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Resistive load / Inductive Load ??

Discussion in 'Electronic Basics' started by news, Nov 21, 2004.

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  1. news

    news Guest

    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 ?

  2. Hi,

    Yes, a light bulb would be a resistive load. This happens when the
    voltage across and current through it are in phase (in step). Normally
    any component that consists of a pure resistance (i.e. not having any
    inductance or capacitance as a part of it) could be used as a resistive
    load. There is however the special case where there are inductors and
    capacitors in the load but their effect cancels out at certain
    frequencies (e.g resonance).

    A typical inductive load would be an electric motor or a transformer.
    The inductance present alters the phase between the voltage and current
    but no power is lost in it, only in the resistive part. The same holds
    for a capacitive load except that the phase shift is the other way
    round. A typical capacitive load might be some sort of transducer used
    say in an alarm.

    Obviously, unless there is a capacitor in series, all loads are
    resistive at DC although the situation is more complicated until the
    current has settled to its final value.

    Cheers - Joe
  3. news

    news Guest


    I have seen several projects that differentiate between the two and wasn't
    sure what they referred to.
  4. Steve Evans

    Steve Evans Guest

    A resistive load is one in which the current is broadly constant from
    switch-on to switch-off. An inductive load is rather more troublesome.
    Inductances don't like being switched on and off, since they store
    charge and will try to keep the current through them constant. This
    results in undesirable surges during switch on and switch off. The
    usuual suoltion is to fit the applicance with a higher value fuse than
    it needs in it's state of steady ooperation. Failing that, a slow-blow
    fuse of the same value as steady state operation is probably a safer
    option. Anything with a motor in it (fridges/freezers, for esamplle)
    are inductive loads.
  5. 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?
  6. Rich Grise

    Rich Grise Guest

    Just for the sake of pedantry, an inductance doesn't actually store charge
    - that's a capacitor. What the inductance does is store energy in the form
    of a magnetic field which corresponds to the current through it. A
    capacitor opposes a change in voltage, and inductor opposes a change in
  7. Steve Evans

    Steve Evans Guest

    Your absolultely right, of course. It was a careless error to use the
    word "charge."
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