For voltage variation with length, in using a slide meter, the wire resistance varies linearly with length as Rx = resistivity/Area. Since the voltage and resistance are related with ohm's law, Vx= iRx, the current is the same alog all pts. along the slide wire. The current is set by the battery's voltage Vin and the total resistance of the wire. Putting all that together write a proof to show Vx = Cx where C is a position along the slide meter.
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To prove that Vx = Cx, where C is a position along the slide meter, we can start by applying Ohm's law to the circuit.
Ohm's law states that V = IR, where V is the voltage, I is the current, and R is the resistance.
In this case, the wire resistance varies linearly with length, so we can express it as Rx = (ρx) / (Area), where ρ is the resistivity and x is the length of the wire.
The total resistance of the wire can be obtained by integrating the resistance over the length of the wire:
R_total = ∫(ρx) / (Area) dx
Now, since the current is the same at all points along the slide wire, we can express it as:
I = Vin / R_total
Substituting R_total from the previous equation:
I = Vin / ∫(ρx) / (Area) dx
Now, let's consider the voltage Vx at a specific position x along the slide meter. Using Ohm's law:
Vx = I * Rx
Substituting the expression for Rx:
Vx = I * [(ρx) / (Area)]
Replacing I with Vin / R_total:
Vx = (Vin / R_total) * [(ρx) / (Area)]
Substituting R_total:
Vx = (Vin / ∫(ρx) / (Area) dx) * [(ρx) / (Area)]
We can see that the terms Vin and (ρ / Area) are constant, so let's define a constant C:
C = Vin * (ρ / Area)
Simplifying the equation:
Vx = (C / ∫x dx) * x
Vx = C * (x / ∫x dx) * x
Now, let's integrate ∫x dx:
Vx = C * (x / (1/2)x^2) * x
Vx = C * (2x / x^2) * x
Vx = C * (2 / x)
Therefore, we have shown that Vx = Cx, where C is a constant that depends on Vin, ρ, and the cross-sectional area of the wire.