, based on the total resistance and based on 10 Volt 136.9 mA. this is our benchmark of much power can be delivered. W will use that in later calculations. Taking the total resistance we have her and multiply this amount with the measured voltage. Will give us the total delivery of power into the system. My assumption was that based on the lower voltage the power delivery will be lower than 1.3 Watt. However, based on the measure current, that is not the case. Next step will be I go via each resistor and measure the voltage drop and take a note
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The first voltage drop I measure is across the 10 Ohm resistor. I have 1.3621 Volt. We take a note and move to the first 15 Ohm resistor.
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We are now across the first 15 Ohm resistor. Let’s measure that. I read 2.0527 Volt. Now we measure the second 15 Ohm resistor.
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Remember that this 15 Ohm resistor was a little bit higher in value for resistance. Let’s see how that pans out. I read 2.0741 Volt. Now we will e=]measure the last resistor with 33 Ohm.
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We are now across the 33 Ohm resistor. I read 4.4489 Volt. We now will add all this voltages together and see what we come up with.
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Before we go to the calculation I like to do one last measurement for current with an precise value. The bench power supply did indicate 0.13 A. I read 0.1346 A. That is a higher value but within the rounding value of the bench power supply. We have now all values for our calculation of Kirchhoff’s second law of voltage.
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We integrate our measurements into the formula.
Voltage drop for each resistor
R1, Vd = 1.362 Volt
R2, Vd = 2.053 Volt
R3, Vd = 2.074 Volt
R4, Vd = 4.449 Volt
Solve for power for each resistor, P=I*E
R1 = 10 Volt*1 A = 10 Watt
R2 = 10 volt*0.664 A = 6.64 Watt
R3 = 10 Volt*0.657 A =6.57 Watt
R4 = 10 volt*0.306 A = 3.06 Watt
Vd = Voltage drop
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We integrate our measurements into the formula.
Voltage drop for each resistor
R1, Vd = 1.362 Volt
R2, Vd = 2.053 Volt
R3, Vd = 2.074 Volt
R4, Vd = 4.449 Volt
KVL = Vd1 + Vd2 + Vd3 + Vd4 – V = 0
1.3620 + 2.053 + 2.074 + 4.449 – 9.994 = 0.002
The law has been with high accuracy confirmed.
Vd = Voltage drop
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The End