CAD CAM EDM DRO - Yahoo Group Archive

Re: [CAD_CAM_EDM_DRO] Electrical Help

Posted by Tony Jeffree
on 2000-08-06 00:53:31 UTC
At 20:15 05/08/00 -0500, you wrote:
>This is not correct. If you have some voltage Vin, and you have a
>resistor, R1
>connected to Vin and Vout, and another resistor R2 connected from Vout
>to
>ground, then the equation is Vout = Vin * (R2 / (R1 + R2)). So, it is
>not a simple
>ratio. If the resistors are of equal value, then the output will be one
>half of the
>input.
>
>(The explanation is that the dividing ratio is that of the load (R2)
>resistance to
>the entire resistor chain (R1 + R2).)

My apologies - what I should have said is that the ratio of the voltages
dropped across each resistor is the same as the ratio of the resistor values.

Regards,
Tony

Discussion Thread

Ejay Hire 2000-08-04 07:47:50 UTC Electrical Help Tony Jeffree 2000-08-04 08:02:28 UTC Re: [CAD_CAM_EDM_DRO] Electrical Help Ejay Hire 2000-08-04 12:27:07 UTC Re: [CAD_CAM_EDM_DRO] Electrical Help David Howland 2000-08-04 13:52:16 UTC RE: [CAD_CAM_EDM_DRO] Electrical Help JanRwl@A... 2000-08-04 18:13:32 UTC Re: [CAD_CAM_EDM_DRO] Electrical Help Tony Jeffree 2000-08-05 01:21:11 UTC Re: [CAD_CAM_EDM_DRO] Electrical Help Jon Elson 2000-08-05 18:11:13 UTC Re: [CAD_CAM_EDM_DRO] Electrical Help Tony Jeffree 2000-08-06 00:53:31 UTC Re: [CAD_CAM_EDM_DRO] Electrical Help