| Voltage, Current, and Power Interview Questions |
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I have an infrared LED and an infrared receptor (which looks similar to an LED).
Now, my idea is to create an obstruction sensor such that when the obstruction is removed from the path of the light from the IR LED, the sensor should pick up the light from the LED and should let current pass (to a transistor, which then opens up to power an alectromagnet).
Can u please help on how to realise this.
Please also help me with information on supply voltage and the polarity.
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CALCULATE THE VOLTAGE NEEDED TO RUN AN AIR COINDTIONER THAT USES 1800 WATTS OF POWER, HAS A RESITANCE OF 24.5 OHMS AND DRAWS 8.57 AMPS OF CURRENT.
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has 1 bar in ceramic block which glows when current feeds it. 24v? i tried hooking ignitor to battery charger [12v] & got nothing, not even a spark when touching poles. i tested power feed lines to ignitor,on furnace, & got power but i dont know if enough to warm ignitor at all. 3v continuity tester also coulnt flow thru ignitor . furnace codeflash indicates: ignitor failure, low voltage, bad ground.
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consider the circuit here
http://i140.photobucket.com/albums/r29/f...
What will be the current passing thru the 4 ohm resistor if the voltage passing thru the 6 ohm resistor is 24V?
a. 5 A
b. 1 A
c. 2 A
d. 4 A
e. 8 A
What is the power dissipated on the 4 ohm resistor?
a. 1 W
b. 2 W
c. 4 W
d. 8 W
e. 16 W
If I use Kirchhoff's loop rule
0 = 24 +6I - 4I - 8I
I = 4 A
P = I^2*R = 4^2* 4 = 64 (not in the choices)
what is wrong with my solution
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in this paragraph i wrote about why nikola tesla should be the greatest american and i was wondering if you can help me put like an ending sentence to what makes him the greatest american. i can't think of anything that makes a catching ending sentence >.> THANKS
One of Tesla's greatest ambitions, was his determination to make electrical power equally available to everyone, especially those in poverty. Before Tesla, all motors ran on direct current only, and that gave Thomas Edison and his "d.c. only" system the edge. Then Tesla created the alternating current motor, giving a.c. the final edge it needed. A.c. was superior because of the easy transformation of voltages. A.c. could be generated remotely at very high voltages, then "stepped down" to accommodate local distribution and home use. In comparison, Edison's system depended on the generators being close to the users, and depended on very large conductors between the source and destination to avoid large power losses.
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I can get it pretty cheap i just wanna make sure it will work and its not dodgy because im getting it online
Kingston Brand New Original
Kingston 2GB MicroSD (TransFlash) Memory Card with Free SD Adaptor
Capacity :2GB
Operation Voltage : 2.7 ~ 3.6 V.
Support SD & SPI Mode.
Low Battery Consumption to Maximize Battery Life in Small Protable Devices.
No. of Pins : 9pins.
Non-Volatile Solid-State, Data is not Lost When Power is Turned Off.
Security Level Complies with Both Current and Future Security Digital Music
Initiative(SDMI) Portable Device Reqirements.
WithAdapter(Included):Any SD-slot device inc PDAs, digital cameraetc.Plus manyother MP3players, digital cameras, PDA's and mobilephones.
MobileCompatibility List:
Nokia 6131 6136 3250 6233 6234 6300 N95 E65
Just wanna know if its a good bran etc, Thanks remember i have a Nokia 6300
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A 25.0 ohm lamp and a 5.0 ohm lamp are connected in series and placed across a potential difference of 54.0 V.
(a) What is the equivalent resistance of the circuit?
___ ohms
(b) What is the current in the circuit?
___ A
(c) What is the voltage drop across each lamp?
___ V (25.0 lamp)
___ V (5.0 lamp)
(d) What is the power dissipated in each lamp?
___ W (25.0 lamp)
___ W (5.0 lamp)
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Currently the transformer supplies the backup invertor and then the backup invertor supplies to the light phase(5A wiring) in series. This way the voltage correction is not very effective (i live in an area with a lot of fluctuation in voltage ) . What i wud like is to connect the tranformer and the invertor in parrallel to the light phase so that voltage correction is more efficient ..... but the problem with that is ... then when the mains power fails and the backup invertor turns ON then the current will also flow from the invertor to the light phase to the transformer ,back into the supply from the govt. cable and to other ppl's houses!!! pls help
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Do complex numbers have any significance in economics?
