Electrical Engineering

A resistive load is one which does not power a motor. Examples of resistive loads would be light bulbs, toasters, stereo systems, televisions, hot plates and convection heaters. To calculate the wattage required to run these items, simply multiply input amps x volts for each individual item, then add those figures together to get the total wattage required. For example, 5 light bulbs at 60 watts each would be a total of 300 watts. If you want to add a 1500 watt hot plate, your total is now 1800 watts. Adding some music from a radio would add another 100 watts, and so on. • A reactive load is one that is usually associated with some type of electric motor. Examples of reactive loads would be circular saws, furnace motors, water pumps and air conditioners. This type of load may take three times the rated power requirement to start up and perform the required work. An example of this would be an air conditioner that runs on 20 amps at 120 volts. The running watts would be calculated.

One of the several uses of an RC coupled amplifier is to amplify the given input signal. It makes use of a sinusoidal input signal.

This is a crystal which, when squeezed, will generate a potential difference across opposite faces. A practical example is an inexpensive phonograph/gramophone pickup.

* resistance increases voltage. Adding more resistance to a circuit will alter the circuit pathway(s) and that change will force a change in voltage, current or both. Adding resistance will affect circuit voltage and current differently depending on whether that resistance is added in series or parallel. (In the question asked, it was not specified.) For a series circuit with one or more resistors, adding resistance in series will reduce total current and will reduce the voltage drop across each existing resistor. (Less current through a resistor means less voltage drop across it.) Total voltage in the circuit will remain the same. (The rule being that the total applied voltage is said to be dropped or felt across the circuit as a whole.) And the sum of the voltage drops in a series circuit is equal to the applied voltage, of course. If resistance is added in parallel to a circuit with one existing circuit resistor, total current in the circuit will increase, and the voltage across the added resistor will be the same as it for the one existing resistor and will be equal to the applied voltage. (The rule being that if only one resistor is in a circuit, hooking another resistor in parallel will have no effect on the voltage drop across or current flow through that single original resistor.) Hooking another resistor across one resistor in a series circuit that has two or more existing resistors will result in an increase in total current in the circuit, an increase in the voltage drop across the other resistors in the circuit, and a decrease in the voltage drop across the resistor across which the newly added resistor has been connected. The newly added resistor will, of course, have the same voltage drop as the resistor across which it is connected.

a step down transformer is used to lower the voltage from the powerlines into your home. a common slang term for this type of transformer is a pole pig.

Michael Faraday 1821

You will want to connect them in "parallel" (positive to positive, negative to negative). If you connect them in "series" (positive to negative) you will add one voltage to the other, making 24 volts.

In its simplest form, the starter of a dc motor is a variable resistance in series with the armature circuit of dc motor to reduce the high starting current so that the armature winding does not get overheated and burnt while the motor isgetting started. As the rotating armature of dc motor picks up speed, the starter resistance is gradually reduced so that the motor is able to attain its full speed when the starter is not expected to offer any additional resistance in series with the armature winding of the dc motor. At full speed the motor starts running normally, of course, without the help of starter. In other words, the starter offers resistance to armature current during starting of dc motor only. Under normal working condition of dc motor , the starter is electrically out of armature circuit of the motor. The starter protects the armature of dc motor from getting damaged. The electromotive force (emf) induced in the armature winding during starting builds up from zero value to max value to restrict the armature current within the permissible value at full speed. As the speed of armature/motor build up, armature induced emf also starts building thus reducing the role resistance offered by the starter, hence requiring it to gradually reduce as the motor picks up full speed.

The device that increases or decreases the voltage impressed across a power line is known as a voltage regulator.

A voltage regulator is a type of transformer where the primary and secondary turns ratio are fairly close; one (primary or secondary) often has a tap changing ability to add or remove several windings, allowing more dynamic control of voltage.

The law was named after the German physicist Georg Ohm who, in 1827, described measurements of applied voltage and current through simple electrical circuits containing various lengths of wire.

The battery has 6 volts across its terminals. The way to discover it is to apply Ohm's law. It (Ohm's law) comes in 3 "flavors" that look a bit different but all say exactly the same thing. Here they are: E = I x R [Voltage equals current times resistance.] I = E/R [Current equals voltage divided by resistance.] R = E/I [Resistance equals voltage divided by current.] In these equations, voltage is E, current is I and resistance is R. They are measured in units of volts, amperes (or amps) and ohms, respectively. In your problem, we have the resistance (R) and the current (I). We need to find the voltage (E), and the formula E = I x R is the logical choice to discover the voltage. As E = I x R here, E = 0.75 x 8 = 3/4 x 8 = 6 volts. Piece of cake.

Increase primary voltage but decraease amerage - eg double voltage and half amperage. This will double your secondary voltage. "Transformed" voltage is easy to understand by following this simple math rule.

secondary voltage divided by primary voltage = no of turns on secondary and no of turns on primary.

eg;

240v \ 12v = 20 (factor)

this means the secondary winding is 20x the primary winding. Basically thats all I know Stick to this rule for step up & step down tranforming & you'll be sweet.

the formula for electric current is VI ,where v is voltage then I is the current.

the unit used for current is ampere and volts for voltage. multiply the total I to the Voltage

The formular of electric current is given by I=V/R ,I=P/V

I diode allows current to flow in only one direction. Therefore, if a lamp is "on" in a DC circuit, and the diode in series with the lamp is reversed, the light will be turnned off due to the diode blocking current flow (unless the voltage is above the breakdown voltage of the diode - if this is the case, the diode will fail). If this is an AC circuit, every half cycle the diode will turn on, then the next half cycle it will turn off. To your eye, the bulb will most likely appear slightly dim due to this on then off cycling. If the diode is reversed, there will be no apparent change. The difference is the half cycle the diode would have been off before reversing, it will now be on, etc.

(i) EMF (Electro Motive Force) the source of voltage can increase or decrease voltage.

(ii) Load connected can increase or decrease voltage (In case of overload voltage reduces.)

(iii) Resistance of wire - I R loss ( Voltage drop V = IR)

if you have 2 bulbs and one burns out, the other wont shine,

if you remove 1 bulb, you are breaking the circuit, and the current

cant flow to the second bulb, so no it wouldn't shine.

Yes, Of Course The lifetime depends on the wattage. The wattage is the amount of Power the light bulb uses, and the life time depends on the power. Thank you

By adding more light bulbs

Voltages used are 1.5kV DC and 25kV AC for mainline trains

V = I.R

R = V/I

R = 110/0.5

R = 220 Ohms