Electronics Forums > Beginner's question -- voltage drop

# Beginner's question -- voltage drop

Rheilly Phoull
Guest
Posts: n/a

 02-26-2004, 12:29 PM

news:(E-Mail Removed)...
> What's the difference between the following:
>
> R
> ------\/\/\/------
> | |
> | + |
> - |
> | - |
> |-----------------
>
> and this one:
>
>
> ------------------
> | |
> | + \ R1
> --- Vcc /--------
> | - / R2 |
> |-------------------------
>
>
> How is that different from the first circuit. Say Vcc is set at 9V. Make R
> in the first circuit drop 4V. That means 5V are left for the load. In the
> second one, R1 drops 4V, Right? So what's the R2 for?
>
>
> Thanks!
>
> --Viktor
>
>

R2 is part of the load, so if there is no load there will always be a
minimum volt drop across R1 R2. As the load increases the volt drop on R1
will increase. This circuit is called a "Voltage divider" and is OK for
Regards ........ Rheilly Phoull

electricked
Guest
Posts: n/a

 02-26-2004, 12:45 PM
What's the difference between the following:

R
------\/\/\/------
| |
| + |
- |
| - |
|-----------------

and this one:

------------------
| |
| + \ R1
--- Vcc /--------
| - / R2 |
|-------------------------

How is that different from the first circuit. Say Vcc is set at 9V. Make R
in the first circuit drop 4V. That means 5V are left for the load. In the
second one, R1 drops 4V, Right? So what's the R2 for?

Thanks!

--Viktor

James W
Guest
Posts: n/a

 02-26-2004, 02:31 PM
First, I assume you understand that voltage drop across a resistor is
caused by the current flow through the resistor.(V=IR)

Now, consider what would happen in your second circuit if the value of
R2 was ZERO ohms... i.e. it is a piece of wire.

The Voltage across ZERO ohms is ZERO volts.(again, V=IR where R=0, gives
V=0, regardless of I). Since your load and R2 are in parallel, they have
the same voltage across each, so the voltage across your load is zero.

Now, consider that R2 has infinite resistance... Then we have a circuit

For more reasonable values of R2 ( say 0<R2<inifinity ) we have your
load in parallel with R2. For two resistors in parallel, the combined
resistance is lower than the resistance of the least resistive of the
two ( Rtotal for TWO resistors in parallel can be found by divding the
product of the resistances by the sum of the resistances).

As their comibined resistance drops, more current flows through the
circuit. As more current flows, more voltage is dropped across R1.

Since the Vsupply is fixed, as more voltage is dropped across R1, less
voltage is 'available' to the load.

I hope that helps a bit.

- jim
electricked wrote:
> What's the difference between the following:
>
> R
> ------\/\/\/------
> | |
> | + |
> - |
> | - |
> |-----------------
>
> and this one:
>
>
> ------------------
> | |
> | + \ R1
> --- Vcc /--------
> | - / R2 |
> |-------------------------
>
>
> How is that different from the first circuit. Say Vcc is set at 9V. Make R
> in the first circuit drop 4V. That means 5V are left for the load. In the
> second one, R1 drops 4V, Right? So what's the R2 for?
>
>
> Thanks!
>
> --Viktor
>
>

John Popelish
Guest
Posts: n/a

 02-26-2004, 05:42 PM
electricked wrote:
>
> Ok, so what's the difference between the two circuits? How are they
> different and what are the circumstances they'd be applied?
>
> Thanks!
>
> --Viktor
>
> "Rheilly Phoull" <(E-Mail Removed)> wrote in message
> news:c1kor3\$1k3pot\$(E-Mail Removed)-berlin.de...
> >
> > "electricked" <no_emails_please> wrote in message
> > news:(E-Mail Removed)...
> > > What's the difference between the following:
> > >
> > > R
> > > ------\/\/\/------
> > > | |
> > > | + |
> > > --- Vcc O LOAD
> > > - |
> > > | - |
> > > |-----------------
> > >
> > > and this one:
> > >
> > >
> > > ------------------
> > > | |
> > > | + \ R1
> > > --- Vcc /--------
> > > - \ O LOAD
> > > | - / R2 |
> > > |-------------------------

The first circuit discards the extra voltage with no extra current.
The second one discards both voltage and current, so it is less
efficient. However, the second circuit produces a more stable voltage
output with changes in load current because the load current is only
part of the total current passing through R1.

