TUTORIAL 1 Fundamentals 1. An inductive load consisting of R and X in series feeding from a 2400-Vrms supply absorbs 288 kW at a lagging power factor of 0.8. Determine R and X. 2. An inductive load consisting of R and X in parallel feeding from a 2400-Vrms supply absorbs 288 kW at a lagging power factor of 0.8. Determine R and X. 3. Two loads connected in parallel are supplied from a single-phase 240-Vrms source. The two loads draw a total real power of 400 kW at a power factor of 0.8 lagging
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bulb 1.2 volts 5. screw base or socket for light bulb 6. wires 7. aligator clips 8. board for mounting the base or board 9. 2 lemons 10. 2 grapefruit 11. 2 bananas 12. 2 limes PROBLEM/QUESTION To determine which fruits will generate enough electricity to light a light bulb and to see which fruit will light a bulb the longest. Can fruits produce electricity to light a light bulb? I think that citrus fruit can produce electricity but not enough to light a light bulb of 1.2 volts. Research
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CHAPTER 1 ELECTRICAL FORMULAS OHM’S LAW/POWER FORMULAS R x I2 E x I P R P E E2 RxI R P I P x R E I P I E R E R P I2 E2 P P = Power = Watts R = Resistance = Ohms I = Current = Amperes E = Force = Volts 1-1 OHM’S LAW DIAGRAM AND FORMULAS E I E = I x R I = E ÷ R R = E ÷ I R Voltage = Current x Resistance Current = Voltage ÷ Resistance Resistance = Voltage ÷ Current POWER DIAGRAM AND FORMULAS P E I = P ÷ E E = P ÷ I P = I x E I Current = Power ÷ Voltage
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ELEC 202 LAB 1 REPORT INTRODUCTION TO LABORATORY INSTRUMENTS AND RESISTIVE CIRCUITS Objectives: The aim of the first experiment is to become familiar with lab instruments‚ get an idea about their working structure and how to use them when necessary. In addition to that in the first part of the experiment our aim is to read color codes of resistors and get idea about how to use multimeter for resistor measurements. In the second part of the experiment our aim is to see the operation of
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EE-123076 HAMZA BIN TARIQ EE-123086 Section PHY 7 Submitted To Ma’am Javeriah Iftikhar Abbasi Physics lab project Introduction: In this project we have created a device which takes input of 220 volts of ALTERNATING SOURCE and it gives us out put of 5 volts of DIRECT CURRENT for this purpose we have used various electronic components as mentioned below and made its circuit on PCB we have learned the concept of RECTIFICATION in this project. Components used: Following are
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the secondary terminals. * Standard tap arrangements are at two-and-one-half and five percent of the rated primary voltage for both high and low voltage conditions. * For example‚ if the transformer has a 480 volt primary and the available line voltage is running at 504 volts‚ the primary should be connected to the 5% tap above normal in order that the secondary voltage be maintained at the proper rating. 3. What is the difference between “Insulating‚” “Isolating‚”and“Shielded Winding” transformers
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To design a 12-volt DC power supply with a maximum output current of 200mA. To reduce the ripple factor to a value less than 0.1%. Apparatus 120-volt rms 60Hz input 60 ohm to 1k ohm load resistor 10 to 1 step-down transformer FWB bridge Voltage Regulator 2 100nF capacitors 10uFcapacitor 4.7 uF Capacitor Overview The goal of this experiment was to construct a power supply that would convert a 120-volt 60 Hz input ‚such as that received from a wall socket‚ to a 12-volt dc voltage output
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independent Silver Impregnated Graphite to Silver contacts‚ this relay conforms to BRB Spec. 930. The following table shows the available range Product Code Nominal Working Volts Contacts Coil Resistance Ohms Volts Maximum Full Operating 40 40 40 40 40 19.2 19.2 19.2 19.2 19.2 9.6 9.6 9.6 9.6 9.6 Minimum Release Volts 7.5 7.5 7.5 7.5 7.5 3.6 3.6 3.6 3.6 3.6 1.8 1.8 1.8 1.8 1.8 Interlocking 4841153 4841165 4842719 4842722 4842734 4841114 4841126 4842746 4842758 4842761 4841141 4841138
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5.7 Necessity of D.C. Motor Starter At starting‚ when the motor is stationary‚ there is no back e.m.f. in the armature. Consequently‚ if the motor is directly switched on to the mains‚ the armature will draw a heavy current (Ia = V/Ra) because of small armature resistance. As an example‚ 5 H.P.‚ 220 V shunt motor has a full-load current of 20 A and an armature resistance of about 0.5 . If this motor is directly switched on to supply‚ it would take an armature current of 220/0.5 = 440 A which
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then we have Number of turns per volt = N_T/V =7/A =7/3.75=1.87turns/volts The power Rating for the Inverter transformer (KVA) = 2.0KVA‚ E2= 24V Assuming the efficiency of transformer= 85%: Then Input rating =output/Efficiency= 2000VA/0.85=2353VA …………..equation 17 I_p=P_1/V_P VP = 280V …………..equation 18 I_P=2353/280=8.4A
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