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Electrical Engineering
Electrical engineering is a field of engineering that deals with the study and application of electricity, electronics and electromagnetism.
23,056 Questions
You get high voltage and burn your appliances due to neutral failure?
It would happen in case of 3phase domestic use where consumer side load balancing is not being done and there is loose connection in the neutral wire.
Due to the unbalance and poor neutral, the system neutral will be shifted and un balanced voltages will be recorded in all the 3 phases of the house. The phase loaded with much utilities will indicate less voltage and other will have high voltage.
Gadgets connected to this phase will be damaged.
the voltage which is given for creating magnetic field in a generator is known as excitation voltage.
Current has less of an effect on the performance of a speaker than wattage?
acording to the rule of transactional resiprocity the scintilators do not allow current to over power wattage
What is the standard wire color for PLC inputs and outputs?
There is no specific standard for PLC inputs or outputs, as existing standards wire color codes are based on voltage rather than by function.
Different standards apply in the USA, Europe or other countries, and it depends if you intend to wire a PLC for a domestic application or it is going to another place where the requirements may be different to the local ones. Check it.
These standards my help you:
NFPA 79
UL 508A
IEC 204.1 / EN 60204-1
If you mention where you are located or where is that piece of equipment going, maybe I can give you further info.
Which of the following op-ed article ideas is not in line with the rule of keeping it current?
An op-ed reviewing the WWII accomplishments of American General Douglas MacArthur
What is mean by power frequency withstand voltage and impulse withstand voltage?
Power frequency volatge is applied for a duration of one minute and as such there would be many cycles of sinusoidal wave form applied during the tets; hence, RMS value is considered.
Whereas LI is a surge and is typically 1.20 micro sec. wave; hence peak is considered.
Difference between inverter and pwm control?
One invert the function or a signal. the other is a pulse width modulator controller which are usually used to convert one source of power to another. As in 12v dc in 5v dc out or vice versa
a device designed to measure small electric currents; the simplest electric meter
A galvanometer is an instrument for detecting and measuring small electric currents.
Or use this long answer: galvanometer , instrument used to determine the presence, direction, and strength of an electric current in a conductor. All galvanometers are based upon the discovery by Hans C. Oersted that a magnetic needle is deflected by the presence of an electric current in a nearby conductor. When an electric current is passing through the conductor, the magnetic needle tends to turn at right angles to the conductor so that its direction is parallel to the lines of induction around the conductor and its north pole points in the direction in which these lines of induction flow. In general, the extent to which the needle turns is dependent upon the strength of the current. In the first galvanometers, a freely turning magnetic needle was hung in a coil of wire; in later versions the magnet was fixed and the coil made movable. Modern galvanometers are of this movable-coil type and are called d'Arsonval galvanometers (after Arsène d'Arsonval, a French physicist). If a pointer is attached to the moving coil so that it passes over a suitably calibrated scale, the galvanometer can be used to measure quantitatively the current passing through it. Such calibrated galvanometers are used in many electrical measuring devices. The DC ammeter , an instrument for measuring direct current, often consists of a calibrated galvanometer through which the current to be measured is made to pass. Since heavy currents would damage the galvanometer, a bypass, or shunt, is provided so that only a certain known percentage of the current passes through the galvanometer. By measuring the known percentage of the current, one arrives at the total current. The DC voltmeter , which can measure direct voltage, consists of a calibrated galvanometer connected in series with a high resistance. To measure the voltage between two points, one connects the voltmeter between them. The current through the galvanometer (and hence the pointer reading) is then proportional to the voltage.
How do you install a resistor in an DC circuit?
It depends on the purpose for installing the resistor. If the intent is to decrease current flow, the resistor must be connected in series with the load. If the purpose is to increase current flow, the resistor must be connected in parallel with the load.
To connect a resistor in series, connect the resistor to one side of the power source, in line with the load. This will decrease circuit current flow.
To connect a resistor in parallel, connect the resistor between the positive and negative sides of the power source, which will effectively connect the resistor across the load . This will increase current flow through the circuit. However, before connecting a component in parallel, make sure the increase in current flow will not exceed the current rating of the circuit or fuses/breakers will blow.
What is the function of microphone?
A microphone converts the pressure waves caused by sound into vibrations within a coil which transforms the vibrations into electric signals. These electric signals can then be sent by wire or wireless to an amp or mixer of some sort for output to speakers or perhaps a recording device.
Why is the gain and bandwidth of a ce-cb cascode amplifier greater than a ce amplifier?
Gain of ce-cb cascode is nearly equal to the gain of ce amplifier, because in a ce-cb cascode, the gain of the ce stage is equal to 1, and the gain of the cb stage is nearly equal to an isolated ce amplifier. Hence, gain of both are nearly equal.
On the other hand, Bandwidth of ce-cb cascode is much higher than the bandwidth of ce amplifier because the cb stage in the cascode configration is not subjected to any Miller effect, thereby improving the high frequency response. The absence of Miller effect is due to the fact that the base of the cb stage is grounded thus, shielding the collector signal from being fed back into the emitter input.
To be more clear, the gain of CE stage in cascode is nearly 1, which reduces the miller effect on the cb stage greatly.
What are the peak and RMS values of the voltage of a pulse-width-modulated signal?
A pulse-width modulated (PWM) signal is basically a signal that is turned "on" and "off" at a fixed rate, but is left "on" for varying amounts of time. In the scheme, a fixed maximum value for the pulse is selected, usually by circuit design. That will determine the peak value of the voltage (Vp) of the pulse. The pulse will have some kind of period associated with it. That's the pulse repetition rate (PRR), and it can tell us the frequency of the pulse. We have a time base or clock to set the PRR. The period of a wave (p) which is the time it takes for one complete cycle of the wave, is the inverse of the frequency of the wave. Recall that period equals one divided by frequency (f). We now look at the last factor: the pulse width.
