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Power in electric circuits.

In addition to voltage and current, there is another measure of free electron activity in a circuit: power. First, we need to understand just what power is before we analyze it in any circuits.

Power is a measure of how much work can be performed in a given amount of time. Work is generally defined in terms of the lifting of a weight against the pull of gravity. The heavier the weight and/or the higher it is lifted, the more work has been done. Power is a measure of how rapidly a standard amount of work is done.

For American automobiles, engine power is rated in a unit called "horsepower," invented initially as a way for steam engine manufacturers to quantify the working ability of their machines in terms of the most common power source of their day: horses. One horsepower is defined in British units as 550 ft-lbs of work per second of time. The power of a car's engine won't indicate how tall of a hill it can climb or how much weight it can tow, but it will indicate how fast it can climb a specific hill or tow a specific weight.

The power of a mechanical engine is a function of both the engine's speed and its torque provided at the output shaft. Speed of an engine's output shaft is measured in revolutions per minute, or RPM. Torque is the amount of twisting force produced by the engine, and it is usually measured in pound-feet, or lb-ft (not to be confused with foot-pounds or ft-lbs, which is the unit for work). Neither speed nor torque alone is a measure of an engine's power.

A 100 horsepower diesel tractor engine will turn relatively slowly, but provide great amounts of torque. A 100 horsepower motorcycle engine will turn very fast, but provide relatively little torque. Both will produce 100 horsepower, but at different speeds and different torques. The equation for shaft horsepower is simple:

Notice how there are only two variable terms on the right-hand side of the equation, S and T. All the other terms on that side are constant: 2, pi, and 33,000 are all constants (they do not change in value). The horsepower varies only with changes in speed and torque, nothing else. We can re-write the equation to show this relationship:

Because the unit of the "horsepower" doesn't coincide exactly with speed in revolutions per minute multiplied by torque in pound-feet, we can't say that horsepower equals ST. However, they are proportional to one another. As the mathematical product of ST changes, the value for horsepower will change by the same proportion.

In electric circuits, power is a function of both voltage and current. Not surprisingly, this relationship bears striking resemblance to the "proportional" horsepower formula above:

In this case, however, power (P) is exactly equal to current (I) multiplied by voltage (E), rather than merely being proportional to IE. When using this formula, the unit of measurement for power is the watt, abbreviated with the letter "W."

It must be understood that neither voltage nor current by themselves constitute power. Rather, power is the combination of both voltage and current in a circuit. Remember that voltage is the specific work (or potential energy) per unit charge, while current is the rate at which electric charges move through a conductor. Voltage (specific work) is analogous to the work done in lifting a weight against the pull of gravity. Current (rate) is analogous to the speed at which that weight is lifted. Together as a product (multiplication), voltage (work) and current (rate) constitute power.

Just as in the case of the diesel tractor engine and the motorcycle engine, a circuit with high voltage and low current may be dissipating the same amount of power as a circuit with low voltage and high current. Neither the amount of voltage alone nor the amount of current alone indicates the amount of power in an electric circuit.

In an open circuit, where voltage is present between the terminals of the source and there is zero current, there is zero power dissipated, no matter how great that voltage may be. Since P=IE and I=0 and anything multiplied by zero is zero, the power dissipated in any open circuit must be zero. Likewise, if we were to have a short circuit constructed of a loop of superconducting wire (absolutely zero resistance), we could have a condition of current in the loop with zero voltage, and likewise no power would be dissipated. Since P=IE and E=0 and anything multiplied by zero is zero, the power dissipated in a superconducting loop must be zero. (We'll be exploring the topic of superconductivity in a later chapter).

Whether we measure power in the unit of "horsepower" or the unit of "watt," we're still talking about the same thing: how much work can be done in a given amount of time. The two units are not numerically equal, but they express the same kind of thing. In fact, European automobile manufacturers typically advertise their engine power in terms of kilowatts (kW), or thousands of watts, instead of horsepower! These two units of power are related to each other by a simple conversion formula:

So, our 100 horsepower diesel and motorcycle engines could also be rated as "74570 watt" engines, or more properly, as "74.57 kilowatt" engines. In European engineering specifications, this rating would be the norm rather than the exception.

