frequency and voltage drop across inductors

Think of reactance as a kind of resistance (more properly, resistance and reactance are impedances).

If you had a circuit with two resistances and one increased in value, what would you expect to happen to the current in the circuit and the voltage dropped across that resistor?

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 =

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

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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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