voltages to make the experiment fair then I will find the average. I will record the Voltage and the Current of the wire in every 10 meters. I will record my results in a format of a table like this. Length (cm) Current (I)(amperes) Voltage (V) (volts) Resistance ® (Ohms) 10m 20m 30m 40m 50m 60m 70m 80m 90m 100m For the Current and the Voltage there will be 3 results then I will find the Resistance and get the average from that. To find the
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by R3 and C3. The distance between each set‚ is defined by the C2 and R1. The Voltage input to the circuit‚ used 12 volt DC power from a car at all. The section capacitor C1 as a filter to smooth the flow. Transistor Q2 may be any number SM3180. The schematic shown here can be used as a circuit for Car brake light or headlight flasher for flashing two 10 watt 12 volt lamps in car or any vehicle. The heart of the circuit is a 555 timer IC‚ wired as an astable multivibrator in the circuit
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ammeters. Potential Difference (voltage) Electrical charge tends to move from points of high potential to points of low potential. The difference in potential between two points is called the potential difference or voltage and is measured in units of volts (V). Resistance For many devices‚ it is found that the potential difference appearing across a device is proportional to the current flowing through it. The connecting relationships is known as Ohm’s law and is written:V = IR where R is the resistance
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along X-axis (in microseconds) & the voltage is represented along Y-axis (in Volts). Figure: 5.4(b) Output voltage of the square inverter and driven signals for switches S1a‚ S3a‚ and S5a Output voltage of combined synchronous rectifiers and driven signals for switches S1a‚ S1b‚ and S5a are shown in Figure 5.4(c). Here time is represented along X-axis (in milliseconds) & the voltage is represented along Y-axis (in Volts). Figure: 5.4(c) Output voltage of combined synchronous rectifiers and
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Calculate the power dissipated in the 10kW. V. DATA Voltage and Current Measured Values By deactivating the 12V power source and replacing it by a wire I got: V5V = 2.0 Volts I5V = 0.20 mA By deactivating the 5V power source‚ replacing it by a wire‚ and in turn restoring the 12V source I got: V12V = -2.4 Volts I12V = -0.24 mA By leaving both power sources ON
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begins with the specification‚ which states the functionality that the finished design must provide‚ but does not indicate how it is to be achieved. The initial specification for this project is a dc voltmeter with an input voltage range of 0 to 300 volts. Although there are many parameters that a design specification should contain‚ such as size‚ weight‚ moisture resistance‚ temperature range‚ thermal output‚ vibration tolerance and acceleration tolerance but we shall only pursue electronic circuit
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of gas in the air decreases to a safe level. The circuit uses an MQ6 gas sensor‚ which is designed to sense LPG‚ propane and isobutene gases. Circuit and working Fig. 1 shows the circuit of the LPG sensor. The circuit is built around 5V volt age regulator 7805 (IC1)‚ gas sensor MQ6 (GS1)‚ counter IC 4060 (IC2) and a few discrete components. GS1 is a six-pin gas sensor that can detect very small traces of LPG in the air and has a swift response time. However‚ it Has very less sensitivity
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Why is it important to remove loose jewelry before working inside a computer case? Jewelry might fall into and/or get caught on any of the components of the machine and corrupt or damage it including potential for static electricity as low as 10 volts. 3. When assembling a system‚ which do you install first‚ the drives or the motherboard? Motherboard 4. What is the purpose of raised screw holes or stadoffs installed between the motherboard and the case? If the bottom of the motherboard touches
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losses (mainly voltage dependent) and Variable losses (current dependent). The expression for the efficiency of the transformer operating at a fractional load “x” of its rating‚ at a load power factor of Cosθ2‚ can be written as ……… (1) Here S in the volt ampere rating of the transformer (V2‚ I2 at full load)‚ Pconst being Constant losses and Pvar the variable losses at full load. Pvar = (IFL)2 R If I is the current at any load‚ then we can define the fractional load “x” as under: x = I/ IFL
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........................................................................................................ coated unipotential Heater Voltage AC or DC ....................................................................................... 5.0 ! 10% Volts Heater Current ....................................................................................................... 1.9 Amps MECHANICAL Base ...................................................................................................
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