Can someone help me figure out this circuit

[etech]

Harvey has the solution.

R1 = 3162.109
R2 = 500
R3 = 523.4504
R4 = 73.84215

Assume a value for R2, Remember ohms law. E=I/R. You have a voltage and the assumed resistance, you can calculate the current. Find the value of R1 for the voltage that must be dropped across it with the current.

Then start plugging from there.

 

[dym]

rite, I picked 100 for R1 and got 15.8, 16.5, 2.2 for r2,R3,R4 respectively
I guess I got this problem when I was in 7th grade...

 

[Qacer]

Tackling the High Voltage part:

When Switch_R3 is closed, you can ignore resistor R3 because current will always take the path with the less resistance (it's a short circuit).
So, when Switch_R4 is open, there is a series resistance (R4 + R2).
Therefore, you can use voltage division:

Voltage sensor = 3.6864 V

Voltage sensor = V_source ( (R4 + R2) / (R4 + R2 + R1) )
3.6864 V = 24 V ( (R4 + R2) / (R4 + R2 + R1) )

At normal voltage, R3 and R4 can be ignored.

Only R1 and R2 can be used in the voltage division:

3.2768 V = 24 V ( R2 / (R1 + R2) )


At low voltage, R4 can be ignored.
There is a series resistance with R1 + R3.
Voltage division:

2.8672 V = 24 V ( R2 / ( R2 + R3 + R1) )


I would suggest assuming a value for R2 (1 K) and solve for R1, R3, and R4.

According to MathCAD,

R1 = 6.3242187500000000000*R2
R3 = 1.0463169642857142857*R2
R4 = .14768431001890359168*R2

So if R2 = 1K

R1 = 6324 ohms
R3 = 1046 ohms
R4 = 148 ohms

Now, double check the values and plug them back in the voltage division equations to see if they are correct.
According to MathCAD, they are.

 

[Qacer]

Tackling the High Voltage part:

When Switch_R3 is closed, you can ignore resistor R3 because current will always take the path with the less resistance (it's a short circuit).
So, when Switch_R4 is open, there is a series resistance (R4 + R2).
Therefore, you can use voltage division:

Voltage sensor = 3.6864 V

Voltage sensor = V_source ( (R4 + R2) / (R4 + R2 + R1) )
3.6864 V = 24 V ( (R4 + R2) / (R4 + R2 + R1) )

At normal voltage, R3 and R4 can be ignored.

Only R1 and R2 can be used in the voltage division:

3.2768 V = 24 V ( R2 / (R1 + R2) )


At low voltage, R4 can be ignored.
There is a series resistance with R1 + R3.
Voltage division:

2.8672 V = 24 V ( R2 / ( R2 + R3 + R1) )


I would suggest assuming a value for R2 (1 K) and solve for R1, R3, and R4.

According to MathCAD,

R1 = 6.3242187500000000000*R2
R3 = 1.0463169642857142857*R2
R4 = .14768431001890359168*R2

So if R2 = 1K

R1 = 6324 ohms
R3 = 1046 ohms
R4 = 148 ohms

Now, double check the values and plug them back in the voltage division equations to see if they are correct.
According to MathCAD, they are.