# Homework Help: Potential divider

1. Mar 6, 2005

### Mo

If a current were taken from the output of the potential divider, what would happen to Vout ??

Assuming i have 2 resistors in series with a cell...y'know the simple normal setup!

Regards,
Mo

Last edited: Mar 6, 2005
2. Mar 6, 2005

### The Bob

Personally I do not understand what you are asking. A potential divider circuit allows two different places for an output depending on the resistances of the resistors and if they change (i.e. variable resisitor, thremistor). Are you asking if a current reading was taken or if the current was removed???

3. Mar 6, 2005

### Mo

There are two resistors in series.The top one has a resistance of 7000 ohms, the other one, the bottom one, a resistance of 3000 ohms. They are connected to a 9 volt battery.

I managed to work out the voltages across both.Now the next question asks me "Predict what would happen to Vout if a current were taken from the output of the potential divider.Explain your reasoning."

Vout is across the bottom resistor.

Regards,
Mo

4. Mar 6, 2005

### The Bob

So if some of the current was removed then the voltage would decrease as well because of Ohm's Law. The overall voltage would not be 9 volts anymore though so I am unsure what I have said is right.

5. Mar 6, 2005

### Janitor

I agree with The Bob. Before a current is drawn, the voltage at the divider would be 9 * 3000/(3000+7000) if I understand the description properly. Drawing a current will cause more current to run through the 7000 ohm resistor, such that there will be more of a voltage drop across that resistor than before.

6. Mar 6, 2005

### kevinalm

Absolutely, the Vout will drop. The Vout proportional rule is only valid when equal currents pass through both resistors. In real world electronics, you always have to figure in the load of the circuit you're powering. And throw in a filter cap or inductor for good measure.

7. Mar 6, 2005

### The Bob

Finally, I am right about something.

8. Mar 7, 2005

### Mo

Thanks for the replies. Ok ,here is an image i found which is very similar to what i am trying to describe. (except that the top resistor has 7000 ohm and the bottom 3000 ohm)

So let me get this right, as i draw a current from Vout (by say attaching a light bulb), then the voltage will go down??

How come? I thought more volatge would = more current.
This is one of the main things i dont quite get, Potential dividers

Regards,
Mo

9. Mar 7, 2005

### kevinalm

The image link isn't working, but I'll try to explain. Rl the load resistance, Rt top resistor, Rb bottom, Vin, Vdrop (the voltage drop across Rt) and Vout.

Vin is constant
Vout = Rb / (Rt + Rb)
When add your "lightbulb" Rb is replaced by a parallel resistor pair whose equivalent resistance is always less that the original Rb. (1/Requiv = 1/Rb + 1/Rl) This increases the current through Rt so Vdrop will increases. Vin = Vdrop +Vout

10. Mar 7, 2005

### Janitor

As a special case of what kevinalm wrote, consider a light bulb having zero resistance. (OK, a real-world light bulb works by heating the filament, so it can't have zero resistance. So think of it being a simple piece of copper wire instead.) This will effectively ground the pick-off point, and bring its voltage all the way down to zero. In this extreme case the current will be zero in the bottom resistor, and it will be 9/7000 in both the upper resistor and in the light bulb.

11. Mar 7, 2005

### kevinalm

Janitor, good example. Here's another.

In real world electronics, small signal circuitry often operates best at a supply voltage lower than the rest of the tv, radio, etc. Since the load resistance of such a circuit is often nearly constant, Rb can be omitted, Rl becoming Rb in effect. Rt is then called a "dropping resistor". To compensate for the slight inconstancy in Rl (usually as a response to signal variations) filter components (a cap from our Vout point to ground for example) are customarily included. Which was what I alluded to in my first post.

12. Mar 8, 2005

### Mo

Thanks both for your replies.I have also fixed up the image, it can be found here (the numbers are a bit different though as you know)

Ok then, let me try and explain this from what you have said.By adding another resistance in parallel to the bottom resisitor, the total resistance from the bottom will decrease.This is because there are now more paths for the current to go through.

This inturn creates a larger current passing through the top resistor, so the voltage through the top resistor increases.

Because the voltage through the top has increased, the bottom voltage will decrease

Is that right?? If so, im happy to get that far!!

But i also have some further questions about that if you dont mind:

1) You said above that the current through Rb will increase if we add another resistor in parallel to Rb.So surely the Vout should also increase (cause there is more current flowing), as well as the Vdrop increasing. (Vdrop is voltage across Rt i have been assuming since i havent come across that term before.)

Come to think of it, is the reason why the current increases purely from the fact that the resistance decreases? (and nothing to do with the voltage)

If you dont mind helping me understand this, i would be very gratefull.
Sorry for the long post as well!!

Regards,
Mo

Last edited: Mar 8, 2005
13. Mar 8, 2005

### kevinalm

When you add Rl in parallel with Rb, the current through Rt increases, and It = Ib+Il (the currents through Rt, Rb and Rl respectively) Also, Vt = It * Rt , Vb = Ib*Rb , and Vl =Il*Rl (Vt,Vb,Vl voltage across Rt,Rb,Rl )

I think you may also need clarified:

Vin = constant
Vb = Vl
Vin = Vt + Vb = Vt + Vl

14. Mar 8, 2005

### Ouabache

Sometimes it is easier to understand if you solve this analytically.

You can use Kirchoffs current law
Ib+Il = It

Using Kirchoff's voltage law, you can write equations for the paths from
Vin to ground.

through the path of Rt and Rb resistors that would be:
9 - It(7K)-Ib(3K) = 0

through the path of Rt and Rl resistors that would be:
9 - It(7K)-Il(Rl) = 0

you have 3 equations and 3 unknown currents.

Algebraically you can solve for Il (in terms of Rl)
For Vout = Vl = Il*Rl
So you can write an expression for Vout (in terms of Rl)

So now you have an analytically expression for Vout,
substitute some values for Rl and see what happens to Vout

For the case where no load is attached Rl = infinity.
What happens to Vout in your analytic expression.

Then try Rl = small (1ohm)
and another in between (100ohm)
and another a bit larger (100Kohm)

see what happens to Vout..

Remember Il = Vout/Rl (so you can also see what happens to current thru Rl)