Hi,

You have declared a frequency of 1Hz (i.e. one cycle per
second) for your source.

OK, now draw a sine wave with a triangular wave of the same
peak amplitude inside it. You should be able to see that the
initial slope (di/dt) of the triangle is much greater than one
amp per second, in fact it is four times that since it reaches
its peak in a quarter of a second. Now look at the sine wave - at
the crossing it is steeper than the triangle.


Cheers - Joe

Hi,

You have declared a frequency of 1Hz (i.e. one cycle per
second) for your source.

OK, now draw a sine wave with a triangular wave of the same
peak amplitude inside it. You should be able to see that the
initial slope (di/dt) of the triangle is much greater than one
amp per second, in fact it is four times that since it reaches
its peak in a quarter of a second. Now look at the sine wave - at
the crossing it is steeper than the triangle.


Cheers - Joe

Well, I wish I could do a good drawing here... but
If we assume that at 0deg, voltage is zero, then, at 90 degrees, voltage
is max. With a perfect inductor, current lags voltage by 90 degrees, so
current is 0. Current is a sine wave. So.. why isn't di/dt==1 at this point?

Joe McElvenney wrote:
Hi,

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


Cheers - Joe

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

John and Joe are obviously correct... and I... am an idiot.

Gentlemen, thanks for helping... it took me a while.. but now I've got it..

- 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:
....
Gentlemen, thanks for helping... it took me a while.. but now I've got it..

Glad to help.
--
John Popelish

James W wrote:
....
Gentlemen, thanks for helping... it took me a while.. but now I've got it..

Glad to help.
--
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.

John and Joe are obviously correct... and I... am an idiot.

Gentlemen, thanks for helping... it took me a while.. but now I've got it..

- 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.