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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 else can I help you with?
What is loading effect of op amp?
An 'ideal' op-amp has infinite input impedance, and real ones that you can buy for anickel come pretty close. That means that the input impedance or "loading effect" ofan op-amp all by itself is nearly zero, and the loading effect of a circuit that youbuild with an op-amp is completely determined by the components you choose toconnect to it. You design the circuit to load the previous stage in any way you want.
What is voltage regulation?
Voltage regulation:(from point of view of electrical machines or generator): It is the change in voltage in between the full loaded and no loaded condition. When there are no loads connected the terminal voltage is equal to the generated voltage in the generator. But when load is connected the terminal voltage is found to be lass than the no loaded condition, due to armature resistance leakage reactance.This phenomena is expressed as, % reg=(Vnl-Vfl)/Vfl * 100%.Which is Voltage regulation. ************************************************************ An ideal voltage source has zero internal impedance. A practical one, even a good one, has internal impedance. With no load on the source, the terminal voltage will have a given value. Once a load current is drawn there will be a voltage drop across the source's internal impedance, and the terminal voltage will therefore drop. The higher the load current, the higher the voltage drop. A regulator circuit, added after the source, can counter the effect of the source's impedance and maintain an output voltage which is more constant than the source itself can achieve.
The ability of conductor to induce voltage in itself when the current changes?
The ability of conductor to induce voltage in itself when the current changes is called inductive reactance.
What are the limitations on forward bias?
A: That will be the power dissipation of the device itself
What is amplifier roll off?
A practical amplifier will contain several components of a "shunt" capacitance inherent in the transistor and physical wiring of the amplifier circuit. As the frequency of the input signal increases, the reactance of these shunt-capacitances will decrease until, at a frequency determined by the value of the shunt-capacitance and the circuit impedance, signal attenuation begins to take place. Thus the shunt capacitances limit the high-frequency response of the amplifier (note that the transistor itself also has inherent limits to it's high frequency amplifying capability). In the case of operational amplifiers, many operational amplifiers are internally compensated by a small capacitor (e.g. about 30pf for a 741). The internal frequency compensation capacitor prevents the operational amplifier from oscillating with resistive feedback.
What conditions could the total impedance of two impedances in series be less than that of either one by itself?
A:resonant
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.
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.
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.
Can you run a 4 ohm speaker at 8 ohms?
The amplifier will have an output impedance of around 0.04 ohms. There will be no 8 ohms. In hi-fi we have always impedance bridging. Zout
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.
Serious nasel problem that will either Aresolve itself B get an operation which should you choose?
let it resolve itself
What conditions are required to make a prime number?
Divisible only by 1 and itself.
