...BATTERY SAFETY AND OHMS LAW Battery Safety Basics- Do not overcharge or over-discharge. When the charger indicates that the batteries have been fully charged, you can remove them from the charger. Leaving batteries in a charger all night long or longer can cause them to be overcharged, which can result in battery failure (depending on the charger). Charging your battery over 4.25 volts can shorten its life-cycle and going over 4.5 volts can cause it to burst. Cease using your charger if this ever happens. Our recommendation on a good charger: Nitecore D2 Charger (2 Bay) or Nitecore D4 Charger (4 Bay) Efest LUC (2 Bay) Efest LUC (4 Bay) Efest Soda (2 Bay) Recharge batteries with a resting voltage below 3.6V as soon as possible. Leaving LiIon batteries in a discharged state will incur irreversible damage – creating a loss in capacity and a loss in cycles. Determining the exact voltage can be tricky, unless you have a Tank-O-Meter/Volt Meter. If you are building your own coils a Tank-O-Meter is great device to have because you can use it to test your coils and your batteries. Sure, you can always use a battery tester, but the majority of battery testers are not equipped for the types of batteries that are used in mods, or even test batteries under load condition. Do not short circuit your batteries. Short circuiting can cause a huge surge of current that will potentially burn out your battery, damage your mod, or even your face! Short circuits happen when the voltage...
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...------------------------------------------------- The aim of this experiment is to know how does the length of the wire affects the wires resistance. Also measuring the voltage and current. A thin and a thick wire are used to get the results needed. There are three factors that influence the resistance of w wire which are, the thickness of a wire, temperature and length. Knowing those factors and measuring them will show how the experiment went. Introduction: ------------------------------------------------- The resistance of a material is the extent to which is oppose the flow of current. “Electronics for today or tomorrow, 2nd Edition, Tom Duncan, page 8” Where as conductors have low resistance and inductors have high resistance. Further more resistance is measures with ohms. The main concept of the experiment is to investigate the difference in the resistance when having several lengths of a wire. The reason why different lengths of a wire affect the resistance is because the length of the will is increased which will also make the resistance increase as well, therefore electrons will have a longer distance to travel. Because of this the length of a wire should be proportional to the resistance. Conductors are good metals of electricity because they have low resistance which is why electrons get away easily just by applying the voltage. Furthermore, the resistance of a wire and the length of the wire both increase but the thickness of the wire decreases at the same time. Moreover, thin wires have high resistance...
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...EPE is. Arrange problem to solve for EPE: EPE= qv = 2C x 100,000,000 = 200,000,000 VC or 2E9J Page 454: one-step calculations #6 P = IV I = 10 A V = 120 V 10 A * 120 V = 1200W P = IV I = 10 A V = 120 V 10 A * 120 V = 1200W V = 120 V = 120 exercise #9 1 2 3 4 5 1 2 3 4 5 Only #5 will work. Page 457: problems #3 C (current) = I P = IV to get current PI = 1200 W120 V 10 A I = 10 A Ohms Law V = IR VI = IRI = VI = R = 120 V10 A = 12 Ω C (current) = I P = IV to get current PI = 1200 W120 V 10 A I = 10 A Ohms Law V = IR VI = IRI = VI = R = 120 V10 A = 12 Ω P = 1200 W P = 1200 W V = 120 V = 120 HW. 3 3.1 Pg. 435 6. 2x106 1.6/3.2 10-10x10-4 Pg454 4. Power= 2w 5. 0.5x120= 60v Pg. 457 2. Resistance= 6 ohms Pg. 435 #4 me= 9.109x10-31 kg mE= 6x1024 kg q= 1.602x10-19 C E= 10000 vm F= qE F= 1.602x10-19 C x 10000 vm F= 1.602x10-15 N Fg= Gme mE r2 re= 6.4x106 m G= 6.67x10-11 Nm2kg2 Fg = 6.67x10-11 Nm2kg2 x 9.109x10^-31 kg x 6x10^24 kg (6.4x106 m)2= 8.89x10-30 N Pg. 454 #1 R= 15 Ω V= 120 V I=V/R I= 120 V15 Ω= 8 A Pg. 454 #2 R=90 Ω V=9 V I= V/R I= 9 V90 Ω= .1 A Pg. 454 #3 R=1000 Ω V=6 V I= V/R I= 6 V1000 Ω= .006 A Pg. 456 #44 The brightness of Bulb A will remain the same because the voltage of the battery will be equal to voltage of all bulbs...
