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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
Difference between gunn diode and pin diode?
gunn diode is transfered electron device & PIN diode is semiconductor device
Using blavier test to check for cable faults?
This is use to locate the ground fault of a single cable. This test is performed with the aid of a low voltage supply with an ammeter or voltmeter. In this test the resistance between one end of the cable C1 and the earth is first measured with the other end, C2 isolated to earth and then with the C2 earthed. Let the two readings be represented by R1 and R2 rep! If r1 and r2 are the conductor resistance of the length of cable at C2 to fault and at C1 to fault rep!, and "r" is the resistance of the fault to earth
Which transistor configuration having 180 degree phase shift?
... in CE config.. as Vo=Vcc-Ic Rc nw as Vcc is const. v cn say Vo is directly proportional 2 (const.-Ic Rc) so nw when d base voltage is increased Ib will inc. thus, Ic will inc. cuz Ic=beta Ib thus Ic Rc drop increases... as Vo is prop 2 (const.-Ic Rc) thus, Vo will decrease for inc. in Ic...thus d curve for d +ve half cycle in which Ib is increasing drawn in 180 deg phase showing Vo is dec. n vice versa 4 -ve half cycle...
nw in CC... as d o/p is taken frm emitter... Vo= Ie Re so der's no -ve sign..n for d +ve cycle Ie will incr(dat's leakage current) as der's no-ve sign it'll b in phase wid i/p..hence no phase shift as leakage current doesnt incr. so much thus,d o/p voltage Ie Re doesnt inc much thus, d o/p doesnt amplify much thus dis config knwn as emiter follower...
nw in CB,... d I/P is appiled at emitter n o/p is taken frm collector... so, 4 d +ve half cycle d EB junc will b less FB thus causing dec in Ib n thus, in Ic... so dec in Ic causes inc. in o/p voltage(cuz Vo=const.-Ic Rc)..n during -ve half cycle d junc will b more FB(inc in Ic) so o/p voltage will b decreasing..hence der's no phase shift...
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Can the rpm on a sears 1 hp electric motor be changed?
yes, you can. in A/C motor rps(rev/sec) is N(rps)= f/p where f is the frequency and p is the number of pair poles. By increasing or decreasing the frequency the rps will change accordingly. By having less or more pair poles the rps will be increased or reduced. the more pair poles the lower will be the rps.
What are performance parameters of rectifier?
Rectifiers have two parameters of interest - maximum forward current - and maximum reverse voltage. A rectifier is generally chosen based on these two parameters.
Other parameters include turn-on/turn-off time, forward voltage drop, and capacitance. The importance of these and other parameters depends on the circuit design requirements.
Wire 1 has radius R, and wire 2 has radius 2R. Wire 1 has length L and wire 2 has length 2L. Since resistance is inversely proportional to cross sectional area and proportional to length resistance of wire 1 is:
c*L/pi*R^2, where c is a constant based on the material.
The resistance of wire 2 is:
c * 2L/pi *(2R)^2, or squaring the 2R term we have c * 2L/pi * 4R^2, and simplifying we have: c*L/2pi*R^2.
Now, to get the ratio of wire 1 to wire 2 we divide equation 1 by equation 2 and simplify:
c*L/pi*R^2 / c*L/2pi * R^2. When you divide a fraction by another fraction you invert and multiply, so this becomes
c*L *2piR^2 / c*L *pi * R^2. Each of the variables cancel leaving the ratio of resistance of wire 1 to wire 2 is 2:1, or wire 1 has twice the resistance of wire 2 even though wire 2 is twice as long.
An Alternative AnswerResistance is directly proportional to length, and inversely proportional to cross sectional area. The cross-sectional area of a circle is proportional to the square of the diameter. So, conductor 2 is twice as long, so you can double its resistance; but its diameter is doubled, so its resistance is then reduced by a quarter. So the resistance of conductor 2 is half that of conductor 1!
Does the power supplied to trip shunt trip breaker have to be applied momentarily or can it latch?
A shunt-trip breaker trips when voltage is applied to the coil. It does not self-reset when the voltage is removed, the breaker must be reset manually. The trip voltage can either latch or be applied momentarily, but must be de-energized before the breaker will reset.
What does it mean when motor ticks?
i have a metal sleeve over each spark plug wire at the motor.....cant get the wire out to get to the plugs?? how do i get the wire off the plug??
A 100W 250 V 200W 250V is connected in series across a 500V AC which bulb will fuse first?
both will be fused.
What are the input and output energies of a fridge?
These are the energies that are used at different times for the fridge. There will be different energy levels depending on how hard it works and if the door is open or closed.
