news said:
Can someone tell me what a Resistive Load and Inductive Load are ??
Im thinking a light bulb is a resistive load. After that Im not sure ?
Thanks.
Ideal resistors have a fixed ratio between the voltage across them and
current through them. The value of this ratio is called the
resistance of the device. So volts per ampere is also know as ohms.
This applies to any waveform, instantaneously, from DC to radio
frequencies. There are no actual ideal resistors, but many metals and
oxides are pretty good approximations over some temperature range.
Light bulb filaments are only described by a single value of
resistance over a very small range of temperature, such as the
variations within a single AC line cycle. They may have an operating
temperature resistance as high as 10 times their room temperature
resistance. The important difference between resistive and inductive
loads is that resistances do not store electrical energy, so when you
interrupt the voltage across them, the current falls to zero,
essentially instantaneously.
Inductive loads store energy in magnetic fields that is proportional
to the square of the current, or
E= 1/2 * L * I^2
Where E is the stored energy is in joules or watt seconds,
L is the inductance in henries
and I is the current in amperes.
For a pure inductance (most inductive loads, except for super
conducting ones, also have resistive losses, so are not pure
inductance) there is a different relationship between voltage and
current than the simple ratio that resistors have.
V = L* (di/dt)
Where V is the volts across the inductor,
L is the inductance in henries
and di/dt is the rate of change of the current through the inductor in
amperes per second
Note that this formula tells you nothing about the current except for
how it will change value as voltage is applied.
When an inductor is connected to a DC voltage, the current starts at
zero and ramps up at a constant rate determined by the voltage (more
increases the rate of change) and the inductance (more decreases the
rate of change).
When a real inductive load that also includes some series resistance
is connected to a DC voltage, the inductance initially is the limiting
factor and sets the current rate of change. But as the current ramps
up, the internal resistance starts using up some of the applied
voltage so the effective voltage across the inductive part of the load
gets smaller, so the rate of change of the current slows, also. The
actual current rises as an exponential with a time constant (which is
L/R) to a final current that doesn't involve the inductance at all,
but is just determined by the resistance and Ohm's law. This pretty
accurately describes a relay coil being energized by DC.
The practical upshot of this characteristic of an inductive load is
that the switch turning it on does not have to handle a sudden current
rise, since the current ramps up smoothly, after the switch is closed,
making it easier on the switch than if it was switching on
instantaneous full current into a resistor or a big inrush current
into a cold bulb filament.
Turning that current off, is another matter. The only way to get an
infinite rate of change from steady state (or any other) value to zero
in zero time, requires an infinite reverse voltage to be applied ot
the inductor for zero time. This, of course is very difficult to
arrange. But if you just open a set of contacts or turn off a
transistor in series with the inductance, the inductance generates a
very large voltage to ramp the current back down very fast. There is
no other way the current can change quickly. This high generated
voltage can be very hard on both mechanical contacts and transistors.
So practical switches have to make provisions to limit this voltage by
allowing a path for the decaying current, or handle the arching the
voltage produces.
If the inductance is driven by an AC voltage, things get even more
interesting, since the switch on and off can take place at different
points in the AC cycle and the current, of course does not reach a
steady state value, except as a steady state AC amplitude. But
throughout any case, the V=L*(di/dt) applies. Are you ready for AC?