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...when the straw is longer. • Cut the halves again in half. • With the four pieces, blow through one, then blow through all four made into a larger, square-shaped straw. • Describe the effect of straw size (diameter) on ease to blow air through the straw. Less resistance • Now, open the PhET Simulation Electricity, Magnets, and Circuits ( Resistance in a Wire[pic] As wire length (cm) increases, the resistance (Ω) INCREASES As wire area (cm2) increases, the resistance (Ω) DECREASES As wire density (Ωcm) increases, the resistance (Ω) STAYS THE SAME Procedure Part II: Ohm’s Law: Electricity, Magnets, and Circuits ( Ohm’s Law [pic] mA is milliamps, and 1000 milliamps equals one Ampere. • Move the potential (volts) and resistance (ohms) sliders and observe the current (amps) As voltage increases, current INCREASES As resistance increases, current DECREASES. Fill out the tables below and check your work in the simulation. ( ½ pt each ) • Remember, the simulation shows milliamps. • You should show Amperes V = I * R |8.0 V |.01 A |800 Ω | |2.0 V |.044 A |.50 Ω...
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...OHM’s Law Although there are literally thousands of formulas, I only try to remember two: Ohm's Law and the Power Formula. You can deal with most anything built before 1980 (except radio and TV) with these two formulas. The first, Ohm's law, states that E = I x R (Or, electromotive force in volts = intensity in amps, times the resistance). So those of you that are good at math games already know that we can change it to read R = E/I or I = E/R. That first class back in the sixties taught it as a magic circle, with E over I x R and you just took the one you wanted to know out of the circle and did what was left . I.e. If you want to know amps, you take the I out which leaves E/R, so you divide the volts by the resistance to find the amps. The power formula states that P = E x I or Power (in Watts) = Volts times Amps. It can also be remembered as a magic circle.Simply plug the resistance that you measure in to Ohm's Law and solve. More on this later - I’m getting ahead of my self. You may see these written with different symbols, depending upon when a book was written, but this is the way it was taught when my Scout was new. We will get back to formulas later. Now I want to go into electricity’s alter ego. | Ohm's Law: E = I x RPower Formula: P = E x I Magic triangle or circle (click for full-size view with comments) | Magnetism is the force possessed by some materials which enables them to attract or repel certain other materials. Magnets fall into...
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...Silver band is always placed to the right. The resistor value is read from the left to right. If there is no tolerance band, then find the side that has a band closest to a lead and make that the first band. Equation 2 0 x 10,000 1,000 = 1K Resistor = 200 K with a + 10 % Tolerance = 200,000 © Copyright 2006 Blue Point Engineering All Rights Reserved Page 1 Quick Guide Color Black Brown Red Orange Yellow Green Blue Violet Gray White Gold (G) Silver (s) None Band 3 Gold Silver Value 0 1 2 3 4 5 6 7 8 9 + 5% + 10 % + 20 % - Resistor Color Code 4 Band Resistance .22 ohm .27 ohm .33 ohm .39 ohm .47 ohm .56 ohm .68 ohm .82 ohm 1.0 ohm 1.1 ohm 1.2 ohm 1.3 ohm 1.5 ohm 1.6 ohm 1.8 ohm 2.0 ohm 2.2 ohm 2.4 ohm 2.7 ohm 3.0 ohm 3.3 ohm 3.6 ohm 3.9 ohm 4.3 ohm 4.7 ohm 5.1 ohm 5.6 ohm 6.2 ohm 6.8 ohm 7.5 ohm 8.2 ohm 9.1 ohm 10 ohm 11 ohm 12 ohm 13 ohm 15 ohm 16 ohm 18 ohm 20 ohm 22 ohm 24 ohm 27 ohm Notation R22 R27 R33 R39 R47 R56 R68 R82 1R0 1R1 1R2 1R3 1R5 1R6 1R8 2R0 2R2 2R2 2R7 3R0 3R3 3R6 3R9 4R3 4R7 5R1 5R6 6R2 6R8 7R5 8R2 9R1 10R 11R 12R 13R 15R 16R 18R 20R 22R 24R 27R Band 1 Red Red Orange Orange Yellow Green Blue Gray Brown Brown Brown Brown Brown Brown Brown Red Red Red Red Orange Orange Orange Orange Yellow Yellow Green...
