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### Network # Beginner's question -- voltage drop

R

#### Rheilly Phoull

Jan 1, 1970
0
electricked said:
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

E

#### electricked

Jan 1, 1970
0
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

J

#### James W

Jan 1, 1970
0
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

J

#### John Popelish

Jan 1, 1970
0
electricked said:
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

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.

K

#### Keith R. Williams

Jan 1, 1970
0
This one:

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

Is equal to this one:

R1*R2
-----
R1+R2
------\/\/\/-------
| |
| + |
--- Vcc*R2 O LOAD
- ------ |
| R1+R2 |
|------------------

J

#### John Fields

Jan 1, 1970
0
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?

---
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.

Likewise, for the load,

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
load, so now what?

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... E

#### electricked

Jan 1, 1970
0
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

E

#### electricked

Jan 1, 1970
0
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

E

#### electricked

Jan 1, 1970
0
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 said:
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?

---
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.

Likewise, for the load,

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
load, so now what?

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... J

#### James W

Jan 1, 1970
0
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

J

#### James W

Jan 1, 1970
0
I should have added.. this is EXACTLY how amplifier's work.. the
'effective' resistance of the transistor ( for example ) changes, which
leads to a change in the current through the circuit, which changes the
voltage across the Collector resistor ( for example )

-jim

E

#### electricked

Jan 1, 1970
0
Hi James,

It makes sense now. I was trying to look at two quantities
(voltage,resistance) while forgetting about the third one (current). I'm
"getting" it now. I know the laws but I'm trying to visualize what's going
on in a circuit.

So if the load's resistance increases (say due to high temperature) and we
have 5V coming after the first resistor drop, then the current will
decrease, right? It's starting to make sense now.

Thanks!

--Viktor

R

#### [email protected]

Jan 1, 1970
0
electricked said:
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

The first circuit is used as the speed controller for Scalelectrix
cars by varying R. The load is *small*.

The second circuit is used as a volume control with the centre tap as
the wiper. The load is *big*.

The first circuit is passing a lot of *energy* into the load, so it is
important to try and make it efficient.

The (volume-control) second circuit is passing virtually no energy at
all (because the load is virtually non-existant) this allows regular
control of the output from maximum to zero (volume).

The first circuit cannot do this, go "down" to zero, (it can if "zero"
== car not moving).

The first circuit is "non-linear".
The second circuit is "linear".

Robin

B

#### Bill Vajk

Jan 1, 1970
0
electricked said:
Hi James,

It makes sense now. I was trying to look at two quantities
(voltage,resistance) while forgetting about the third one (current). I'm
"getting" it now. I know the laws but I'm trying to visualize what's going
on in a circuit.

So if the load's resistance increases (say due to high temperature) and we
have 5V coming after the first resistor drop, then the current will
decrease, right? It's starting to make sense now.

Thanks!

When you're first learning parallel/series networks (that's
what these are) you need to ignore real world effects such
as heat until later when you've mastered the basics.

J

#### James W

Jan 1, 1970
0
Yes, and no..

Remember, Ohm's law and Kirchhoff's Voltage law must be true at all times.

You cannot just say "We have 5v coming after the first resistor's drop".
As your loads resistance increase, so does the total circuits
resistance, since you have a series circuit, and resistances in series
ADD to give you total resistance.

So, if you loads resistance increases, and hence the total resistance
increases, then the circuit current decreases ( per Ohms law ). Now,
since the total current decreases, and the current is the same in all
parts of a series circuit, we'll have a lowering of the voltage across
the series resister R. Again, because of ohms law. So, if the voltage
acrros the resistor drops, and Kirchhoffs law requires that the total
voltages around the circuit add to ZERO, the voltage across your load
will increase.

i.e. A change in resistance through any component effects ALL of the
components.

Get your hands on twp potentiometers and a battery. Play with varying
the resistance of the two pots and take lots of voltage and resistance
measurements ( be sure to disconnect the battery each time you measure
the resistance of the two pots). Make a chart of your data, and think
about it until it all makes sense.

BTW, in a series circuit, the higher the resistance in any component,
the higher the voltage across that component and the lower the current
through that component. But don't forget, since we have a fixed voltage
source, if the voltage across one component increases then there must be
an offsetting decrease in the voltage across some collection of other
components (KVL)

J

#### John Fields

Jan 1, 1970
0
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?

---
Somewhere in between... Looking at the voltage divider first, classically it's configured like
this, which is the same as your first circuit but with the battery
replaced with E1 and 0V, which represent the output voltage from the
battery and the return to the battery.

E1
|
|
[R1]
|
+---->E2
|
[R2]
|
|
0V

The reason it's called a voltage divider is because if E1 is sitting at,
say, 9V, then E2 must be somewhere between 9V and 0V, so the two
resistors are "dividing" the total voltage into two pieces; one from E2
to E1, and one from E2 to 0V.

It's the same as if you had a hose hooked up to a faucet and the
pressure at the faucet was at 9 PSI, and where the water was spouting
out of the other end of the hose the pressure was 0 PSI; If you took a
pressure gauge and ran it up and down the hose, the pressure would vary
from 0 PSI to 9 PSI depending on where along the hose it was. Having
two resistors is like having two hoses screwed together with a pressure
gauge where they're screwed together. If they're the same length and
the same diameter and the pressure at the faucet is 9 PSI with the water
running, then the pressure gauge between the hoses will read 4.5 PSI.

