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How do you get half of the voltage from a DC supply circuit?
By using a voltage divider, that is two resistors of the same value in series across the DC supply. Half of the supply voltage will be at the point where the two resistors is connected. But how much wattage of those resistors is also an issue.
Two technicians are discussing how to measure the voltage outputs of a power supply. Technician A says that the power supply must be connected to the system and turned on before voltage can be measu?
It does have to be turned on. A highly regulated PS will be the same under load as without a load. But it may not be able to supply the amps. Best connect it and test it. WHO WON?
How do you reduce power consumption of a Wheatstone Bridge without changing the bridge resistances?
It seems to me that there are two methods. Reduce the supply voltage to the bridge, or reduce the time that the supply voltage is connected (you could also apply both methods simultaneously). Which is best depends on the details of the application. 1. Reducing the supply voltage: The power consumption of the bridge will be proportional to the square of the supply voltage so halving the voltage will quarter the power consumption. It will also halve the voltage at the detector for any given imbalance so, to maintain the same resolution, the sensitivity of the detector needs to be increased by the same factor as the voltage. Clearly there are limits to this as the imbalance signal will eventually be lost in noise. 2. Reducing the time the supply is connected: It is possible to pulse the supply voltage and the power consumption will be reduced in proportion to the duty factor. Clearly the detector must be fast enough to take a reading during the time that the supply is connected. This might be done using a microcontroller - an output controlling the supply to the bridge and an A/D input to measure the imbalance signal.
How can power be calculated from current and voltage?
Power is the product of voltage and current -in other words, mutliply the two together.
Which bulb will glow brighter two bulbs of rating 100w-220vand200w-220v are connected in a series to a supply of 22ov?
For a lamp to operate at its rated power, it must be subject to its rated voltage. The lower the voltage, the lower the resulting power. In fact, a small drop in voltage will cause substantial drop in power. Higher 'wattage' lamps have lower resistance values than lower 'wattage' lamps.So, if you put two lamps in series, the greater voltage drop will appear across the lamp with the greater resistance. In your example, that means the 100-W lamp will be subject to the greater voltage drop, and its loss of power will be less than that of the 200-W lamp. So the lower power lamp will be the brighter of the two.
What is voltage divider circuit?
The two resistor voltage divider is used often to supply a voltage different from that of an available battery or power supply. In application the output voltage depends upon the resistance of the load it drives.
How do you get half of the voltage from a DC supply circuit?
By using a voltage divider, that is two resistors of the same value in series across the DC supply. Half of the supply voltage will be at the point where the two resistors is connected. But how much wattage of those resistors is also an issue.
What supplies the voltage to the USB?
The two outside contacts (of four contacts) are connected to the power supply in the computer.
Basides proviing the PC with the correct voltage the power supply also performs what additional task?
The power supply unit provides the PC with "clean" power at the correct voltage levels.Related Information:The PC has two main power requirements. Its power must be:a) at the correct voltage.b) smooth (the typical maximum deviation is +-5%).
What is ratiometric instrumentation architecture?
Ratiometric instrumentation architecture is common in the automotive industry, specifically for analog sensors attached to an engine. The idea is to help divorce a measurement system from the influences of circuit aging, temperature, power supply voltage change, etc. In a ratiometric system A to D converters are referenced to a power supply voltage, the same power supply used for sensors (e.g. a Throttle Position Sensor), making the power supply a scalar (reference) for the ADC. The net result is a system that has two inputs, a sensor voltage and a power supply voltage. Mathematically, the system test for ratiometricity is to vary an input voltage and supply voltage by the same factor (percentage), and then verify the result is invariant. For example, a thermistor at 25°C produces, through signal conditioning, a voltage of 2.500V while using a 5.000V supply. That same thermistor at 25°C must produce 2.550V when the supply voltage is raised to 5.100V (+2%) in order for the system to be considered ratiometric.
Two technicians are discussing how to measure the voltage outputs of a power supply. Technician A says that the power supply must be connected to the system and turned on before voltage can be measu?
It does have to be turned on. A highly regulated PS will be the same under load as without a load. But it may not be able to supply the amps. Best connect it and test it. WHO WON?
. Two technicians are discussing how to measure the voltage outputs of a power supply. Technician A says that the power supply must be connected to the system and turned on before voltage can be measu?
It does have to be turned on. A highly regulated PS will be the same under load as without a load. But it may not be able to supply the amps. Best connect it and test it. WHO WON?
What is the power of a ciruit with 10 ohmns 10 volts and 2 amps?
Well, first of all, if the resistance of the circuit is 10 ohms and you connect 10 volts to it,then the current is 1 Amp, not 2 . So either there's something else in your circuit thatyou're not telling us about, or else the circuit simply doesn't exist.-- If you connect some voltage to some resistance, then the resistance heats up anddissipates (voltage)2/resistancewatts of power, and the power supply has to supply it.-- If there is some current flowing through some resistance, then the resistance heats up anddissipates (current)2 x (resistance)watts of power, and the power supply has to supply it.-- If there's a circuit with some voltage connected to it and some current flowingthrough it, then the resistance of the circuit is (voltage)/(current) ohms, the partsin the circuit heat up and dissipate (voltage) x (current) watts of power, andthe power supply has to supply it.There's no such thing as "the power of a circuit". The power supply supplies thecircuit with some amount of power, the circuit either dissipates or radiates someamount of power, and the two amounts are equal.
Why does not wattmeter measure the reactive power?
A wattmeter is designed so that it measures the supply voltage and the in-phase component of the load current. The product of these two quantities is the true power of the load.
How do you find the power factor with true and apparent power?
to put out the power fector you have to divided apparent power with true power.AnswerYou can determine the true power of any load using a wattmeter. To find the apparent power, you use a voltmeter to measure the supply voltage and an ammeter to measure the load current, and multiply the two readings together.If you then want to go on to find the power factor, then you divide the true power by the apparent power. If you want to find the reactive power you use the following equation:(reactive power)2 = (true power)2 x (apparent power)2
How do you reduce power consumption of a Wheatstone Bridge without changing the bridge resistances?
It seems to me that there are two methods. Reduce the supply voltage to the bridge, or reduce the time that the supply voltage is connected (you could also apply both methods simultaneously). Which is best depends on the details of the application. 1. Reducing the supply voltage: The power consumption of the bridge will be proportional to the square of the supply voltage so halving the voltage will quarter the power consumption. It will also halve the voltage at the detector for any given imbalance so, to maintain the same resolution, the sensitivity of the detector needs to be increased by the same factor as the voltage. Clearly there are limits to this as the imbalance signal will eventually be lost in noise. 2. Reducing the time the supply is connected: It is possible to pulse the supply voltage and the power consumption will be reduced in proportion to the duty factor. Clearly the detector must be fast enough to take a reading during the time that the supply is connected. This might be done using a microcontroller - an output controlling the supply to the bridge and an A/D input to measure the imbalance signal.
Which two pieces of information are needed before selecting power supply?
1.the form factor of the case 2. the voltage requirements of peripheral devices