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...Brandon Jenner ECT 125 CHP 11 34.) calculate component current and voltage values. XL(1)= 2 pie fl = 2 pie (60Hz)(470mH) = 177 ohms XL(2) = 2 pie fl = 2 pie (60Hz)(1H) = 376 ohms XL(T) = XL1 + XL2 = 553 ohms Z = _/XL(T)^2 + R^2 = _/(553)^2 + (1000 ohms)^2 = 1.1k ohms IL(T) = Vs/Zt = 120/1.1kohms = 105.2mA IR = Vs/R = 120/1k ohms = 120mA VL(1) = XL(1)/ZT = 177ohms/1.1k ohms = 160.9mV VL(2) = XL(2)/Zt = 376 ohms/1.1 kohms = 341.8m V VLR = R/Zt = 120/1.1k ohms = 909.mV CHP 13 34.) Calculate current and voltage. Xc = 1/ 2 pie fc = 1 / 2 pie (220Hz)(0.47uF) = 1.5kohms Zp = XcR2/_/Xc^2 + Rt^2 = (1.5kohms)(1.5kohms)/_/(1.5kohms^2)+(1.5kohms^2) = 1.06kohms 0 = tan^-1 (R2/-Xc) = tan^-1 1.5kohms/1.5kohms = -45 degrees R = Z cos 0 = (1.06kohms) cos (-45degrees) = 749 ohms Xc = Z sin 0 = (1.06kohms)sin(-45degrees) = -749 ohms Zt = _/Xc^2 + Rt^2 = _/1.5kohms^2+1.3kohms^2 = 1.9kohms 0 = tan^-1 (-Xc/Rt) = tan^-1 (-1.5kohms/1.3kohms) = -49 degrees It = Vs/Rt = 6V/1.69kohms /_-49degrees = 3.15mA /_-49 degrees Vr1 = It*Rt = (3.15mA)(560ohms) = 1.76V Vp = 3.3V /_-45degrees Ic = 3.3V/_-45 degrees/1.5kohms/_-90 degrees = 2.2mA /_-45 degrees Ir2 = Vp/R2 = 3.3V /_ -45 degrees/1.5kohms /_0 degrees = 2.2mA /_-45...
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...also more conductive than copper. When an electric current is applied to a conductor, the delocalized electrons assume a directed motion depending on the polarity of the potential difference. As the electrons move, they collide and bombard with atoms and molecules of the conducting material, thereby creating heat (T. B. Theraja). Materials which offer higher hindrance to the flow of electrons on application of a potential difference are called poor conductors of electricity. Examples of poor conductors are: Polyvinyl chloride (PVC), dry wood, rubber, glass and mica. Insulators are used in the coating of conductors to prevent electrical shocks .Resistance is measured in ohms denoted Ω. Depending on the degree of resistances other commonly used units are; mega-ohms (M Ω), kilo-ohms (k Ω), mill-ohms (m Ω) and micro-ohms (μ Ω) (T. B. Theraja). Resistance of a material is derived from the factors affecting it. According to the laws of resistance, resistance is directly proportional to length, l, and inversely proportional to the cross-sectional area of the conductor. It also depends on the nature of the material and the temperature of the material. Thus resistance of a conductor is given by: R= ρlA ρ is the resistivity of the conductor material, l is the length of the conductor, and A is the cross-sectional area of the conductor in this formula. In...
