Need some help applying the voltage divider rule in this circuit.

[Jeremy]

I am supposed to apply the voltage divider rule to solve for R1, R2, R3 and R4. I am given what you see on the schematic below:

If you can't see the image, try: http://www.use.com/841d8bec35dbffc53b6d

I can figure out what R1 and R2 are, just by using the current and acknowledging that for R1 there is a 52v drop and R2 has a 36v drop. Then I used I=V/R to find their resistiance.

R1=3.36k ohm
R2=2.25k ohm

Now for R3 and R4 I am stumped. I also don't understand how there could possible be a -20v potential where its placed on the diagram and further have no idea how to deal with the ground place between the two resistors.

This textbook I am reading for school has not gone over voltage divider circuits (only the voltage divider rule) and we haven't read passed series circuits. However this is a question in the chapters practice problem's section.

Thanks

 

[CocaCola]

Image is broken...

 

[Jeremy]

Image is broken...

Here give this a try (weird because it works on my end):

http://www.use.com/841d8bec35dbffc53b6d


Thanks.

 

[duke37]

You know the current through the loop.
Start at the ground, R4 will need to drop 20v, R3 12v and R2 36V.
R1 will be needed to be a resistance to get 16mA through the loop (drop 100-68V).

 

[Jeremy]

Ah, that makes sense. I have one last question though.

Wouldn't my potential reading in front of my source be 100v? Then if my potential reading after the resistor is 48v, my drop accross the resistor would be 52v opposed to 32v? I just want to understand where I went wrong in solving it.

Thanks.

 

[john monks]

No. Your 100 volts is across the battery, not referenced to ground. Let's take another look. You have 16 milliamps and only 16 milliamps going through each resistor. The confusion you face is that the other voltages you see are referenced to ground. Do not let this confuse you.you know how much voltage is across R4 because you have -20 volts and a ground reference or zero volts on the other side. This is defined in the problem. Likewise you have 12 volts on one side of R3 and zero volts on the other side so you know the voltage drop. You have 48 volts on one side of R2 and 12 volts on the other side. Now on the battery you have -20 volts on the negative and a voltage of 100 volts on the other. So now you can figure out the voltage on the positive terminal. Now you have that voltage on one side of R1 and 48 volts on the other side. Now you can figure out the voltage across R1 and thus figure out the resistance because you have the current, 16 milliamps. I suggest you build a similar circuit using a 10 volt source and four resistors and use a voltmeter to see what is going on.

 

[Jeremy]

No. Your 100 volts is across the battery, not referenced to ground. Let's take another look. You have 16 milliamps and only 16 milliamps going through each resistor. The confusion you face is that the other voltages you see are referenced to ground. Do not let this confuse you.you know how much voltage is across R4 because you have -20 volts and a ground reference or zero volts on the other side. This is defined in the problem. Likewise you have 12 volts on one side of R3 and zero volts on the other side so you know the voltage drop. You have 48 volts on one side of R2 and 12 volts on the other side. Now on the battery you have -20 volts on the negative and a voltage of 100 volts on the other. So now you can figure out the voltage on the positive terminal. Now you have that voltage on one side of R1 and 48 volts on the other side. Now you can figure out the voltage across R1 and thus figure out the resistance because you have the current, 16 milliamps. I suggest you build a similar circuit using a 10 volt source and four resistors and use a voltmeter to see what is going on.

That makes perfect sense to me. Thanks for the descriptive explanation.