Consider a simple inductive cicuit with a 1v(p-to-p) AC source at
1Hz,and an inductor with Z=1ohm.

The inductors value would be 1ohm=2*pi*1Hz*L, so L= 1/2pi

Looking at the standard Voltage and Current drawings, we see current
lagging voltage by 90degrees.

Here's my problem. At 90 degrees, the applied voltage is 1volt. The
current is zero. di/dt is 1, and we know that the cemf V=(di/dt)L=1/2p
Volts.

So... why if applied V is 1v and cemf is less than 1v do we have a
current of zero?

I've got the sneaky feeling that I'm trying to add apples and oranges.

My assumption was that the reason the current lags the voltage is the
back emf is being added to ( subtracted from ) the applied voltage to
give us the instantaneous voltage that then drives the instantenous
current, but perhaps this whole theory is wrong.

HELP!!

James W wrote:

Consider a simple inductive cicuit with a 1v(p-to-p) AC source at
1Hz,and an inductor with Z=1ohm.

The inductors value would be 1ohm=2*pi*1Hz*L, so L= 1/2pi

Looking at the standard Voltage and Current drawings, we see current
lagging voltage by 90degrees.

Here's my problem. At 90 degrees, the applied voltage is 1volt. The
current is zero. di/dt is 1,
(snip)

Tell me how you arrived at this di/dt.

--
John Popelish

James W wrote:

Consider a simple inductive cicuit with a 1v(p-to-p) AC source at
1Hz,and an inductor with Z=1ohm.

The inductors value would be 1ohm=2*pi*1Hz*L, so L= 1/2pi

Looking at the standard Voltage and Current drawings, we see current
lagging voltage by 90degrees.

Here's my problem. At 90 degrees, the applied voltage is 1volt. The
current is zero. di/dt is 1,
(snip)

Tell me how you arrived at this di/dt.

--
John Popelish

at the 90deg point, the I curve is crossing the zero point. It's a sine
wave. Sine waves of amplitude 1 have a slope of 1 at the zero crossing, no?

- jim

John Popelish wrote:
James W wrote:

Consider a simple inductive cicuit with a 1v(p-to-p) AC source at
1Hz,and an inductor with Z=1ohm.

The inductors value would be 1ohm=2*pi*1Hz*L, so L= 1/2pi

Looking at the standard Voltage and Current drawings, we see current
lagging voltage by 90degrees.

Here's my problem. At 90 degrees, the applied voltage is 1volt. The
current is zero. di/dt is 1,

(snip)

Tell me how you arrived at this di/dt.

at the 90deg point, the I curve is crossing the zero point. It's a sine
wave. Sine waves of amplitude 1 have a slope of 1 at the zero crossing, no?

- jim

John Popelish wrote:
James W wrote:

Consider a simple inductive cicuit with a 1v(p-to-p) AC source at
1Hz,and an inductor with Z=1ohm.

The inductors value would be 1ohm=2*pi*1Hz*L, so L= 1/2pi

Looking at the standard Voltage and Current drawings, we see current
lagging voltage by 90degrees.

Here's my problem. At 90 degrees, the applied voltage is 1volt. The
current is zero. di/dt is 1,

(snip)

Tell me how you arrived at this di/dt.

James W wrote:

at the 90deg point, the I curve is crossing the zero point. It's a sine
wave. Sine waves of amplitude 1 have a slope of 1 at the zero crossing, no?

No. If the period is 1 and the peak to peak amplitude is 1, the only
way the slope could be 1 would be if it were a saw tooth wave.

--
John Popelish

James W wrote:

at the 90deg point, the I curve is crossing the zero point. It's a sine
wave. Sine waves of amplitude 1 have a slope of 1 at the zero crossing, no?

No. If the period is 1 and the peak to peak amplitude is 1, the only
way the slope could be 1 would be if it were a saw tooth wave.

--
John Popelish

Hi,

John is right, di/dt isn't equal to one here. You are mixing
up seconds and degrees.


Cheers - Joe

Hi,

John is right, di/dt isn't equal to one here. You are mixing
up seconds and degrees.


Cheers - Joe

John,

It's been a while (20 years) since I took calculus, but I'm pretty sure
the slope of a sine wave at the zero crossing is +-1.

What value do you believe the slope to be?



John Popelish wrote:
James W wrote:

at the 90deg point, the I curve is crossing the zero point. It's a sine
wave. Sine waves of amplitude 1 have a slope of 1 at the zero crossing, no?


No. If the period is 1 and the peak to peak amplitude is 1, the only
way the slope could be 1 would be if it were a saw tooth wave.