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...With the materials gathered, set-up VOM to its Ohmmeter function and check the continuity of all the wire connectors that will be used. Next step would be to set resistors 1 and 3 to 500 ohms and resistor 2 to 1000 ohms where one terminal of each battery will remain disconnected to the circuit until notice. After which, one has to connect the terminals of the batteries and gather then data of the voltage measurements of V1 and V2 using the voltmeter (fig.3). After the experiment proper, the student must have to start computing for the necessary values and correct polarities of the resistors through Kirchhoff’s Laws of both current and voltage then evaluate the necessary corrections (if any) in the proposed set...
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...Name: Date: 13.3 Ohm's Law A German physicist, Georg S. Ohm, developed this mathematical relationship, which is present in most circuits. This relationship is known as Ohm's law. This relationship states that if the voltage (energy) in a circuit increases, so does the current (flow of charges). If the resistance increases, the current flow decreases. Voltage (volts) Current (amps) = --------------------------------------------Resistance (ohms, ) To work through this skill sheet, you will need the symbols used to depict circuits in diagrams. The symbols that are most commonly used for circuit diagrams are provided to the right. If a circuit contains more than one battery, the total voltage is the sum of the individual voltages. A circuit containing two 6 V batteries has a total voltage of 12 V. [Note: The batteries must be connected positive to negative for the voltages to add.] If a toaster produces 12 ohms of resistance in a 120-volt circuit, what is the amount of current in the circuit? Given The resistance (R) is 12 ohms. The voltage (V) is 120 volts. Looking for The amount of current (I) in the circuit. Relationships V I = -R Solution 120 volts I = V = --------------------- = 10 amps -R 12 ohms The current in the toaster circuit is 10 amps. If a problem asks you to calculate the voltage or resistance, you must rearrange the equation I=V/R to solve for V or R. All three forms of the equation are listed below. V V II = V-V IR R R R I In this section...
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...trigger pin 2 is at lower then 1/3rd level of the supply voltage. Conversely the output goes low (0) when it is above 1/3rd level. So small change in the voltage of pin-2 is enough to change the level of output (pin-3) from 1 to 0 and 0 to 1. The output has only two states high and low and can not remain in any intermediate stage. It is powered by a 6V battery for portable use. The circuit is economic in power consumption. Pin 4, 6 and 8 is connected to the positive supply and pin 1 is grounded. To detect the present of an object we have used LDR and a source of light. LDR is a special type of resistance whose value depends on the brightness of the light which is falling on it. It has resistance of about 1 mega ohm when in total darkness, but a resistance of only about 5k ohms when brightness illuminated. It responds to a large part of light spectrum. We have made a potential divider circuit with LDR and 100K variable resistance connected in series. We know that voltage is directly...
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...Score:_____________________ DIRECTIONS: SOLVE EACH PROBLEM COMPLETELY. CLEARLY INDICATE ALL STEPS OF YOUR SOLUTION. BOX YOUR FINAL ANSWER. ATTACH THIS QUESTIONNAIRE WHEN SUBMITTING YOUR WORK. 1. Given the resistive network below with the given values. R1 = R8 = 1000 Ohms, R2 = R3 = 400 Ohms, R4 = R6 = 100 Ohms and R5 = 300 Ohms. Find Rab and Rac. (20 points) [pic] 2. For the given circuit below, find the following: [pic] 3. For the given figure below, find the following: [pic] 4. The resistance of a certain device is equal to 1000 Ohms at a temperature of 15OC. Its temperature coefficient is 0.001/OC and displays a positive response towards a change in temperature. Find the following under the assumption that a linear behavior is observed: a. The slope of the line. (5 points) b. The temperature and resistance intercepts. (10 points) c. The resistance at the temperature T = 80OC. (5 points) 5. The charge flowing in a wire is given by the figure below. Determine the following a) current flowing through the wire at each time interval and the average current through the wire. (15 points) [pic] ----------------------- 1. The current i and v. (10 points) 2. The current through the 8-Ohm resistor. (5 points) 3. The power dissipated by each resistive element. (10 points) 4. The power supplied by the 9-A source. (5 points) 5. How many nodes, branches and loops are in the circuit? (10 points) 1. The total current...