In pulse width modulation, what we concentrate on is the width of the pulse we generate. (Peak pulse voltage and the PRR are almost always fixed.) All we have to do is decide how long to leave the pulse "on" before we turn it "off" at the end of one cycle. The pulse width (the amount of "on" time) could be quite narrow compared to the amount of "off" time. As we increase the amount of "on" time, the "off" time decreases. ("On" time plus "off" time equal the period of our PRR.)
RMS is root mean square, and it's a methematical way to "average" a sine wave. In a voltage sine wave, the area under the curve of a cycle (the RMS value of the voltage for a cycle) is tricky to calculate in a "regular" way. Hey, it's an irregular curve. Not so with a simple PWM signal. It's a modified square wave. Forget RMS. Forget integration. Go with simple math.
A PWM signal is modulated by changing the pulse width (naturally), so a fixed width (tw) must be chosen to find the average value of the output voltage. It's really very simple. Start with the peak voltage. It will have to be known. Then look at the amount of time the pulse is "on" compared to the period of the pulse. A short "on" time will result in a low average voltage. A little longer pulse width will result in a little higher average output voltage. If we reach a point where the pulse is "on" for half the period of the cycle, it will be "off" for the same amount of time. It is "on" for a full half a cycle and "off" for a full half a cycle. If the peak voltage is 100 volts, and the pulse is "on" for half a cycle and "off" for the other half, average voltage is 50 volts. It's a snap to pin it down. The actual average voltage can be calculated by using the following formula:
Vav = Vp x (tw / p)
Let's take a 50 volt peak, 10kHz clocked PWM device. The period of the clock is .1 milliSeconds. (p = 1/f = 1/10kHz = 0.1 mS)
If we have a pulse width modulated signal that has a pulse width (tw) of 1/50,000th of a second (0.02 milliSeconds) we have:
Vav = 50 volts x (0.02 mS / 0.1 mS) = 10 volts average
Finding the average value of a simple PWM device is easy if it's for a fixed pulse width. In a complex application of varying pulse widths, integration is needed, and the calculus is called for. But in the simple application, simple math works.
Not True!
"RMS" is not an "average" by any stretch of the imagination. If it were, a 50% square wave with peaks at 1 and -1 would be 0. (The RMS is actually 1). If it were, the AC from your house receptacles would be 0 volts! (The RMS is actually 120V).
RMS is represents a periodically varying waveform with its equivalent effective DC value. It can be a voltage (V) or a current (I).
For PWM:
VRMS = ( ( C x V12 ) + ( (1-C) x V22) ).5
Where: V1 and V2 are the two peak (top and bottom) voltages in respect to 0, C is the duty cycle in respect to V1 (ie C = .2, duty cycle is 20%). If you plug in the numbers for the 50% square wave above, the answer will agree. It's 1.
The physical interpretation of voltage leading current (by 90o) in an inductor is just that. The instant voltage is applied, the inductor "feels" the applied voltage and responds by "beginning to generate" a reverse voltage (reverse electromotive force or emf, or back emf) which prevents current from flowing for the first instant of time. At time zero, the inductor is infinitely resistive. As time goes on, current is increasing. It is trying to "catch up" to voltage, which is continuing to climb. The back emf is trying to limit current, but it climbs to try to "follow" the increasing voltage. But voltage peaks and starts down. The magnetic field that was building while voltage was climbing (and was "holding back" current flow at the same time) will begin to collapse. The collapsing field generates emf that will try to keep current flowing, so current will continue to increase while voltage is decreasing. The current peaks later on (like 90o later) and then starts to decrease. In a purely inductive circuit to which AC is applied, the current is "chasing" the voltage peaks and is always 90o degrees behind. [There are other ways to look at the situation, but this one is a basic way to interpret the physics.]
Why dont the electrical power lines trip when it rains if rain is a conductor of electricity?
Water is a very poor conductor of electricity and even if it has enough solutes in it to conduct electricity then it will still have a high resistance.
If rainwater were to conduct the electricity then that water would heat up and evaporate almost instantly.
But you need to remember that it would need a continuous channel of water to conduct electricity and the surface tension of water makes this unlikely to happen.
Additional AnswerOverhead conductors are supported using insulators, the shape of which is designed to (a) shed water, and (b) to provide a long conducting path for any moisture between the top of the insulator and the bottom. For high-voltage lines, several insulators may be used in the form of 'strings', providing a sufficiently-long conducting path for any moisture path to have a sufficiently high resistance to prevent conduction.
Haw to calculate the parameters of distance protection for REL 670
2 The air gap between stator and armature of an electric motor is kept as small as possible?
a. To get a stronger magnetic field
b. to improve the air circulation
c. To reach the higher speed of rotation
d. To make the rotation easier
How do you solve circuit analysis using thevenin's and norton's theorem?
its very simple to solve a thevinien circuit first open the terminal across which we have ti calculate the voltage
suppose in a circuit if we have to cakculate across 5ohm then open it
then calculate the rth by open all the independent current source and close all the voltage source...........
after then calculate the voltage
by using nodal
take one as zero and another as vth
calculate the vth
then equivalent circuit is drawn.. in which the load reristance and the equivalent reristance is put in series....
Are there any restrictions for the use of circuit elements in Thevenin's theorem verification?
no thevenins theorem works for every type of element. for a.c. analysis of a circiut consisting of capacitors inductors etc. a different method is followed to find thevenins equivalent but it is valid...
Use of single phase and three phase variac in electrical engineering?
single phase variac can give 0-270 voltage in ac forms. So it is used to know the voltage in any circuit.
