Question

# A resistance of $$R$$ drawn current from a potentiometer . The potentiometer wire, $$AB$$ has a total resistance of $$R_0$$ A voltage V is supplied to the potentiometer . Derive an expansion for the voltage across $$R$$ when the sliding contact is in the middle of potentiometer wire.

Solution

## Given, $$Resistance =R,Potentiometer =AB,Voltage=V$$When the slides are in the middle of the potentiometer only half of its total resistance i;e $$\dfrac{R_0}{2}$$ will between A and B,So, $$\dfrac{1}{R}=\dfrac{1}{R}+\dfrac{1}{\dfrac{R_0}{2}}\Rightarrow R_1=\dfrac{R_0R}{R_0+2R}$$The total resistance between the A and B is  will be the sum of the resistance between A and C or C and B,$$R_1+\dfrac{R_)}{2}$$The current is following through the potentiometer will be:$$I=\dfrac{V}{R_1+\dfrac{R_0}{2}}=\dfrac{2V}{2R_1+R_0}$$The voltage $$V_1$$taken from the potentiometer will be product of the current $$I$$ and the resistance $$R_1$$,$$V_1=IR_1=\dfrac{2V}{2(\dfrac{R_0R}{R_0+2R})+R_0}\times(\dfrac{R_0R}{R_02R})$$$$V_1=\dfrac{2VR}{2R+R_0+2R}=\dfrac{2VR}{R_0+4R}$$Physics

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