Both circuits are poor ways to drop voltage if stability is important
and load current varies. The first one may be practical if load
current is very nearly fixed, and the second one may be practical if
the load current is very small.

--
John Popelish

Keith R. Williams
Guest
Posts: n/a

 02-26-2004, 06:01 PM
In article <(E-Mail Removed)>, "electricked"

This one:

------------------
| |
| + \ R1
--- Vcc /--------
| - / R2 |
|-------------------------

Is equal to this one:

R1*R2
-----
R1+R2
------\/\/\/-------
| |
| + |
- ------ |
| R1+R2 |
|------------------

--
Keith

John Fields
Guest
Posts: n/a

 02-26-2004, 06:45 PM
On Thu, 26 Feb 2004 04:45:05 -0800, "electricked" <no_emails_please>
wrote:

>What's the difference between the following:
>
> R
> ------\/\/\/------
> | |
> | + |
> - |
> | - |
> |-----------------
>
>and this one:
>
>
> ------------------
> | |
> | + \ R1
>--- Vcc /--------
> | - / R2 |
> |-------------------------
>
>
>How is that different from the first circuit. Say Vcc is set at 9V. Make R
>in the first circuit drop 4V. That means 5V are left for the load. In the
>second one, R1 drops 4V, Right? So what's the R2 for?
>

---
If we redraw your circuits to look like this:
(View with a fixed-pitch font like Courier New)

+-----------+
| |
| [R1]
|+ |
[BATTERY] |
| |
| |
+-----------+

+-----------+
| |
| [R1]
|+ |
[BATTERY] +------+
| | |
| | |
+-----------+------+

It's easy to see that if R2 is removed from the second circuit both
circuits will be identical. That is, if R1 and LOAD are the same in
both cases.

Now, with R2 out of the circuit, if we throw some numbers in there we
can get an idea of what's going on in the circuit. Remembering that
Ohm's law states that:

E = IR

where E = voltage in volts,
I = current in amperes, and
R = resistance in ohms

And knowing that the battery voltage = 9V and that we want R1 to drop
4V, we'll either have to know the resistance of the load or how much
current it draws before we can figure out what R1 needs to be to drop
4V.

Just to make it easy, let's say that the load draws 1 amp and that since
we want to drop 4V across R1, the 5V that'll be left will be dropped
across the load. Since R1 and the load are in series, the same current
will be flowing through both of them, and we can determine what R1 needs
to be rearranging E = IR like this:

R = E/I

and solving for R:

R = 4V/1A = 4 ohms.

R = 5V/1A = 5 ohms

Since the load and R1 are in series, the total resistance in the circuit
will be the sum of both resistances, so we can represent the circuit
like this:

+-----------+
| |
| |
|+ |
[BATTERY] [9 OHMS]
| |
| |
| |
+-----------+

And if we want to check our calculations we can say:

E = IR = 1A * 9 ohms = 9V

So everything works out fine, and we can conclude that by knowing what
the battery voltage, the voltage across the load, and the current
through the load are, we can figure out what the resistance of the
series resistor needs to be in order to drop a particular voltage at
that current.

Now, if we look at the second circuit:

+-----------+
| |
| [R1]
|+ |
[BATTERY] +------+
| | |
| | |
+-----------+------+

And redraw it so that it looks like this, for convenience,

+---------------+
| |
| [R1]
|+ |
[BATTERY] +---+---+
| | |
| [R2] [R3]
| | |
+-----------+-------+

we can see that we have the load (R3) and R2 in parallel, and that that
parallel combination is in series with R1.