What is the difference between circular relay and shuttle relay?
the difference is that in a shuttle relay, you run back and for(e.g. you run 100m to the next person and they run back to where you started, where the next person is waiting e.t.c.). A circular relay is around a track e.g. a 400m track you each run 100m going round the track. hope this has helped.
What do i do if my Outboard motor wont go forward but will reverse?
Kind of a broad question: what are you attempting to accomplish? get back to shore? fix the thing? How large is the boat, how large the motor? What materials are at hand to use? If attempting to get to shore and the water is calm, simply go backwards all the way. If not, and such is possible, remove the motor from the transom and clamp it on the side of the boat, you should be able to use reverse to go forward. Really, I think you are going to have to re-write this question.
a: 3.5v/0.014= 250 or 252 a 8 bit conversion will cover that.
7200 volt 25kw transformer how many amps at 240 single phase?
The formula you are looking for is I = W/E. Amps = Watts/Volts.
Current and voltage readings taken on the far right of a meters scale provide greater?
Current and voltage readings taken on the far right of a meter's scale provide greater inaccuracy that readings taken from mid scale on the meter.
Do I need transformer for 800 feet 240v electric power line?
You may need to step up the voltage to minimize the power loss / voltage dip at the end of the 800 feet depending on:
1. the type of cable used (the higher resistance, the large cables would be required to not do this)
2. the amperage that you will need to draw.
The voltage drop in the wires is due to the I^2*R losses in the cable. If your cable has a resistance of 1 ohm over 800 feet, and you draw 5 amps (so your load resistance is R = V/R = 240 / 5 = 48 ohms), the actual current delivered will be (I = V / R = 240 / 49) 4.9 amps, and the voltage at the end of the 800 feet will be (240 - 4.9*1) 235 volts. voltage is probably fine for whatever equipment you're using. If used continuously, the 800 ft of cable would waste 210kWh of energy/yr (15-35 dollars/year).
If you need 20 amps, the load resistance would be (240 / 20) 12 ohms, current delivered to the load would be (240/13) = 18.5, and the voltage at the end of the 800 feet would be (240 - 18.5*1) 222 volts. This might be too low, depending on the equipment you're using and if you're area has a naturally low voltage to begin with. If used continuously, the 800 ft of cable would waste ~3,000kWh of energy/yr (180-450 dollars/year).
As your question is written, a straight yes or no cannot be given. I hope the above explains enough to give you the knowledge you need to determine this on your own.
A kilowatt-hour is a unit of energy. The amp is a unit of electric current. Current and Energy are fundamentally different things, and cannot be compared directly. To determine the energy dissipated by a current one would need to know the time for which it flowed, and the potential difference through which the associated charge was moved. the product of current by potential difference by time would give the energy. If one assumes the current flows for one second through a potential difference of one volt then the energy in joules would be numerically equal to the current in amps. One kilowatt-hour (kWh) is 3,600,000 joules, so to get the answer in kWh divide the energy, expressed in units of joules, by 3600000 .
Or you can use an easier method, its called Ohms law however one element is missing, the voltage is required. according to Ohms law Watts= Volts x Amps so if the circuit is 120 Volts the answer would be found by 1000 watts (1kW) divided by 120 volts and you get 8.333 Amps forget the Joules unless your a master electrician and fully understand it, even though we use Ohms law because it is sooooo much easier. if you Google the Ohms law chart remember Power is represented by a P Current is Amps and will be represented by C or R, Resistance is Ohms and is represented by a R and Voltage is represented by E
What is a difference between hv motor and lv motor?
the motor having voltage rating above 480 volt is hv motors and motor having rating below 480 volt is lv motors.
generally motor above 185 kw is hv or mv motor and motor below 185 kw is lv motor.
What are the limitations of z-transform?
The main limitation is that the z-transform is appropriate only if the underlying data are normally distributed. It is also important that the estimates for the mean and variance are sufficiently accurate.
What is difference between ofc and copper wire?
The main difference is in the price. Oxygen-free copper is sold to people with plenty of money for loudspeaker wire, but provided the wire has less resistance than one tenth of the speaker resistance (usually 4 or 8 ohms), the resistance of the wire is immaterial, and ordinary copper wire is perfectly all right.
Highly refined copper has about 1% better conductivity than the usual variety. That difference in conductivity is insignificant for audio use and is also produced by a 3 degree C temperature rise in copper.
no voltage will be induced on the secondary side of the motor as the windings will become saturated.
Which electrical test requires the supply to be energized?
Testing of an electrical circuit for voltage requires the supply to be energized.
