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Generally speaking, a relay is an electrically operated switch. As such, there are two sets of ratings associated with a relay. The operating voltage will be specified, and the ratings of the contacts are also set down.
As regards exceeding the operating voltage, this will cause excessive current to flow in the relay when it is energized. If the voltage is not significantly higher than the rating, and if the relay is not operated often or for long, it will function normally, but probably with a shorter life. At some point, excessive operating voltage will cause the coil in the relay to fail.
Operating a relay to energize an electrical circuit with a higher voltage or current than the relay is designed for will shorten the life of the contacts in the relay. Higher voltage and current will quickly burn the contacts and destroy them. This is particularly true if that voltage or current is significantly higher than the ratings set by those who designed and constructed the relay. Things might work for a short time, but not for long.
What else can I help you with?
Why does the current in a transformer fall when voltage increases?
What happens to the voltage as the load is increased is determined by the characteristics (e.g. impedance) of the source. Let's look at Ohm's Law: E=IR Increasing load means lowering the resistance. Imagine putting a second R load in parallel with the first. The current will double. But it is not necessarily true that the source is capable of providing the increased current while maintaining its output voltage. Generally this information is provided by the Power rating of the supply. A supply where E=12 volts and the power rating is 120 watts, will deliver 10A continuously 12V X 10A = 120W. Beyond 10A the voltage cannot be maintained. Hope this helps!
What does the voltage regulator do?
it controls the amount of voltage that is flowing in a circuit. To be more accurate, a voltage regulator provides a limiting mechanism to an electrical or electronic power source. In the case of modern automotive alternators the voltage regulator "regulates" the output from the alternator by limiting the current flowing through the rotating field assembly inside the alternator. The weaker the electromagnetic field of this "rotor" the lower the output (both current and voltage) of the alternator. When the electrical load on the alternator is low the regulator reduces the field strength based on it's set upper voltage limit. When the load increases beyond the capacity of the "limited" output the voltage drops (the load is kind of "shorting out" the power from the alternator) and the regulator allows the field current (and therefor the output of the alternator) to rise until the voltage reaches that "upper limit" again and so on. Hope this helps
What happens when the switch is open in a circuit containing wires a light bulb a switch and a battery?
When you open a switch to a light for example, you cause a break in the flow of current to the light. Opening a light switch means you are turning the light off by stopping the flow of current to it. Closing a switch means you have completed a circuit (basically a loop) and current now flows causing the light to go on. Think of electricity like water. To cause your bath tub to fill up (light on) you have to also complete the circuit by turning the tap. Now we have water flow. (current) To stop the flow of water, we want to open the circuit by turning the tap in the opposite direction causing a break in the flow of water.
How many days to get to Pluto and beyond?
That would depend a lot on the speed. With current technology, it takes several years.
When calculating kilowatts from known amps on a 3 phase circuit do you use 415 volts or the 240 line volts in the equation?
You can use either. I assume the 240 volts is the line to neutral voltage, and the 415 is the line to line voltages. They will both give you the correct answer. It is important to note these values are RMS, not 0 - peak voltages, but this may be beyond your question. The equations below are for calculated from RMS values (both voltage and current).If you are using a L-L voltage, P = I*V*sqrt(3)If you are using the L-N voltage, 1-phase power P = I*V (for the power in a single phase, for all three, multiply by 3), or 3phase power P = 3*I*VYou will get the same answer, since the L-N voltage is (1/sqr(3)) times the L-L voltage.
On what part of the curve is a reverse biased diode is normally oprated?
A reverse-biased diode is typically operated in the reverse breakdown region of its current-voltage characteristic curve. In this region, the diode allows a small reverse current to flow, which is generally negligible until a certain breakdown voltage is reached. Beyond this breakdown voltage, the diode can conduct significant current, but in normal applications, it operates below this threshold to prevent damage. Thus, the diode primarily remains in the reverse bias region with minimal current flow.
What happens if you use a fuse that's overrated for voltage but has the correct current rating?
Using a fuse correctly rated for current but "overrated" for voltage does not present a problem. Current ratings are critical safety issues, and fuses should be replaced with those of the same current rating. But using a fuse with an identical current rating but a higher voltage rating is not a problem. The reason for that lies in what the voltage rating of a fuse is. Fuses are given a voltage rating to state a maximum voltage in a circuit that they are designed to protect. And the voltage rating has nothing to do with the "normal" operation of the fuse. The fuse carries current when it operates normally, but when something happens and excessive current flows, the fusible link heats up and opens. This is where the voltage rating comes into play. It is possible that a fuse can arc through when it fails. It is the voltage rating that stands in the way of this. As long as the voltage rating of a circuit is not beyond the voltage rating of the fuse, that fuse will fail safely when it fails. It is acceptable to use a fuse of an equal current rating but a higher voltage rating when replacing a fuse that has failed.