If we remove R2 and R3 from the rest of the circuit and relabel them for
a moment:

A
|
+---+---+
| |
[R1] [R2]
| |
+---+---+
|
B

There will be a resistance between point A and point B which be due to
the parallel combination of R1 and R2, and that resistance will be less
than the resistance of either R1 or R2.

That resistance can be found by using:

1
Rt = -------------------
1 1 1
--- + --- ... + ---
R1 R2 Rn

Which, in the case of two resistors, simplifies to:

R1*R2
Rt = -----
R1+R2

Going back to our former labeling, then, we'll say:

R2*R3
Rt = -----
R2+R3

and write it:

Rt = R2R3/R2+R3

Going back to our solved problem and relabeling it for convenience, we
have:

+---------------+
| |
| [R1] 4 ohms
|+ |
[BATTERY] +---+---+
| | |
| [R2] [R3] 5 ohms
| | |
+-----------+-------+

everything is the same except we now have R2 in parallel with our 5 ohm

OK, let's say that R2 is also 5 ohms. Then the total resistance of R2
and R3 will be:

Rt = R2R3/R2+R3 = 5*5/5+5 = 25/10 = 2.5 ohms.

Since the parallel combination of R2 and R3 is the equivalent of a
single resistor, we can redraw our schematic:

+---------------+---> 9V
| |
| [R1] 4 ohms
|+ |
[BATTERY] +---> 3.5V
| |
| [Rt] 2.5 ohms
| |
+---------------+

Solving for the current in the circuit we can say, from Ohm's law:

I = E/R = 9V/4R+2.5R = 1.38 ampere

Now, since there's 1.38 amps flowing in the circuit, the voltage dropped
across R1 will be:

E = IR = 1.38A*4R ~ 5.5V

Which means that, since we started with 9V and R1 is eating up 5.5V of
it, there'll only be 3.5V left over for the load. The reason is that
the parallel resistor caused more current to flow through R1, which
caused it to drop more voltage than it would have if the parallel
resistor wasn't there.

An easier way to think about it might be to consider R2 and R3 to be
light bulbs. With only one of them in the circuit it will be fully
bright, but when you put another one in there they will both be dimmer
than the single lamp. And R1 will get hotter, but that's for another
time... :-)

--
John Fields

electricked
Guest
Posts: n/a

 02-26-2004, 06:54 PM
One question. I'm looking at the first diagram. Say R drops 4 volts. So 5 is
left to the load. Say 4 volts are dropped by the load so 1 volt is left.
What would happen if I really connected and executed this circuit? Would the
circuit run? If so what would the effects be?

--Viktor

"James W" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> First, I assume you understand that voltage drop across a resistor is
> caused by the current flow through the resistor.(V=IR)
>
> Now, consider what would happen in your second circuit if the value of
> R2 was ZERO ohms... i.e. it is a piece of wire.
>
> The Voltage across ZERO ohms is ZERO volts.(again, V=IR where R=0, gives
> V=0, regardless of I). Since your load and R2 are in parallel, they have
> the same voltage across each, so the voltage across your load is zero.
>
> Now, consider that R2 has infinite resistance... Then we have a circuit
> that matches your circuit #1.
>
> For more reasonable values of R2 ( say 0<R2<inifinity ) we have your
> load in parallel with R2. For two resistors in parallel, the combined
> resistance is lower than the resistance of the least resistive of the
> two ( Rtotal for TWO resistors in parallel can be found by divding the
> product of the resistances by the sum of the resistances).
>
> As their comibined resistance drops, more current flows through the
> circuit. As more current flows, more voltage is dropped across R1.
>
> Since the Vsupply is fixed, as more voltage is dropped across R1, less
> voltage is 'available' to the load.
>
> I hope that helps a bit.
>
> - jim
> electricked wrote:
> > What's the difference between the following:
> >
> > R
> > ------\/\/\/------
> > | |
> > | + |
> > --- Vcc O LOAD
> > - |
> > | - |
> > |-----------------
> >
> > and this one:
> >
> >
> > ------------------
> > | |
> > | + \ R1
> > --- Vcc /--------
> > - \ O LOAD
> > | - / R2 |
> > |-------------------------
> >
> >
> > How is that different from the first circuit. Say Vcc is set at 9V. Make

R
> > in the first circuit drop 4V. That means 5V are left for the load. In

the
> > second one, R1 drops 4V, Right? So what's the R2 for?
> >
> >
> > Thanks!
> >
> > --Viktor
> >
> >

>

electricked
Guest
Posts: n/a

 02-26-2004, 06:55 PM
Ok, so what's the difference between the two circuits? How are they
different and what are the circumstances they'd be applied?

