rheostat in voltage divider circuit

ScottWang

Joined Aug 23, 2012
7,501
The Vout in #13 will be as -

The Vout of the first circuit is:
Vout = Vin(R1/(R1+R2))
= Vin(1K/(1K+[0~100K]))


The Vout of the second circuit is:
Vout = Vin((R1+R2)/R2)
= Vin(([0~100K]+1K)/1K)
 

djsfantasi

Joined Apr 11, 2010
9,237
sorry, i meant why isn't the arrow connected to the horizontal line?
I'm sorry. I just cannot see the diagram the way you are seeing. It's my fault that I perceive the answer to your question is so obvious. You ask. "...why isn't the arrow connected to the horizontal line?" And all I can come up with is, "Because it isn't!"

Sorry I can't be more help.
 

kalu

Joined Dec 23, 2011
11
i need to determine the voltage ranges for voltage divider circuits that contain rheostats. here are what i got for each circuit (i've just learned about these so please correct me if i'm wrong)

this is the first circuit. here's what i think
1) if R2 = 0 ⇒ Vout = Vin / R1
2) if R2 = R2max ⇒ Vout = Vin / (R1 + R2)
View attachment 120929

this is the second circuit and my results
1) if R1 = 0 ⇒ Vout = Vin / R2
2) if R1 = R1max ⇒ Vout = Vin / (R1 + R2)
View attachment 120931

for the last circuit i need to prove why is the range 0-Vin
1) R1 = 0 ⇒ Vin and Vout are connected so Vout = 0
2) R1 = R1max ⇒ (i can't prove this, i don't think it is Vout = Vin / R1)
View attachment 120932

any help would be appreciated. thanks
Hi, check for open circuit, it may look un likely but it's the only reason to your questions
 

kalu

Joined Dec 23, 2011
11
Yes @ScottWang, I agree. I was making that assumption that the horizontal green line was ground. This graphic illustrates what I was assuming. It's similar to yours and differs only in that it uses the red dot as a junction, similar to the TS diagrams where the "rheostat" connections are made.

View attachment 120956

Basically, the question comes down to how accurate the original drawings were? Which one below is correct?
View attachment 120957
The one with common gnd
 
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