- #1
iainfs
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Hi folks,
We have a voltage divider (more specifically, a resistive divider), with [tex]V_i[/tex] volts input and resistors [tex]R_1[/tex] and [tex]R_2[/tex] in series. The output voltage, [tex]V_o[/tex], is measured over [tex]R_2[/tex]. [tex]R_1[/tex] is a fixed ohmic resistor; [tex]R_2[/tex] is a potentiometer. I'm looking for an algebraic proof that, for a constant [tex]V_i[/tex] and given range of resistances on [tex]R_2[/tex], the sensitivity (range of [tex]V_o[/tex]) is maximised when the middle range of [tex]R_2 = R_1[/tex].
I have read this in a textbook but have yet to be satisfied by a proof!
Well, I suppose we have [tex]V = IR[/tex] and [tex]V_o = \frac{V_i \cdot R_2}{R_1+R_2}[/tex].
I've guess that this requires calculus, which I should be OK with as long as we don't get into anything too complicated. I'm not entirely sure how to approach this, but I'll give it a shot anyway!
As far as I can see, I want to maximise the rate of change of [tex]V_o[/tex] with respect to [tex]R_2[/tex].
[tex]\frac{dR_2}{dV_o}\;({\frac{V_i \cdot R_2}{R_1+R_2}})[/tex]
I can't differentiate that because I don't know how. The constant seems to be all wrapped up with the variable. Some help here would be appreciated.
I would then go on to maximise this by finding [tex]\frac{d^2V_o}{d{R_2}^2}\;({\frac{V_i \cdot R_2}{R_1+R_2}}) = 0[/tex]; hopefully solving the problem.
As you can see, I think I have a viable method, but I'm not able to follow it through. Any help would be greatly appreciated!
Many thanks,
Homework Statement
We have a voltage divider (more specifically, a resistive divider), with [tex]V_i[/tex] volts input and resistors [tex]R_1[/tex] and [tex]R_2[/tex] in series. The output voltage, [tex]V_o[/tex], is measured over [tex]R_2[/tex]. [tex]R_1[/tex] is a fixed ohmic resistor; [tex]R_2[/tex] is a potentiometer. I'm looking for an algebraic proof that, for a constant [tex]V_i[/tex] and given range of resistances on [tex]R_2[/tex], the sensitivity (range of [tex]V_o[/tex]) is maximised when the middle range of [tex]R_2 = R_1[/tex].
I have read this in a textbook but have yet to be satisfied by a proof!
Homework Equations
Well, I suppose we have [tex]V = IR[/tex] and [tex]V_o = \frac{V_i \cdot R_2}{R_1+R_2}[/tex].
The Attempt at a Solution
I've guess that this requires calculus, which I should be OK with as long as we don't get into anything too complicated. I'm not entirely sure how to approach this, but I'll give it a shot anyway!
As far as I can see, I want to maximise the rate of change of [tex]V_o[/tex] with respect to [tex]R_2[/tex].
[tex]\frac{dR_2}{dV_o}\;({\frac{V_i \cdot R_2}{R_1+R_2}})[/tex]
I can't differentiate that because I don't know how. The constant seems to be all wrapped up with the variable. Some help here would be appreciated.
I would then go on to maximise this by finding [tex]\frac{d^2V_o}{d{R_2}^2}\;({\frac{V_i \cdot R_2}{R_1+R_2}}) = 0[/tex]; hopefully solving the problem.
As you can see, I think I have a viable method, but I'm not able to follow it through. Any help would be greatly appreciated!
Many thanks,