- Yes, what is this voltage divider good for?

It wouldn't be any good to power another circuit that requires current, for

the circuit would load down the voltage. It would be good, however, at

the input of a circut with high imput impedance, such as an input of an

opamp.

Ed

<html></html>

>From: mrostamy@...

_________________________________________________________________

>Reply-To: Electronics_101@yahoogroups.com

>To: Electronics_101@yahoogroups.com

>Subject: [Electronics_101] Re: newbie question of voltage dividers

>Date: Sat, 29 Sep 2001 16:37:20 -0000

>

>Hi!

>Tako placed a very basic question, and Larry very nicely put the

>answer...where do you get the art for designing circuits in TXT??

>

>However I would suggest the following question to Tako and all other

>begginers. We have a voltage divider like the one Larry designed. So

>the 3 different voltages (refered to GND) are present. But what if we

>connect something to them, thus drawing current from the circuit.

>Would we still have a voltage divider? Maybe a current divider? Surely

>if we darw current from any of the three voltages their voltage would

>change! So what is this Voltage divider stuff good for???

>

>have a nice weekend!

>

>

>--- In Electronics_101@y..., Larry Hendry <hendrysr@y...> wrote:

> > Tako, What you have read is true. What you have drawn below is not a

> > parallel circuit. In a circuit of parallel resistors, both ends of

> > each resistor would be tied together. voltage is the same across

> > all, and current divides based on resistance.

> >

> > Like this:

> >

> > +9V ---+----+----+

> > | | |

> > 9K 9K 18K

> > | | |

> > Gnd ---+----+----+

> >

> >

> > In this parallel circuit, each resistor has 9 volts across it. the 9

> > K resistors have .001 amp or 1 milliamp each. The 18 K resistor has

> > 0.5 milliams. Total current is 2.5 milliamps. 9/9,000=.001 and

> > 9/18,000 = .0005

> >

> > A series circuit is used to divide voltages. The current is the same

> > in all resistors. like this:

> >

> > +9V ----1k---2k---6k---ground

> >

> > total current flow = 9/(1000+2000+6000) = 0.001 amps

> > Voltage across each individual resistor

> > = 1,000 * .001 amps = 1 volt

> > = 2,000 * .001 amps = 2 volts

> > = 6,000 * .001 amps = 6 volts

> >

> > The sum of the voltage drops (1+2+6) must equal the total voltage

> > across the resistors. The both equal 9 volts.

> >

> > So, if I took a voltage reading in reference to ground starting at

> > the 9 volt point, I would read 9V, then 8V then 6V.

> >

> > Hope this helps.

> > Larry H.

> >

>

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