I know that in electronics, there is such a thing as "j" amps: it means a current of 1 amp, delayed with respect to the voltage waveform so that the peaks and troughs of one waveform align with the zeros of the other. You still get a voltage drop along the wiring because of its resudual resistance, but there is no actual power dissipated in the load because of the misalignment. Likewise, in some subdisciplines of mechanical engineering, you get complex numbers cropping up.
Do complex numbers ever occur in economics? And if so, what is the significance of "j" pounds or dollars?
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While looking at circuit schematic of a rectifier, i recalled that i once came across a fine article on diodes that read "diodes offer almost zero resistance to current hence theres a greater possibility that it can burn out at high voltage operation, hence its advisable to use resistors before every power diode" So, does it mean that we should place resistors before diodes or is it just for high voltage operation? if so, how should resistance be matched with every diode?
I'm really confused because the schematics for rectifiers dont show a resistor placed before any of the diodes :-s
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I have two diagrams with all bulbs at equal wattage( equal resistance). Each circuit has 3 bulbs, first circuit bulb 1 bulb 2 bulb 3! Second circuit bulb 4 bulb 5 bulb 6. The only difference is bulb 5 has a short circuit. The voltage is 4.5V.
Please Help Me Answer These Questions.
a.In which circuit is the current greater?
b.In which circuit are all three bulbs equally lighted.
c.What bulbs are the brightest?
d.What bulb is the dimmest?
e.What bulb have the largest voltage drop across them?
f. Which circuit dissipates more power?
g.What circuit produces more light?
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SIMPLE CIRCUIT:
In the circuit , a 10 V battery is attached to a light bulb of resistance R = 4 ohms.
a.) What will be the current flowing out of the battery?
b.) How much power is dissipated in the resistor? (Tells you how bright it glows.)
c.) Which labelled point in the circuit is at a higher voltage, A or B?
SERIES CIRCUIT:
In the circuit, the same 10 V battery is now attached to two light bulbs in series, each of the same resistance R = 4 ohms.
a.) What is the total resistance of the entire circuit?
b.) What will be the current flowing out of the battery?
c.) How much power will be dissipated in each resistor? (Tells you how bright each glows.)
PARALLEL CIRCUIT:
In the circuit, the same 10 V battery is now attached to two light bulbs in parallel, each of resistance R = 4 ohms.
a.) What is the total resistance of the entire circuit?
b.) What will be the current flowing out of the battery?
c.) How much power will be dissipated in each resistor? (Tells you how bright each glows.)
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I tried solving this and I'm not sure what I've done wrong. See work below. Thanks.
Determine the power dissipated in the 6.0 resistor in the circuit shown in the drawing. (R1 = 3.0 , R2 = 6.0 and V1 = 14 V.)
? Watts
http://www.webassign.net/CJ/20_67alt.gif
I attempted to solve this by first calculating the resistance of the parallel resistors. I came up with 2.11 ohms. I then treated these 4 resistors as in series with the 3 ohm resistor to come up with a 5.11 ohms of total resistance. I calculated the total current as 2.739 A
I then calculated the voltage across the 6 ohm resistor using V= (2.739 A)(2.11 ohms) which was equal to 5.779 V which I divided by 6 ohms to get .96 A. I used P=I^2 R and got 5.526 W as my answer but it is incorrect and I have no idea what I've done wrong. Any ideas?
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Determine the power dissipated in the 6.0ohms resistor in the circuit shown in the drawing. (R1 = 4.0ohms , R2 = 6.0ohms and V1 = 15 V.)
W?
http://6.uploadhouse.com/fileuploads/115...
Req=4+(7*3)/10
6.1 ohm
Current is
15/6.1
2.46 amp
The voltage across the paralell structure is
15-4*2.46
5.164 volts
The current throgh the 6 ohm resistor is
5.164/7
0.7377 amp
Power = V*I and V=I*R
so power is I^2/R
Power=0.7377^2/6
0.09 watts
but .09 is not right, do you see where i made a mistake?
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