Electronics Engineering

Passive Infra Red. Units that detect movement.

the quick answer is you cant. squirrel cage induction motors need AC power to generate the magnetic field in the rotor...if however you have a motor that has brushes in it (drills, mixers,etc) then this motor can be run off ac or dc..the commutator brushes do the work of moving the current to the proper coils at the proper time.

A: Overload is a term used to describe a condition whereby a circuit is overstressed by additional loads imposed to it. Overload can be as small quantity or a very big quantity it depends on the circuit as it is stressed.

Digital codes are represented by strings of binary digits, zeroes and ones. There are a wide variety of coding schemes for transmission of codes.

If we take a quick look at the binary codes for zero through 15 it would look as follows, with the least significant bit being at the right.

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

The least significant bit represents 1's, the next 2's, the next 4's and the most significant bit on the left represents 8's. Therefor for 15 we have 8+4+2+1 and so forth for the other codes.

You can use the same scheme for representing characters. Google for ASCII codes to see an example.

The four bit code above is how one represents hexadecimal numbers. Another favorite form is Octal representation which looks at 3 bits at a time. Hence, 111111 would be Octal 77 or decimal 63.

ic voltage regulator is good for line or load regulation

Tape recording is a magnetic process; sound is translated into a pattern of magnetic fields which is recorded by a magnetic material (usually ferric oxide) that coats a plastic tape.

switching on the tourch for example is when as you switch it on it will move the switch connecting it to a circuit that lights up the bulb and when you switch the switch off the switch will move breaking the circuit

A capacitor can charge to its' maximum OR the voltage applied to it, whichever is LESS.

yes of coarse but use some modulating technique such as pam,pwm or ppm this will increase distance between reception and transmission without much distortion

A desoldering pump is a device that removes molten solder during the process of soldering. The molten solder cools in the pump and can then be removed.

one pushes while the other pulls and vice versa.

One can buy a comparator cheaply from both eBay and Amazon online. One can download software for them from sites such as 'ApSIC Localization Solutions.

If the menu is also upside down then get it serviced by the manufacturer. If the menu is ok but the picture is upside down, then the issue could be with set top box, or transmission service. You may have to do factory reset as per the manual.

To charge a battery, the charger voltage must be higher than the battery voltage. If a AA battery (or any other) has a normal voltage of 1.2V the charger voltage must be at least 1.2V. The type of voltage supply and its current capability is immaterial. No, the charging voltage have to be larger than the battery voltage, to charge 2 AA batteries, that is 3 volt if connected in series, so a voltage of at least 4 volt is needed

If you're going to use a function generator, presumably you'll be natural sampling a sinusoid for SPWM. In order to generate unipolar spwm, you'll need two sinusoids that are 180 degrees apart. So take the output of your sinusoid into two unity gain buffers, one inverting and one non-inverting. This will also allow you to set your reference for the sinusoids as compared to the (ONLY ONE!) sampling waveform, which can either be a triangle or a sawtooth. A sawtooth can easily be generated using a single comparator, by the way. One side of the bridge driven from the buffered sample ouptut and the other from the inverted and buffered sample output. It is my understanding that unipolar spwm has the advantage of higher order harmonics, as compared to bipolar, which makes them easier to filter out. The tradeoff is greater complexity because of the required extra input waveform. Good luck.

Darlington is configured differently than most NASCAR tracks, with it's distinctive egg-shape. It is especially known for it's exciting and often attrition-filled races.