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...Current Probe Vernier Differential Voltage Probe Wires Clips to hold wires Light bulb (6.3 V) Adjustable 5 volt DC power Supply 2 resistors (10 and 51 Ohm) Procedure: 1. Connect the Current Probe and Differential Voltage Probe to LabQuest and choose New from the File menu 2. Connect the power supply, 10 Ohm resistor, wires, and clips as shown in the diagram. 3. Zero both probes with no current flowing 4. Using the 10 Ohm resistor, find the value of the current that flows in the circuit for various values of the potential. Start with 0 volts, increasing by .5 volts each time; record in a data table. 5. Repeat with the lightbulb and the 51 Ohm resistor Data: 10 Ohm Resistor: Voltage (watts) | Current (amperes) | 0 | 0 | .533 | .0551 | .880 | .0911 | 1.692 | .1752 | 2.735 | .2833 | 3.188 | .3303 | 4.642 | .418 | 51 Ohm Resistor Voltage (watts) | Current (amperes) | 0 | 0 | .513 | .0101 | 1.007 | .0201 | 1.969 | .0398 | 2.940 | .0595 | 3.904 | .0795 | 4.905 | .0997 | Lightbulb Voltage (watts) | Current (amperes) | 0 | 0 | .264 | .0472 | .362 | .0570 | .480 | .0570 | .532 | .0564 | .734 | .0648 | .958 | .0746 | 1.839 | .1059 | 2.784 | .1332 | 3.748 | .1579 | 3.951 | .1624 | 4.151 | .1672 | Evaluation of Data: For the 10 Ohm resistor, to determine the relationship between voltage and current, we graphed the voltage on the y axis and current on the x axis and found the slope with...
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...current = direct__________________________ Go to ⎝Ohm's Law. Click on the green “Run Now” button. The simulation should look like the picture to the right. 2. What is the current through a resistor with the following resistances? Let voltage = 6 V a. R = 100 ohms I = _60__ mA(current) b. R = 300 ohms I = 20____mA(current) 3. Now, determine the current through the wire with the following volts. Let resistance = 500 ohms a. Volts = 3 V I = 6___mA(current) b. Volts = 6V I = 12____mA(current) 4. Think about the formula (V=IR), does this make sense according to this formula? Explain! (Be sure to include the relationship between resistance and current, and the relationship between voltage and current in your answer) Yes, it does make sense according to the formula (V=IR). In question 2 where the voltage was set at 6V, the current was higher when the resistance was 100 ohms than when the resistance was 300 ohms. This is due to the fact that resistance and current have an inverse relationship, when one value goes up the other one goes down. In question 3 where the resistance was set at 500 ohms the current was...
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...Instructions: Complete the three parts—Part 1, Part 2, and Part 3 of Lab 1. When you have completed each part, answer the questions and transcribe/transfer the test results recorded in the lab manual’s tables to the tables provided. ------------------------------------------------- Part 1—Common Emitter (CE) Amplifier Theory: 1. What type of bias is used in the common-emitter amplifier shown in Figure 3-1(a)? * Voltage Divider 2. Explain the reason(s) why the AC equivalent circuit appears as it does in Figure 3-1(b). * When an AC input signal is applied to the amplifier, the capacitors will act as shorts. This is why we call this Figure the “AC Equivalent”. 3. What values must be calculated to determine the DC operating parameters? * VB * VE * IE * VC * VCE * VCC 4. What information can be obtained by determining the DC load line of an amplifier? * We use the DC load line to graph all possible current and voltage characteristics of a biased transistor. This includes any measurements that are between saturation and cutoff, including the Q-point. 5. What values must be calculated to determine the AC operating parameters? * Vin = Vb * r’ e * Ve * Av * Vout = Vc * Rin(total) * Rac 6. What information can be obtained by determining the AC load line of an amplifier? * We use the AC load line to graph the maximum/minimum current...
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