Thanks!

--Viktor

"Rheilly Phoull" <(E-Mail Removed)> wrote in message
news:c1kor3\$1k3pot\$(E-Mail Removed)-berlin.de...
>
> "electricked" <no_emails_please> wrote in message
> news:(E-Mail Removed)...
> > What's the difference between the following:
> >
> > R
> > ------\/\/\/------
> > | |
> > | + |
> > --- Vcc O LOAD
> > - |
> > | - |
> > |-----------------
> >
> > and this one:
> >
> >
> > ------------------
> > | |
> > | + \ R1
> > --- Vcc /--------
> > - \ O LOAD
> > | - / R2 |
> > |-------------------------
> >
> >
> > How is that different from the first circuit. Say Vcc is set at 9V. Make

R
> > in the first circuit drop 4V. That means 5V are left for the load. In

the
> > second one, R1 drops 4V, Right? So what's the R2 for?
> >
> >
> > Thanks!
> >
> > --Viktor
> >
> >

> R2 is part of the load, so if there is no load there will always be a
> minimum volt drop across R1 R2. As the load increases the volt drop on R1
> will increase. This circuit is called a "Voltage divider" and is OK for
> Regards ........ Rheilly Phoull
>
>

electricked
Guest
Posts: n/a

 02-26-2004, 10:44 PM
Thanks John! Very informative indeed. I like how you explain things.

So basically, they can be considered the same in terms of practical
application, right? But the first one will put more current through the
load, and the second one will split the current (if both the load and the
resistor in parallel are same resistance, it should split the current in
half, right?) so less current goes to the load. So why is it called a
voltage divider then? Shouldn't it be called a current divider? Or both
voltage and current divider? The reason I'm asking is that if I wanted to
divide the voltage only I could've used the first circuit. But if I wanted
to divide the voltage and the current (because the load draws less current)
then I'd use the second circuit. Am I close or far from the truth?

Thanks!

--Viktor

"John Fields" <(E-Mail Removed)> wrote in message
news(E-Mail Removed)...
> On Thu, 26 Feb 2004 04:45:05 -0800, "electricked" <no_emails_please>
> wrote:
>
> >What's the difference between the following:
> >
> > R
> > ------\/\/\/------
> > | |
> > | + |
> > - |
> > | - |
> > |-----------------
> >
> >and this one:
> >
> >
> > ------------------
> > | |
> > | + \ R1
> >--- Vcc /--------
> > - \ O LOAD
> > | - / R2 |
> > |-------------------------
> >
> >
> >How is that different from the first circuit. Say Vcc is set at 9V. Make

R
> >in the first circuit drop 4V. That means 5V are left for the load. In the
> >second one, R1 drops 4V, Right? So what's the R2 for?
> >

