Engineering Determine the open circuit voltage for VL

[Josh225]

Homework Statement

See image attached

Homework Equations

Voltage divider law?
Ohms law?

The Attempt at a Solution

I am unsure on how to treat the problem since I have only learned how to find information within parallel circuits and series circuits.
I tried to use the voltage divider law, but I got the wrong answer:
4.7 k Ω / 8 kΩ = .5875 kΩ (9 V) = 5.29 volts ... After this didnt work, I tried adding 3.3 kΩ to .5875 kΩ to get 3.89 kΩ (9v) and got a wild answer of 34.99 V.

The answer is supposed to be 6.13 Volts.

 

[NascentOxygen]

The first step is to identify that path/s that current will flow. Use a coloured pencil and trace out the closed path/s, starting from the + terminal and going back to the - terminal.

 

[CWatters]

If the output is "open circuit" the current through the 3.3k resistor is zero (no where for it to go). So the voltage drop across the 3k3 resistor is also zero. So you can effectively replace the 3.3k with a wire. Is it easier now?

Sorry. Cross posted with Nascent Oxygen.

 

[Josh225]

Alright, so since the 3.3 kΩ does not have current, I can essentially ignore it.
Next I use the voltage divider rule and do 2.2 kΩ/6.9 kΩ (9)= 2.87
This is where I really am unsure if I just got lucky since I had expected to get the correct answer from the voltage divider rule..
I then subtracting 2.87 V from 9V .. 9V - 2.87 V = 6.13 V

If this is the correct way, could you explain why this works? Because from my understanding the voltage divider law states : V2= (R2/R1+R2) V which is why I thought I would get the correct answer from that alone.

 

[The Electrician]

Alright, so since the 3.3 kΩ does not have current, I can essentially ignore it.
Next I use the voltage divider rule and do 2.2 kΩ/6.9 kΩ (9)= 2.87
This is where I really am unsure if I just got lucky since I had expected to get the correct answer from the voltage divider rule..
I then subtracting 2.87 V from 9V .. 9V - 2.87 V = 6.13 V

If this is the correct way, could you explain why this works? Because from my understanding the voltage divider law states : V2= (R2/R1+R2) V which is why I thought I would get the correct answer from that alone.

There's a difference between V2= (R2/R1+R2) V and V2= R2/(R1+R2) V

Dividing R2 by R1 and then adding R2 to that quotient gives a different result than adding R1 and R2, then dividing that sum into R2. Which of these two procedures did you follow?

 

[Josh225]

I used the second one, V2 = R2/ (R1+R2) V

 

[The Electrician]

I used the second one, V2 = R2/ (R1+R2) V

Which resistor are you calling R1? Which is R2?

Using your choices, show the fraction R2/ (R1+R2) with the numerical values substituted in.

 

[Josh225]

Which resistor are you calling R1? Which is R2?

Using your choices, show the fraction R2/ (R1+R2) with the numerical values substituted in.

I assumed 2.2 would be R1 since it is closest to the power supply and 4.7 would be R2.

 

[The Electrician]

I assumed 2.2 would be R1 since it is closest to the power supply and 4.7 would be R2.

Using your choices, show the fraction R2/ (R1+R2) with the numerical values substituted in.

 

[Josh225]

Yeah... It turns out I just plugged the wrong numbers. I put R1 where R2 was supposed to be and vise versa Dumb mistake. Thanks!

 

[Merlin3189]

I don't know if it will help you remember / understand:
what your voltage divider formula is telling you is the fractional voltage across one of the resistors in the chain - the fractional voltage across any resistor in the chain is equal to its resistance divided by the total resistance of the chain (provided no current is drawn from any intermediate point in the chain.)

Since you were looking for the voltage across the 4k7 resistor, then it must be 4k7/ (4k7 +2k2) of the voltage across the whole chain (9V), ie 9V *47/69.

The "open circuit" condition is simply to ensure that no current is drawn from any intermediate node in the chain.