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What conditions could the total impedance of two impedances in series be less than that of either one by itself?
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∙ 13y ago
A:resonant
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∙ 13y ago
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Why input impedance should be high for an amplifier?
The best way to answer this question might be to consider the consequences if the input impedance was low: with a low input impedance, (signifficant) current would start flowing, and the amplifier would draw energy from the signal sources. None of the typical signal sources is designed to deliver energy on its outputs (after all, this is where the amplifier itself comes in). It is certainly possible to think that some of these sources might be changed to deliver some energy, but this is not the case with present-time tuners, CD players, microphones, and so forth. Assuming that the energy supply was not the issue, just to ponder this theoretical scenario a little further, the fact that current would flow from the source to the amplifier would also make the signal more vulnerable to the characteristics of the cable that connects the two. The high impedance of an amplifier input draws no energy, thereby avoiding these issues. It is the amplifier's task to convert a very low energy, voltage-driven signal into an higher energy output signal (driving the speakers which themselves have a very low impedance). ---- The way I typically think about this is to consider connecting a load to a Thevenin equivalent circuit [1]. The voltage across the load is given by the voltage divider formula (Vload = Vsrc * Rload/(Rload+Rthevenin)). If there is a very low load impedance--this means the amplifier has a very low input impedance--most of the source voltage will drop over the Thevenin equivalent resistance. With a very high input impedance, however, the majority of the signal voltage will be transferred from the source to the load because in the above equation, if Rload >> Rthevenin, Vload is approximately equal to Vsrc. if an amplifier has low impedance input the f/b must be low impedance also which make it in practical to use. The hi impedance of a typical amplifier is because the input is one two diodes basically operating on it exponential curve. Making it virtual the same as the other diode. for a differential amplifier. Boltzmann constant will define the impedance of a single diode.
When you prove a voltage to be induced using a Simpson why does the input impeadance on the different voltage scales produce different readings for the amount of voltage?
One possibility is that the accuracy of the Simpson is different on the different scales. Another (more probable) possibility is that the impedance of the Simpson on the different scales is sufficiently different so as to affect the reading. This is a common issue with low impedance multi-meters. Lets say you are using a typical Simpson meter with 20,000 Ohms per Volt. On a three volt scale, that means the meter itself has an impedance of 60,000 Ohms. On a 60 volt scale, however the meter has an impedance of 1,200,000 Ohms. Depending on the circuit impedance, that can have a significant impact on the final reading, which must be taken into consideration. Look at the equation for parallel resistance: RT = R1R2 / (R1+R2). If the meter impedance changes the circuit impedance by more than, say, 5%, that is going to affect the observed value. (You pick the percent limit - it depends on the situation.) Even for the case with a high impedance meter, say a 10,000,000 Ohm Digital Multi-meter, impedance must be considered if the circuit impedance is high enough. (I have a WWVB receiver that requires a 1,000,000,000 Ohm voltmeter to correctly measure the AGC voltage - no ordinary digital multimeter will suffice.)This does not mean that you have to spend lots of money on a high performance, high impedance, meter. You simply have to consider what the impedance of the meter is going to do to the circuit, and calculate that impact, before you state the results.
What is a high impedance state?
A high impedance state is a state where the component hardly draws any current at all. In most cases it also mean that the component is more or less turned off. In memory chips, it will keep the memory but no longer accept commands for changing or reading memory content. In portable measuring devices, it is a state where electronics turn itself off to conserve battery power. I hope this was helpful. Regards.
Why javascript function name and variable name same?
Either a case of bad programming practise or probably the function calls itself.
What is the law of the curve for power against voltage?
P = I x E (where E is voltage) so it's linear. ======================== The power dissipated by a component or circuit depends not only on the voltage across the load, but also on the characteristics of the load itself. If the resistance/impedance of the load is constant, then P = E2/R so it's proportional to the square of voltage.
Under what conditions could the total impedance of two impedances in series be less than that of either one by itself?
If both were reactances instead of resistances.AnswerIf one impedance was resistive-inductive (R-L) and the other impedance was resistive-capacitive (R-C), then the effective impedance could be less than either. For example, towards or at resonance, the inductive reactance will negate the capacitive reactance, leaving resistance as the main (or only) opposition to current flow. At resonance, the impedance of a circuit is simply its resistance.
What stops the current flowing in a circuit?
Current will cease when either or both the potential difference across the load is Zero or when the load, itself, is Infinite in resistance or impedance.
When diode is connected to ohmmeter What happens?
A: Depends on meter leads voltage polarity and the diode itself orientation to these polarity. One way is should show a low impedance + to anode Reverse the diode it should be hi impedance
Can an impedance matching device reduce electrical costs?
