...
> It seems that there is a relationship between
> frequency and the voltage drop across an
> inductor or resistor. If you have a series
> circuit consisting of an inductor and resistor,
> then if frequency increases, the voltage
> drop across the inductor will increase
> while the drop across the resistor decreases.
> When frequency decreases, the voltage drop
> across the inductor decreases and the drop
> across the resistor increases.
> Now frequency is directly proportional
> to inductive reactance. Which opposes
> a change in current. My thinking is that
> if frequency goes up, and inductive reactance
> also goes up as a consequence, then
> that drives current down. Which would
> lead me to think that the voltage drop
> across an inductor would go down. But
> this manifestly does not happen. Why?
properly, resistance and reactance are impedances).
"conrad" <con
@lawyer.com> wrote in message
...
> It seems that there is a relationship between
> frequency and the voltage drop across an
> inductor or resistor. If you have a series
> circuit consisting of an inductor and resistor,
> then if frequency increases, the voltage
> drop across the inductor will increase
> while the drop across the resistor decreases.
> When frequency decreases, the voltage drop
> across the inductor decreases and the drop
> across the resistor increases.
> Now frequency is directly proportional
> to inductive reactance. Which opposes
> a change in current. My thinking is that
> if frequency goes up, and inductive reactance
> also goes up as a consequence, then
> that drives current down. Which would
> lead me to think that the voltage drop
> across an inductor would go down. But
> this manifestly does not happen. Why?
There is such a think as frequency dependent resistance(called reactance).
Inductors have reactance X_L = i*w*L and capacitors have reactance X_C =
1/(i*w*C)
The i is unimportant at this point so you can ignore it. w is the angular
frequency = 2*Pi*f. Obviously C and L are the capacitance and inductance.
Guess what? X_L and X_C are in ohms and they act like frequency dependent
resistors.
So for DC or w = f = 0, X_L = 0 and X_C = infinity. This means that an
inductor acts like a short to DC and a capacitor acts like an open circuit.
with w = inf then X_L = inf and X_C = 0. This tells you that an inductor
acts like an open circuit to an infinite frequency and X_C acts like a
short.
What happens inbetween is governed by the expressions I gave for X_L and
X_C. All you really have to know at this point is that as w goes up the
reactance of an inductor goes up(that is, it becomes more resistive) and the
reactance of a capacitor goes down.
Impedence is the combinations of reactance and resistance and is usually
expressed as a complex quantity. The reason is that when you do the math its
much easier to use complex numbers than differential equations. In any case
all you really have to know is that every component has an impedence and it
can depend on frequency. (as it can depend on many other factors) Just
think of it as a resistance but keep in mind that it can change for
different diffrences.
Jon
-----------------------------------------------Reply-----------------------------------------------
conrad wrote:
> It seems that there is a relationship between
> frequency and the voltage drop across an
> inductor or resistor. If you have a series
> circuit consisting of an inductor and resistor,
> then if frequency increases, the voltage
> drop across the inductor will increase
> while the drop across the resistor decreases.
> When frequency decreases, the voltage drop
> across the inductor decreases and the drop
> across the resistor increases.
> Now frequency is directly proportional
> to inductive reactance. Which opposes
> a change in current. My thinking is that
> if frequency goes up, and inductive reactance
> also goes up as a consequence, then
> that drives current down. Which would
> lead me to think that the voltage drop
> across an inductor would go down. But
> this manifestly does not happen. Why?
> --
> conrad
The simple answer, Conrad, is that the
source voltage is always going to be
dropped across the series resistance and
the series reactance. The sum of those
voltage drops will always equal the
source voltage. So if the reactance
increases, more voltage will be dropped
across the inductor than before, even
though the current is less. Since more
voltage is dropped across the inductor,
less will be dropped across the
resistance. The sum will be unchanged.
OK?
Chuck
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-----------------------------------------------Reply-----------------------------------------------
On May 6, 7:26 pm, "Jon Slaughter" <Jon_Slaugh
@Hotmail.com> wrote:
> "conrad" <con
@lawyer.com> wrote in message
> news:1178501175.379649.231320@l77g2000hsb.googlegroups.com...
> > It seems that there is a relationship between
> > frequency and the voltage drop across an
> > inductor or resistor. If you have a series
> > circuit consisting of an inductor and resistor,
> > then if frequency increases, the voltage
> > drop across the inductor will increase
> > while the drop across the resistor decreases.
> > When frequency decreases, the voltage drop
> > across the inductor decreases and the drop
> > across the resistor increases.
> > Now frequency is directly proportional
> > to inductive reactance. Which opposes
> > a change in current. My thinking is that
> > if frequency goes up, and inductive reactance
> > also goes up as a consequence, then
> > that drives current down. Which would
> > lead me to think that the voltage drop
> > across an inductor would go down. But
> > this manifestly does not happen. Why?
> There is such a think as frequency dependent resistance(called reactance).
> Inductors have reactance X_L = i*w*L and capacitors have reactance X_C =
> 1/(i*w*C)
> The i is unimportant at this point so you can ignore it. w is the angular
> frequency = 2*Pi*f. Obviously C and L are the capacitance and inductance.
> Guess what? X_L and X_C are in ohms and they act like frequency dependent
> resistors.
> So for DC or w = f = 0, X_L = 0 and X_C = infinity. This means that an
> inductor acts like a short to DC and a capacitor acts like an open circuit.
> with w = inf then X_L = inf and X_C = 0. This tells you that an inductor
> acts like an open circuit to an infinite frequency and X_C acts like a
> short.
> What happens inbetween is governed by the expressions I gave for X_L and
> X_C. All you really have to know at this point is that as w goes up the
> reactance of an inductor goes up(that is, it becomes more resistive) and the
> reactance of a capacitor goes down.
> Impedence is the combinations of reactance and resistance and is usually
> expressed as a complex quantity. The reason is that when you do the math its
> much easier to use complex numbers than differential equations. In any case
> all you really have to know is that every component has an impedence and it
> can depend on frequency. (as it can depend on many other factors) Just
> think of it as a resistance but keep in mind that it can change for
> different diffrences.
If I understand correctly, the as the frequency
increases the larger the induced emf. Which
would explain the relationship between
increase in frequency leads to increase in voltage
drop across an inductor. And because the voltage
drop will increase, the overall voltage left for
a resistor will be minimal.
--
conrad
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