Same thing with the resistors. If E1 is 9 volts (9 PSI) and R1 is the
same resistance as R2, (two hoses with the same length and diameter)
then E2 will be 4.5 V (4.5 PSI)!

If you know E1 and you know the values of the two resistors or the
current you can find E2 using Ohm's law, and you'll find that if either
R1 or R2 vary, E2 will also vary. Sometimes this circuit is used to
drop the voltage from a source with too high a voltage to a voltage a
load can use, but it's usually not a good idea unless the load current
is small and the load always draws the same current no matter what and
is never disconnected from the circuit or connected to the circuit with
the power on. The reason for that is if it's disconnected the voltage
at the botton end of R1 will immediately go to whatever the source
voltage is, and then when it's reconnected the load will have that
voltage on it until it starts drawing enough current for the voltage to
drop down to where it's supposed to be. Problem is, it may never get
the chance to work properly if it gets connected to 9V and 5V is the
maximum it's ever supposed to see. This circuit is generally used to
get a voltage to use as a reference for something when the reference
needs to be somewhere between E1 and 0V and the device it's supplying
the reference voltage to (the load) has a resistance much, much, higher
than R2. Here's an example:

E1
|
[R1]
|
E2----+<------+
| |
[R2] [R3]
| |
+<------+
|
0V

If we assume for a moment that R3 isn't connected, we can find the
voltage at E2 using Ohm's law, like we did before, or to make it easier,
we can say:

E2 = E1R2/R1+R2

Let's make E1 = 9V and R1 and R2 both = 1000 ohms. Then we can write:

E2 = (9V*1000R)/1000+1000 = 9000/2000 = 4.5 V

Which is just exactly what we said about the two equal hose segments!

Now, let's say that to be really sure of what E2 is we'll want to
measure it, and what we'll use to measure will be a digital voltmeter
(R3) with a resistance of 10 megohms. Since we know that the combined
resistances of R2 and R3 are going to be less than the resitance of R2
alone, and that if R2 isn't equal to R1 E2 won't be equal to 4.5V, we'll
need to find the value of 1000 ohms paralleled with 10 megohms and use
that to calculate the new R2. So:

Rt = R2R3/R2+R3 = (1000R*10000000R)/1000R+10000000R = 999.9 ohms

Now, using 999.9 ohms for R2 and plugging it into the voltage divider
equation equation we get:

E2 = E1R2/R1+R2 = (9V*999.9R)/1000R+999.9R ~ 4.4998V

so there's only a 200 microvolt error there, which is about 0.004%.

Now, let's say that we measure the voltage with a crappy voltmeter that
has an input resistance of 1000 ohms. For the equivalent resistance
we'll get:

Rt = R2R3/R2+R3 = (1000R*1000R)/1000R+1000R = 500 ohms

and if we plug that into the voltage divider equation well get:

E2 = E1R2/R1+R2 = (5V*500R)/1000R+500R = 1.666V

a HUGE 56.7% error!

So, you can see that the uses for the circuit can vary from a gross way
to set the voltage across, or the current through, a load to a precise
way to generate a particular reference voltage.

But just _how_ to generate the reference, starting from scratch? Just
for fun, let's say that you have a 10V supply, you've decided to use a
10k ohm resistor for R1 and that you'd like to generate a 2.5V reference
for a comparator input. Going to the circuit:

E1
|
|
[R1]
|
+---->E2
|
[R2]
|
|
0V

And rearranging

E2 = E1R2/R1+R2

to solve for R2, (since that's the only piece of the puzzle we don't
have) we get:

R2 = E2R1/E1-E2

And plugging everything in:

R2 = (2.5V*10k)/10V-2.5V = 3333.33... ohms

Now we have a problem because, if we're limited to standard 1% resistors
the closest we can get to 3333.33 ohms will be 3320 ohms on the low side
and 3400 ohms on the high side.

Just to see what we've got, plugging that 3320 ohms into the first
voltage divider equation will get us:

E2 = E1R2/R1+R2 = (10V*3.320k)/10k+3.32k ~ 2.4925V

which is only about 7-1/2 millivolts low. If you can live with that,
you're done. If you can't, you'll need to go back to the table of
resistances and see if you can find a "gear ratio" that works. Or use a
pot!^)

Phew...!

J

#### John Fields

Jan 1, 1970
0
On Fri, 27 Feb 2004 16:28:37 -0600, John Fields

Oops...

If you know E1 and you know the values of the two resistors or the
current you can find E2 using Ohm's law, and you'll find that if either

.... If you know E1 and you know the values of the two resistors or the
value of one and the current you can find E2 using Ohm's law, and
you'll find that if either ...

D

#### DarkMatter

Jan 1, 1970
0
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

Analyse it. Determine the power in each component. If you use a
250mW resistor, will it get hot from being asked to dissipate too much
heat? Is a higher wattage capable resistor needed?

Learn how to refrain from top posting your replies.

D

#### DarkMatter

Jan 1, 1970
0
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

Give up already. You have taken up the wrong vocation.

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