An impedance matching device is used to test the resistance, inductive reactant and capacitive reactant in a circuit. If one component did not match the impedance of the conductor, some of the current will be lost by the conductors itself. In conclusion if electricity is lost, the component needs to meet its regular voltage. It consumes more voltage than expected because of the loss. Impedance matching device can actually reduce electrical cost.
Why is smith chart circular?
It is because it represents 180 degrees of a wavelength. every 180 degrees the impedance on a transmission line repeats itself
What will happen to the input signal when the transmission line is terminated by its characteristic impedence?
When the input signal to a transmission line is terminated by its characteristic impedance then the signal gets absorbed in the terminating impedance itself and is not reflected back along the line. Thus, no standing waves are produced in the transmission line.
Why input impedance should be high for an amplifier?
The best way to answer this question might be to consider the consequences if the input impedance was low: with a low input impedance, (signifficant) current would start flowing, and the amplifier would draw energy from the signal sources. None of the typical signal sources is designed to deliver energy on its outputs (after all, this is where the amplifier itself comes in). It is certainly possible to think that some of these sources might be changed to deliver some energy, but this is not the case with present-time tuners, CD players, microphones, and so forth. Assuming that the energy supply was not the issue, just to ponder this theoretical scenario a little further, the fact that current would flow from the source to the amplifier would also make the signal more vulnerable to the characteristics of the cable that connects the two. The high impedance of an amplifier input draws no energy, thereby avoiding these issues. It is the amplifier's task to convert a very low energy, voltage-driven signal into an higher energy output signal (driving the speakers which themselves have a very low impedance). ---- The way I typically think about this is to consider connecting a load to a Thevenin equivalent circuit [1]. The voltage across the load is given by the voltage divider formula (Vload = Vsrc * Rload/(Rload+Rthevenin)). If there is a very low load impedance--this means the amplifier has a very low input impedance--most of the source voltage will drop over the Thevenin equivalent resistance. With a very high input impedance, however, the majority of the signal voltage will be transferred from the source to the load because in the above equation, if Rload >> Rthevenin, Vload is approximately equal to Vsrc. if an amplifier has low impedance input the f/b must be low impedance also which make it in practical to use. The hi impedance of a typical amplifier is because the input is one two diodes basically operating on it exponential curve. Making it virtual the same as the other diode. for a differential amplifier. Boltzmann constant will define the impedance of a single diode.
When you prove a voltage to be induced using a Simpson why does the input impeadance on the different voltage scales produce different readings for the amount of voltage?
One possibility is that the accuracy of the Simpson is different on the different scales. Another (more probable) possibility is that the impedance of the Simpson on the different scales is sufficiently different so as to affect the reading. This is a common issue with low impedance multi-meters. Lets say you are using a typical Simpson meter with 20,000 Ohms per Volt. On a three volt scale, that means the meter itself has an impedance of 60,000 Ohms. On a 60 volt scale, however the meter has an impedance of 1,200,000 Ohms. Depending on the circuit impedance, that can have a significant impact on the final reading, which must be taken into consideration. Look at the equation for parallel resistance: RT = R1R2 / (R1+R2). If the meter impedance changes the circuit impedance by more than, say, 5%, that is going to affect the observed value. (You pick the percent limit - it depends on the situation.) Even for the case with a high impedance meter, say a 10,000,000 Ohm Digital Multi-meter, impedance must be considered if the circuit impedance is high enough. (I have a WWVB receiver that requires a 1,000,000,000 Ohm voltmeter to correctly measure the AGC voltage - no ordinary digital multimeter will suffice.)This does not mean that you have to spend lots of money on a high performance, high impedance, meter. You simply have to consider what the impedance of the meter is going to do to the circuit, and calculate that impact, before you state the results.
What are the advantages when using a digital voltmeter?
The electronic voltmeter has higher input impedance than other voltmeters, such as traditional VOM's. As a result, it loads the circuit under test to a smaller extent, introducing a smaller error in measurement. Many electronic (or digital) voltmeters have an 11 Megohm or 20 Megohm input impedance, as opposed to a typical 20 Kiloohm per volt impedance of a typical VOM with a 50 microampere movement. Some high end electronic voltmeters have an input impedance well into the thousands or millions of Megohms.
What conditions are required to make a prime number?
Divisible only by 1 and itself.
Are weather conditions cause by the sun earth and air?
Mostly by earth itself
Is either a conjunction?
No, "either" is not a conjunction. It is often used as a determiner or pronoun to refer to one of two things.