• REVIEW:
• Power is the measure of how much work can be done in a given amount of time.
• Mechanical power is commonly measured (in America) in "horsepower."
• Electrical power is almost always measured in "watts," and it can be calculated by the formula P = IE.
• Electrical power is a product of both voltage and current, not either one separately.
• Horsepower and watts are merely two different units for describing the same kind of physical measurement, with 1 horsepower equaling 745.7 watts.

District Okara Educators School wise Merit List 2017

District Okara Educators School wise Merit List 2017

After that short lists will be displayed after the Interview and here we will display the Merit lists about the Educator Jobs 2017 Districts Wise Teachers Details ESE..............................................................................................................................................................................................................................................................................................................................











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Power Factor

In AC circuits, the power factor is the ratio of the real power that is used to do work and the apparent power that is supplied to the circuit.

The power factor can get values in the range from 0 to 1.

When all the power is reactive power with no real power (usually inductive load) - the power factor is 0.

When all the power is real power with no reactive power (resistive load) - the power factor is 1.

Power factor definition

The power factor is equal to the real or true power P in watts (W) divided by the apparent power |S| in volt-ampere (VA):

PF = P_(W) / |S_(VA)|

PF - power factor.

P   - real power in watts (W).

|S|   - apparent power - the magnitude of the complex power in volt⋅amps (VA).

Power factor calculations

For sinusoidal current, the power factor PF is equal to the absolute value of the cosine of the apparent power phase angle φ (which is also is impedance phase angle):

PF = |cos φ|

PF is the power factor.

φ   is the apparent power phase angle.

The real power P in watts (W) is equal to the apparent power |S| in volt-ampere (VA) times the power factor PF:

P_(W) = |S_(VA)| × PF = |S_(VA)| × |cos φ|

When the circuit has a restrictive impedance load, the real power P is equal to the apparent power |S| and the power factor PF is equal to 1:

PF_{(restrictive load)} = P / |S| = 1

The reactive power Q in volt-amps reactive (VAR) is equal to the apparent power |S| in volt-ampere (VA) times the sine of the phase angle φ:

Q_(VAR) = |S_(VA)| × |sin φ|

Single phase circuit calculation from real power meter reading P in kilowatts (kW), voltage V in volts (V) and current I in amps (A):

PF = |cos φ| = 1000 × P_(kW) / (V_(V) × I_(A))

Three phase circuit calculation from real power meter reading P in kilowatts (kW), line to line voltage V_L-L in volts (V) and current I in amps (A):

PF = |cos φ| = 1000 × P_(kW) / (√3 × V_L-L(V) × I_(A))

Three phase circuit calculation from real power meter reading P in kilowatts (kW), line to line neutral V_L-N in volts (V) and current I in amps (A):

PF = |cos φ| = 1000 × P_(kW) / (3 × V_L-N(V) × I_(A))

Power factor correction

Power factor correction is an adjustment of the electrical circuit in order to change the power factor near 1.

Power factor near 1 will reduce the reactive power in the circuit and most of the power in the circuit will be real power. This will also reduce power lines losses.

The power factor correction is usually done by adding capacitors to the load circuit, when the circuit has inductive components, like an electric motor.

Power factor correction calculation

The apparent power |S| in volt-amps (VA) is equal to the voltage V in volts (V) times the current I in amps (A):

|S_(VA)| = V_(V) × I_(A)

The reactive power Q in volt-amps reactive (VAR) is equal to the square root of the square of the apparent power |S| in volt-ampere (VA) minus the square of the real power P in watts (W) (Pythagorean Theorem):

Q_(VAR) = √(|S_(VA)|² - P_(W)²)

The reactive power Q in volt-amps reactive (VAR) is equal to the square of voltage V in volts (V) divided by the reactance Xc:

Q_(VAR) = V_(V)² / X_C = V_(V)² / (1 / (2πf_(Hz)·C_(F))) = 2πf_(Hz)·C_(F)·V_(V)²

So the power factor correction capacitor in Farad (F) that should be added to the circuit in parallel is equal to the reactive power Q in volt-amps reactive (VAR) divided by 2π times the frequency f in Hertz (Hz) times the squared voltage V in volts (V):