Knee pont voltage- briefly distinguish?
The knee point voltage of a CT is the voltage at the "knee" of a I-V characteristic (if you increase voltage, and plot this voltage with respect to the current flow, you will see a logorithmic type response). The knee is usually specified as 10% distortion (ie, the voltage is 10% less than you would expect relative to the current flow). beyond the knee point, the CT is considered in saturation. This applies to amplifiers / transistors as well as CTs. Transistors used as ampifiers are operated in the "linear region", or the region below the knee point of that particular transistors I-V characteristic.
Why the alternator do not operate near the knee of its saturation curve?
Actually, alternators (synchronous generators in other parlance) are normally operated slightly saturated by design. If they were operated completely unsaturated, the machine is not being used to its full capability. We don't attempt to operate them beyond the knee, because the losses increase significantly, and the output voltage & current are distorted. It is important to note that there are two different windings to consider when discussing saturation. The DC field (rotor) winding is usually operated somewhat in the saturated region. The AC armature (stator) winding is operated near - but below - the knee to avoid output distortion.
What is meant by pinch off voltage in fet?
Pinch-off voltage in a Field Effect Transistor (FET) refers to the specific gate-to-source voltage at which the channel conducting current begins to 'pinch off' or constrict. Beyond this voltage, increasing the gate voltage does not significantly increase the drain current; instead, the current becomes relatively constant as the channel narrows and limits the flow of charge carriers. This phenomenon is crucial for the operation of FETs in saturation mode, where the device is used for amplification and switching applications.
What happens if the current is constant and voltage increases?
Generally the voltage is constant and current varies as per the load. Load can vary and hence current can vary. You are stating an abnormal situation, where in voltage increases while current remains constant. I am assuming a constant load situation then normally when voltage increases, the current tends to reduce since over all load remains same. If the voltage goes up beyond a limit the insulation fails and may lead to short circuit, equipment failure, shock and fatality
What is knee poit volt and what is c t saturation?
It is the maximum voltage across the secondary terminals beyond which a CT will saturate when it is loaded.If the CT gets saturated,the ratio of primary current to the secondary current will not be to the designed value.
What is maximum voltage of a reverse bias diode explain and also defined?
The maximum voltage of a reverse-biased diode, known as the reverse breakdown voltage (or reverse voltage rating), is the maximum reverse voltage that can be applied before the diode begins to conduct in the reverse direction, potentially leading to breakdown. Beyond this voltage, the diode may experience a sudden increase in reverse current, which can damage the diode if not controlled. The reverse breakdown can be utilized in certain types of diodes, like Zener diodes, for voltage regulation. In general, reverse bias prevents current flow until this breakdown voltage is reached.
In what manner will the inverter of a variable frequency drive behave during a voltage sag condition?
Within permissible limits the VFD continues to drive the torque. However for a longer duration and beyond a limit voltage drop results in an increase in current, resulting increased heat and finally the motor halts if voltage further deteriorates .
Can a watt meter that has current through its current coil and a potential across its voltage coil indicate zero?
A watt meter that has current through its current coil and voltage across its voltage coil will indicate zero if the power factor between the volts and amps is zero. This condition would be one in which the current will either lead or lag the voltage by 90 degrees and the circuit will have amps, volts, VARS, and VA, but will not have Watts.
Can you saturate a CT with the secondarys shorted?
A CT becomes saturated when the induced secondary voltage is beyond the capabilities of the CT. Yes it is possible, if enough current is forced through the secondary to raise the voltage beyond the CT capability curve. Remember the CT has inherent resistance due to the secondary winding. That said, to saturate strictly based on the CT resistance is unlikely. But it is still theoretically possible.
Draw and explain the UJT characteristics?
The Unijunction emitter current vs voltage characteristic curve figure(a) shows that as VE increases, current IE increases up IP at the peak point. Beyond the peak point, current increases as voltage decreases in the negative resistance region. The voltage reaches a minimum at the valley point. The resistance of RB1, the saturation resistance is lowest at the valley point.Ip and Iv, are datasheet parameters; For a 2n2647, IPpand IVvare 2µA and 4mA, respectively. Vp is the voltage drop across RB1 plus a 0.7V diode drop; see Figure (b). Vv is estimated to be approximately 10% of VBB.