>
> ---
> If we redraw your circuits to look like this:
> (View with a fixed-pitch font like Courier New)
>
>
>
> +-----------+
> | |
> | [R1]
> |+ |
> [BATTERY] |
> | |
> | |
> +-----------+
>
>
>
> +-----------+
> | |
> | [R1]
> |+ |
> [BATTERY] +------+
> | | |
> | | |
> +-----------+------+
>
> It's easy to see that if R2 is removed from the second circuit both
> circuits will be identical. That is, if R1 and LOAD are the same in
> both cases.
>
> Now, with R2 out of the circuit, if we throw some numbers in there we
> can get an idea of what's going on in the circuit. Remembering that
> Ohm's law states that:
>
> E = IR
>
> where E = voltage in volts,
> I = current in amperes, and
> R = resistance in ohms
>
> And knowing that the battery voltage = 9V and that we want R1 to drop
> 4V, we'll either have to know the resistance of the load or how much
> current it draws before we can figure out what R1 needs to be to drop
> 4V.
>
> Just to make it easy, let's say that the load draws 1 amp and that since
> we want to drop 4V across R1, the 5V that'll be left will be dropped
> across the load. Since R1 and the load are in series, the same current
> will be flowing through both of them, and we can determine what R1 needs
> to be rearranging E = IR like this:
>
> R = E/I
>
> and solving for R:
>
> R = 4V/1A = 4 ohms.
>
>
> R = 5V/1A = 5 ohms
>
>
> Since the load and R1 are in series, the total resistance in the circuit
> will be the sum of both resistances, so we can represent the circuit
> like this:
>
>
> +-----------+
> | |
> | |
> |+ |
> [BATTERY] [9 OHMS]
> | |
> | |
> | |
> +-----------+
>
>
> And if we want to check our calculations we can say:
>
>
> E = IR = 1A * 9 ohms = 9V
>
> So everything works out fine, and we can conclude that by knowing what
> the battery voltage, the voltage across the load, and the current
> through the load are, we can figure out what the resistance of the
> series resistor needs to be in order to drop a particular voltage at
> that current.
>
>
> Now, if we look at the second circuit:
>
> +-----------+
> | |
> | [R1]
> |+ |
> [BATTERY] +------+
> | | |
> | | |
> +-----------+------+
>
>
> And redraw it so that it looks like this, for convenience,
>
>
> +---------------+
> | |
> | [R1]
> |+ |
> [BATTERY] +---+---+
> | | |
> | [R2] [R3]
> | | |
> +-----------+-------+
>
>
> we can see that we have the load (R3) and R2 in parallel, and that that
> parallel combination is in series with R1.
>
>
> If we remove R2 and R3 from the rest of the circuit and relabel them for
> a moment:
>
> A
> |
> +---+---+
> | |
> [R1] [R2]
> | |
> +---+---+
> |
> B
>
>
> There will be a resistance between point A and point B which be due to
> the parallel combination of R1 and R2, and that resistance will be less
> than the resistance of either R1 or R2.
>
> That resistance can be found by using:
>
> 1
> Rt = -------------------
> 1 1 1
> --- + --- ... + ---
> R1 R2 Rn
>
> Which, in the case of two resistors, simplifies to:
>
>
> R1*R2
> Rt = -----
> R1+R2
>
>
>
> Going back to our former labeling, then, we'll say:
>
>
> R2*R3
> Rt = -----
> R2+R3
>
>
> and write it:
>
>
> Rt = R2R3/R2+R3
>
>
>
> Going back to our solved problem and relabeling it for convenience, we
> have:
>
>
> +---------------+
> | |
> | [R1] 4 ohms
> |+ |
> [BATTERY] +---+---+
> | | |
> | [R2] [R3] 5 ohms
> | | |
> +-----------+-------+
>
> everything is the same except we now have R2 in parallel with our 5 ohm
>
> OK, let's say that R2 is also 5 ohms. Then the total resistance of R2
> and R3 will be:
>
>
> Rt = R2R3/R2+R3 = 5*5/5+5 = 25/10 = 2.5 ohms.
>
>
> Since the parallel combination of R2 and R3 is the equivalent of a
> single resistor, we can redraw our schematic:
>
>
>
> +---------------+---> 9V
> | |
> | [R1] 4 ohms
> |+ |
> [BATTERY] +---> 3.5V
> | |
> | [Rt] 2.5 ohms
> | |
> +---------------+
>
>
> Solving for the current in the circuit we can say, from Ohm's law:
>
>
> I = E/R = 9V/4R+2.5R = 1.38 ampere
>
> Now, since there's 1.38 amps flowing in the circuit, the voltage dropped
> across R1 will be:
>
>
> E = IR = 1.38A*4R ~ 5.5V
>
>
> Which means that, since we started with 9V and R1 is eating up 5.5V of
> it, there'll only be 3.5V left over for the load. The reason is that
> the parallel resistor caused more current to flow through R1, which
> caused it to drop more voltage than it would have if the parallel
> resistor wasn't there.
>
> An easier way to think about it might be to consider R2 and R3 to be
> light bulbs. With only one of them in the circuit it will be fully
> bright, but when you put another one in there they will both be dimmer
> than the single lamp. And R1 will get hotter, but that's for another
> time... :-)
>
> --
> John Fields

