A battery of e.m.f 3.0V is connected across a potentiometer wire AB of length 10\(\Omega m^{-1}\) as illustrated in the diagram above. If the jockey J makes...

A battery of e.m.f 3.0V is connected across a potentiometer wire AB of length 10\(\Omega m^{-1}\) as illustrated in the diagram above. If the jockey J makes contact at the 30cm mark, determine the resistance of length AJ and voltage across AJ.

Answer Details

The potentiometer wire is a wire of known resistance per unit length, and it acts as a potential divider. The battery of e.m.f 3.0V is connected across the wire AB, and the jockey J makes contact at the 30cm mark, which means that the wire AJ is 30cm long and wire BJ is 70cm long. The resistance of the wire AJ can be calculated as follows: Resistance of AJ = Resistance per unit length * Length of AJ = 10\(\Omega m^{-1}\) * 0.3m = 3.0\(\Omega\) The voltage across AJ can be calculated by using the concept of a potential divider. The voltage across AJ can be given by: Voltage across AJ = Total voltage * (Resistance of AJ / Total resistance) = 3.0V * (3.0\(\Omega\) / (3.0\(\Omega\) + 10\(\Omega m^{-1}\) * 0.7m)) = 0.9V So, the resistance of length AJ is 3.0\(\Omega\) and the voltage across AJ is 0.9V.