Please write in paragraph form and remove hard returns. Please also delete this comment. Thanks! The difference between the measured value and the true value is called as:a) Relative errorb) Absolute errorc) Gross errord) Probable error

First, we assume you are asking for the resistance of the bulb. Second, there can be only one possible resistance. The bulb has one and only one resistance. The equation for solving this problem may be a quadratic or even a cubic equation, which may have multiple roots, but only ONE can be a real-world answer. Third, let's list everything we know about the circuit. Let's label the power source Vcc. We'll label the known resistor R, and the resistance of the bulb Rb. We'll label the power dissipated by the bulb as simply P. We'll also label the current flowing through the bulb as I. What do we know about power? The power dissipated by a resistor is given by three well-known formulas: P = IV, P = I2R, or P = V2/R. (The second and third are derived from the first.) The second one will come in handy for us right now. We can write: [Equ. 1] P = I2Rb What else do we know? Since it's a series circuit, the current flowing through the bulb, I, is the same as the current flowing through the resistor. And we know from Ohm's Law that I = V/RT, where RT is the sum of the two resistances in the circuit: R + Rb. So, we can write: [Equ. 2] I = Vcc/(R + Rb) So, substituting for I in Equ. 1 with the I we found in Equ. 2 we have: P = [Vcc/(R + Rb)]2Rb = [Vcc2/(R + Rb)2]Rb = Vcc2Rb/(R2 + 2RRb + R2b) Phew! The algebra's getting a bit hairy. We need to do a few more manipulations. We can swap the P on the left side of the equal sign with the denominator of the fraction on the right side of the equal sign, so we write: R2 + 2RRb + R2b = Vcc2Rb/P We now multiply both sides of the equation by P and we write: PR2 + 2PRRb + PR2b = Vcc2Rb Stay with me now; we're almost there. If we subtract Vcc2Rb from both sides and gather similar terms, we can write: PR2b+ (2PR - Vcc2)Rb + PR2 = 0 We can now substitute all the known values into the equation above. We know Vcc = 131, R = 133, and P = 22.9. So, we can write: 22.9R2b - 11070Rb + 405078 = 0 So, we have quadratic equation, which will have two roots for Rb. Using the quadratic formula, we determine that there are two real roots: Rb = 443.5 and 39.9. The second result is thrown out since it is not consistent with Equ. 1 and Equ. 2.

deflect more

checking if it is an energy signal

E= integration from 0 to infinity of t gives infinity

so it is not an energy signal

P=limit ( t tending to infinity)*(1/t)*(integration from 0 to t/2 of t)

gives us infinity

so it is not an energy or a power signal

um66 is a melody generator cmos ic

it is a three terminal ic

pin no: 1 is set to ground,

and pin 2 is Vdd

pin 3 is the output melody signal

this ic generally operates from 1.5 to 5 volts

it is generally used in toys, bells etc................

The easiest way is to use a logic family such as CMOS that can operate at supplies up to 15 volts.

If you're using 5-volt TTL, the only way to get more than about 4 volts signal output is to use a transformer.

A sine wave has no harmonics. It only has a fundamental, so the value of the 2nd, 3rd, and 12th harmonics of a sine wave is zero.

The EMF of a cell is the voltage across the terminals at zero current. This is the quoted cell voltage but as soon as a current is dawn from the cell, the voltage will drop. It's due to the internal resistance of the cell.

In a circuit diagram, a cell is often shown as a voltage source (a perfect source) and a resistor in series to represent the internal resistance. Using Ohms Law, it can be seen that as soon as a current flows, a voltage will be developed across the internal resistance, so reducing the voltage that is seen at the terminals of the cell. The higher the current draw, the higher the voltage drop inside the cell.

Normally, the voltage drop is minimal but in most cells, as it loses charge, the internal resistance rises. Eventually it will reach the point where most of the voltage is dropped across the internal resistance, leaving little to drive the intended load. Often, if a battery is removed from a device and measured, the voltage will be measured as equal to or very close to the quoted cell voltage. It is easy to make a judgment that a battery is good when it is almost dead. The only way to confirm the state of the battery is to measure the voltage at the terminals while the load is attached. The results can be very different to the off load voltage.

Alkaline cells have a low internal resistance compared to other dry cells. This makes them well suited for high current drain applications. The internal resistance also rises more slowly than most other cells, so they remain useful far longer than zinc-carbon types.

Kirchoff's Current Law: The signed sum of the currents entering a node is zero.

Assume the top of the battery is a node. The current entering it (from the battery) is equal to the sum of the currents leaving it (to the branches). This adds up to zero.