Inductance = Magnetic Flux/Current = [ML2T-2A-1]/[A] = [ML2T-2A-2]
So, Dimensional Formula of Inductance = [ML2T-2A-2]
Power = Voltage x Current x power factor.
To calculate the inductance of a home made inductor simply take the number of turns,the magnetic flux linkage and the current and use the inductance formula.
The equation of an inductor is ...di/dt = V/L... meaning that the rate of change of current in amperes per second is proportional to voltage and inversely proportional to inductance in henries.If, for example, you connect a 200 millihenry inductor across a 12 volt battery, the current will increase at a rate of 60 amperes per second.Now, the question is, can the inductor, conductors, and/or battery handle that? The answer is no. Something is going to fail. The inductor will rather quickly look like a short circuit across the battery.This example does not take resistance into account. Practical inductors, conductors, and batteries have resistance, and that will place an upper limit on current but, still, this is not an appropriate way to connect an inductor to a battery.DO NOT TRY IT IN THE LAB - THERE IS RISK OF EXPLOSION.
Use the formula: reactance equals 2.pi times frequency times inductance.
The unit of power is watts, the unit of current is amps, and the unit of voltage it volts. Power = Voltage X Current Voltage = Power / Current Current = Power / Voltage In electricity, power is symbolized with a P, current with an I, and voltage with a V. The real formula looks like: P = V x I V = P / I I = P / V
V= 0.05 i + 10 for mmaw welding v= 0.05 i + 20 for tig welding
To calculate the inductance of a home made inductor simply take the number of turns,the magnetic flux linkage and the current and use the inductance formula.
an inductor has inductance(L). its unit is henry. when any change in currentin a inductor occurs it produces an self induced emf equal to e=-Ldi/dt volt. minus(-) sign indicates the direction of the induced voltage which is in opposition to the cause which is producing it. here the case is change in current(di/dt). that's why, whyan inductor opposes any change in voltage and hence current in it.
Since we know that inductance of an inductor depends on the length of inductor by the formula L=muAN*N/l, where l is the length of inductor. So by varying the length of inductor we say that inductance of inductor varies.
Formula
They are called I squared R losses. That is the formula for calculating power (P) in watts. P=I^2*R. I equals current in amps. R equals resistance in ohms. Also if the voltage (E) is known the formula is P=E^2/R. The current of electrons meets the resistance of the coil wire. That results in heat in inductor and transformer coils.
The equation of an inductor is ...di/dt = V/L... meaning that the rate of change of current in amperes per second is proportional to voltage and inversely proportional to inductance in henries.If, for example, you connect a 200 millihenry inductor across a 12 volt battery, the current will increase at a rate of 60 amperes per second.Now, the question is, can the inductor, conductors, and/or battery handle that? The answer is no. Something is going to fail. The inductor will rather quickly look like a short circuit across the battery.This example does not take resistance into account. Practical inductors, conductors, and batteries have resistance, and that will place an upper limit on current but, still, this is not an appropriate way to connect an inductor to a battery.DO NOT TRY IT IN THE LAB - THERE IS RISK OF EXPLOSION.
Yes, the current through an inductor is E / 2 pi f L. This is Ohm's law, I = E / R, where R = XL = 2 pi f L.Doubling the voltage will double the current, doubling the frequency will halve the current, and doubling the inductance will halve the current. In the specific question, if the initial current is 100 mA, then doubling voltage, frequency, and inductance will result in a current of 50 mA.Look at the formula E / 2 pi f L. Current is proportional to voltage, and inversely proportional to frequency and inductance. You don't even have to do a calculation - you can solve this by inspection.
refer your text
A VAR Meter is used to measure Reactive Power in AC Circuits - Pure reactive components dissipate zero power, which makes sense in a DC circuit, as a capacitor passes no DC current and an inductor displaces no voltage. Yet, in an AC circuit, the reactive components "seem" to dissipate power, as current passes through the capacitor and the inductor sees a voltage drop. This counterfeit power is called "reactive power" and is measured not in Watts, but in VARs (Volt-Amps-Reactive). Its mathematical formula symbol is "Q". A VAR Meter is used to measure Reactive Power in AC Circuits - Pure reactive components dissipate zero power, which makes sense in a DC circuit, as a capacitor passes no DC current and an inductor displaces no voltage. Yet, in an AC circuit, the reactive components "seem" to dissipate power, as current passes through the capacitor and the inductor sees a voltage drop. This counterfeit power is called "reactive power" and is measured not in Watts, but in VARs (Volt-Amps-Reactive). Its mathematical formula symbol is "Q".
The formula you are looking for is Watts = Amps x Volts. killo = 1000 1 kw = 1000 watts.
Time taken =Distance/Speed
The relationship between the formulas is that in all the radius is cubed.