C_(F) = Q_(VAR) / (2πf_(Hz)·V_(V)²)

Make an inverter : Power AC devices with a battery

Hi everyone, today we’ll show you how to make an inverter! An inverter provides power to objects working on alternating current with only a 12 V battery. First we’ll test our inverter on different devices then we’ll show you how to create yours.
Let’s start with a simple test. We plug the terminals on the 12 V battery and then we switch on the inverter. The bulb, which normally works on AC with 230V and consumes 25 Watt, is now on as if it was plugged in the power outlet! Instead of using high consumption light bulbs, we can use economic light bulbs, here is a test with 36W, and as you see it works perfectly. Here is another test with our inverter; we will recharge at the same time an iPod, an iPhone and an iPad. When we switch on, there is enough energy remaining to light up a bulb. The inverter is compatible with every switching-mode power supply, even the most powerful ones. Now we will try to supply a stereo. I turn up the volume gradually. So obviously we can’t reach the volume’s maximum because the inverter can give only up to approximately 30 Watt while the stereo reaches 260 Watt when on maximum volume. There is black out? With the inverter we can supply energy to your room’s lamp. The soldering iron’s cable is too small? With the inverter, the cables are never too small !
Our inverter is made out of 3 parts: 
- We have a 50 Hz oscillator 
- 2 MOSFET transistors that amplify the signals 
- And a transformer 
We first have to make the oscillator which can be considered as an integral component. The positive terminal is placed in the upper position, while the negative one is in the lower position. It consumes 30mA under a 12 V current. On the sides it supplies a positive current on one side and a zero current on the other side, and vice-versa, with the frequency that we wish. To have a 50 or 60 Hz frequency, the mains frequency, we will give the specifications of the capacitors and resistances that we used at the end of the video. 
As for the transistors, we can consider it as an amplifier with 3 terminals: base collector and emitter. For the transistor that we are using, when there is a small current between the base and the transmitter, a much powerful current can flow between the collector and the emitter. For example, the current that flows through this resistance would have never been able to light the bulb, but thanks to the transistor, the signal that I’m doing can be reproduced in the bulb.
In our inverter we use MOSFET transistors. They act like the other transistors except that their terminals are named Gate (G), Drain (D) and Source (S). The current that must flow between G and S to activate D and S can be very low. As long as there is more than about 4V between G and S, the other circuit is activated. As you can see the current that can flow through the body is enough to activate the MOSFET transistor!
We use MOSFET transistor for our inverter because they need very low voltage between G and S to be activated which doesn’t affect the oscillator’s frequency. Since we use recycled components only, the MOSFETs even from different objects have relatively the same specifications, which is very important since the transistor have to be in perfect sync: there should be neither any dead time nor a time when both transistors are activated. We connect the two sides of the oscillator to the transistor’s G terminals. Since the frequency is 50 or 60Hz, each hundredth of a second the activated transistor changes.
The transformer must have a 1 to 19 ratio so we can go from 12 V to 230 V, twice less to reach 110V. Since the primary coils are in opposite directions, the direction in which the induced current in the secondary coil flows is opposed to the one from either one of the primary coils. We have a 230 V and 50 Hz current, not really sinusoidal but ideal for supplying mains powered devices.
Other tests:
A last test with 24 V, this way we can have an 85 Watt power tested here with 2 bulbs: a 60 Watt bulb and a 25 Watt bulb fully lit. However, with 24 V the voltage isn’t stable and we can only obtain 230 V when we consume 85 Watt, and less we consume more the voltage is increasing, it can go up to 600 V on open circuit.
We have to supply the circuit with 12 V which consumes 500 mA open and up to 7 A in short circuit with 230 V; this is why we use a big lead battery so that we can have a range of several hours. We have for example a range of 5 hours and a half with this 10 Watt bulb; we can also supply a 230 V laptop stereo with only two 9 V batteries, I switch it on and it works !
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Part List Of a UPS

Part List Of a UPS