Data Analysis: In connection with the familiarization with electrical instruments‚ this experiment allowed us to make a simple circuit and measured some circuit elements such as voltage‚ resistance‚ current and power. From the obtained data we can see the relationship of the resistance and current. The Ohm’s Law was very apparent with the data gathered. In Ohm’s Law we know that when voltage and other factors are held constant‚ the current ( I ) is inversely proportional with resistance ( R
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Answers to Questions 1. Is it desirable for an ammeter to have a high resistance or low resistance? Why? ➢ Yes‚ it is desirable for an ammeter to have a high resistance or low resistance because the lower the resistance‚ the better the ammeter as it is connected in series and so its resistance will affect the current flow and thereby affect any calculations to be obtained. 2. An ammeter of resistance Ra is connected through a rheostat to a cell of negligible internal resistance
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Assignment Week 4 Chapter 11 20. A parallel RL circuit has values of IL = 240 mA and IR = 180 mA. Calculate the magnitude and phase angle of the circuit current. It=√ IL^2 + IR^2= √ 240mA^2 + 180mA^2= 0.3 x 10^-3= 300mA θ= tan^-1(-IL/IR)= (-240mA/180mA)= -53° 300mA∟-53° 22. Calculate the total impedance for the circuit shown in Figure 11.38b. It= √IL^2 + IR^2= √800mA^2 + 120mA^2= 0.808949 x 10^-3= 808.95mA Zt= Vs/It= 120V/808.95mA = 148Ω θ= tan^-1(-IL/IR)= (-800mA/120mA)= -81.5= 82° 148∟82° 24
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Humidity Sensors Humidity Humidity is defined as the water vapor content in air or other gases. Humidity is usually measured in terms of absolute humidity (the ratio of the mass of water vapor to the volume of air or gas)‚ dew point (the temperature and pressure at which a gas begins to condense into a liquid)‚ and relative humidity‚ or RH (the ratio of the moisture content of air compared to the saturated moisture level at the same temperature or pressure). Thermal conductivity humidity sensors
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GCSE Physics Coursework - Resistance of a Wire Coursework Introduction (Theory) In this piece of Coursework I am going to investigate the resistance in a piece of wire. Resistance is the force that slows down the flow of electrons; this can be due to electrons colliding. The total length of the wires will affect the amount of resistance. The longer the wire‚ the more resistance that there will be. This is also true for thickness of wire; the thinner the wire is the more resistance. Resistance
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IED12102 Basic Laws: 1. 2. 3. 4. 5. Electronic terminology Series Circuits Parallel Circuits Ohm’s Law and dc Circuits Simple Electrical Diagrams For single-source parallel networks‚ the source current (I ) is equal to the sum of the individual branch currents. s Is = I1 + I 2 For a parallel circuit‚ source current equals the sum of the branch currents. For a series circuit‚ the applied voltage equals the sum of the voltage drops. V V In this first example‚ we
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DSE BASED ZONE PROTECTION: The dynamic state estimation based protection (also known as setting-less protection) has been inspired from the differential protection function and can be considered as an extension and generalization as illustrated in Figure 1. In current differential protection the electric currents at all terminals of a protection zone are measured and their weighted sum must be equal to zero (generalized Kirchhoff’s current law). Thus the current differential protection function consists
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Ex.No Date Aim: To find resistance of a given wire using metre bridge and hence determine the specific resistance of its material. Apparatus: A meter bridge ‚a Leclanche cell ‚ a galvanometer ‚a resistance box‚ a jockey‚ a resistance wire or resistance coil‚ a metre scale‚ a set square and connecting wires Procedure: 1.Draw the circuit diagram as shown in figure and arrange the apparatus according to the arrangement diagram. 2. Connect the resistance wire or resistance coil whose resistance
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Since the flux is sinusoidal‚ r.m.s = (form factor*average e.m.f )/turns But form factor = (r.m.s value)/(average value) =1.11‚ Then r.m.s value of E.m.f = 1.11*4F*max.Flux = 4.44F*max.Flux‚ But max .flux = B_m*A r.m.s value of E.m .f in primary turn (Tp) = 4.44F*B_m*A*T_p NOTE: Bm is assumed to be 15000Wb/m. F = Frequency of A.C input/ Output F=50 Hz By introducing stacking factor (10-8) and Tp factor (0.9) then we have Number of turns per volt = N_T/V =7/A =7/3.75=1.87turns/volts
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Single Stimulus Activity 1: Identifying the Latent Period 1. How long is the latent period? 2.78 msec 2. Does the latent period change with different stimulus voltages? no Activity 2: Identifying the Threshold Voltage 1. What do you see in the Active Force display? Leveled line 2. What is the threshold voltage? .8 V 3. How does the graph generated at the threshold voltage differ from the graphs generated at voltages below the threshold? They were Level Activity
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