Hi,

I have been racking my brain trying to figure out why the marking key uses 5V in this problem to size R1 when the supply voltage is 12V for the voltage divider.

Any help would be greatly appreciated.

 

Maybe the answer shown is incorrect?
What values did you calculate?

 

Thank you for the reply,
My assumption would be that that the highest current supplied would be when the variable resistor is 0 ohm so R1 = V/I = 12/0.02 = 600 ohms.

This is Mark scheme for a statewide university entrance exam so having difficulty accepting the answer could be wrong

 

Thank you for the reply,
My assumption would be that that the highest current supplied would be when the variable resistor is 0 ohm so R1 = V/I = 12/0.02 = 600 ohms.

This is Mark scheme for a statewide university entrance exam so having difficulty accepting the answer could be wrong

I agree with your answer.

 

The answers they give do not meet the power consumption requirement. Because I = 12V/(250Ω + 178Ω) = 28.03mA. Which is larger than 20mA.

 

Hi,

I have been racking my brain trying to figure out why the marking key uses 5V in this problem to size R1 when the supply voltage is 12V for the voltage divider.

Any help would be greatly appreciated.

At first it looked like they might have simply swapped the roles of R1 and R2 in their answers, but their problems go way beyond that.

Since the current is given by

Io = Vo/(R1+R2) <= 20 mA

You immediately get the constraint that

(R1+R2) >= Vo/Io = 12 V / 20 mA = 600 Ω

Since R2 can be 0 Ω, that means that R1 >= 600 Ω

Giving the benefit of the doubt, it's possible the answers are for an earlier version of the problem and they didn't get updated when the problem changed.

 

The answers they give do not meet the power consumption requirement. Because I = 12V/(250Ω + 178Ω) = 28.03mA. Which is larger than 20mA.

And that's the MINIMUM current that will be drawn. The max will be 48 mA.