You are provided with a potentiometer AB, a \(102\Omega\) standard resistor R, a battery of emf 4.5V, a jockey J, and other necessary materials. Connect a c...

Answer Details

Practical: current and jockey position on a potentiometer wire

With the standard resistor \(R\) and battery in circuit, the ammeter reading \(I\) is taken for each jockey contact position \(x\), and \(x^{-1}\) is evaluated. A specimen table (values illustrative):

Plot \(x^{-1}\) (vertical) against \(I_{1}\) (horizontal) from the origin; the graph is a straight line whose slope \(s = \dfrac{\Delta(x^{-1})}{\Delta I_{1}}\). Extrapolate to find \(l_{o}\), the value of \(I_{1}\) at \(x^{-1} = 0\), then evaluate \(\dfrac{I_{o}}{I}\).

Two precautions: ensure firm, clean jockey contacts and tap (do not drag) the jockey on the wire; check that all connections are tight and the key is opened between readings to avoid heating the wire and running down the cell.

(b)(i) EMF of a battery

The emf of a battery is the total electrical energy it supplies per unit charge driven round a complete circuit (the work done per coulomb by the battery), equal to the terminal p.d. when the battery delivers no current.

(b)(ii) EMF of cell Y

On a potentiometer the balance length is proportional to the emf, so \(\dfrac{E_{Y}}{E_{X}} = \dfrac{l_{Y}}{l_{X}}\).

\(E_{Y} = E_{X}\times\dfrac{l_{Y}}{l_{X}} = 1.018\times\dfrac{75.0}{50.0} = 1.018\times1.5 = 1.527\ \text{V}\).