James W
Guest
Posts: n/a

 02-27-2004, 05:51 AM
You need to study Ohm's law Kirchhoff's Voltage law(KVL) a bit more. I'm
not being difficult here.. you really do need to sit down with a text
book and study, work out some problems, etc..

The sum of the voltages in a circuit MUST equal 0 (KVL). If you have a
9V source ( a battery for example) then the total voltage drop across
the 'loads' will be equal to 9V.

The extra volt you talk about does NOT exist. A resistor does NOT drop a
fixed voltage. It simply follows Ohm's Law (V=IR), for a given current
I, there will be a voltage drop V/R.

So.. trying to respond to your query.. Let's say the circuit was running
, and you had a 4 volt drop across R and a 5 volt drop across Load, and
something happenned to the load such that the voltage drop 'wanted' to
go to 4 volts.. Well, from Ohms law, V=IR, we see that only two things
effect the voltage drop. Either I or R must drop to lower the V across

Let's say the the resistance of the load dropped. Well, now, the total
resistance of the circuit will have dropped, so we apply Ohms law AGAIN,
and find that the current MUST have increased.. so the voltage drop
across each load will 'balance' out to MATCH the 9V source..

Whew... does that make sense?

- jim

electricked wrote:
> One question. I'm looking at the first diagram. Say R drops 4 volts. So 5 is
> left to the load. Say 4 volts are dropped by the load so 1 volt is left.
> What would happen if I really connected and executed this circuit? Would the
> circuit run? If so what would the effects be?
>
> --Viktor
>
>
> "James W" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>
>>First, I assume you understand that voltage drop across a resistor is
>>caused by the current flow through the resistor.(V=IR)
>>
>>Now, consider what would happen in your second circuit if the value of
>>R2 was ZERO ohms... i.e. it is a piece of wire.
>>
>>The Voltage across ZERO ohms is ZERO volts.(again, V=IR where R=0, gives
>>V=0, regardless of I). Since your load and R2 are in parallel, they have
>>the same voltage across each, so the voltage across your load is zero.
>>
>>Now, consider that R2 has infinite resistance... Then we have a circuit
>>
>>For more reasonable values of R2 ( say 0<R2<inifinity ) we have your
>>load in parallel with R2. For two resistors in parallel, the combined
>>resistance is lower than the resistance of the least resistive of the
>>two ( Rtotal for TWO resistors in parallel can be found by divding the
>>product of the resistances by the sum of the resistances).
>>
>>As their comibined resistance drops, more current flows through the
>>circuit. As more current flows, more voltage is dropped across R1.
>>
>>Since the Vsupply is fixed, as more voltage is dropped across R1, less
>>voltage is 'available' to the load.
>>
>>I hope that helps a bit.
>>
>> - jim
>>electricked wrote:
>>
>>>What's the difference between the following:
>>>
>>> R
>>> ------\/\/\/------
>>> | |
>>> | + |
>>> - |
>>> | - |
>>> |-----------------
>>>
>>>and this one:
>>>
>>>
>>> ------------------
>>> | |
>>> | + \ R1
>>>--- Vcc /--------
>>> | - / R2 |
>>> |-------------------------
>>>
>>>
>>>How is that different from the first circuit. Say Vcc is set at 9V. Make

>
> R
>
>>>in the first circuit drop 4V. That means 5V are left for the load. In

>
> the
>
>>>second one, R1 drops 4V, Right? So what's the R2 for?
>>>
>>>
>>>Thanks!
>>>
>>>--Viktor
>>>
>>>

>>

>
>

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