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Inductance

P

Peter

Jan 1, 1970
0
In a circuit with an inductor, the voltage leads the current by 90 degrees.
Why 90 degrees. Does this change as inductance changes to more or less than
90?

Pierre
 
G

Greg Neill

Jan 1, 1970
0
Peter said:
In a circuit with an inductor, the voltage leads the current by 90 degrees.
Why 90 degrees. Does this change as inductance changes to more or less than
90?

The voltage across an ideal conductor is given by
the expression:

V = L*dI/dt

If the current is a sine wave, the voltage will
be a cosine wave (derivative of sine is cosine).
A cosine is equivalent to a sine shifted by 90
degrees. But which way is the shift?

Since the current cannot change instantaneously
in an inductor, we know that the voltage change
will precede the current change.
 
J

John Popelish

Jan 1, 1970
0
Peter said:
In a circuit with an inductor, the voltage leads the current by 90 degrees.
Why 90 degrees. Does this change as inductance changes to more or less than
90?

For sine waves and pure inductance, the voltage always leads the
current by 90 degrees. If you change the value of inductance, the
magnitude of the current changes (assuming you apply the same voltage
and frequency) but the phase relationship stays 90 degrees. This all
comes out of the property of inductance where the voltage and current
relate instantaneously by V=L*(di/dt) which says in words that the
voltage across an inductance of L henries is L times the rate of
change of the current through the inductance in amperes per second.
If the voltage is sinusoidal, this differential equation indicates
that the current is also a sine type wave but it has to be shifted by
90 degrees for the peak voltage to occur where the current is going
through zero, but changing the most rapidly (having the highest
magnitude of di/dt).
 
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