Question

a potentiometer of wire length 1.0m has a resistance of 15ohms .it is connected to a 5volt battery in series with a resistance of 5ohms .determine the emf of the primary cell which gives a balance point at 60cm.

Answer 1

According to Ohm's law the current is:

$$I=\frac{E}{R+r}=\frac{5}{15+5}=0.25\text{ A}.$$

Voltage drop across the potential wire according to the same law:

$$V=IR=0.25\cdot15=3.75\text{ V}.$$

Potential gradient on the wire:

$$\sigma=V/L=3.75/1=3.75\text{ V/m}.$$

Electromotive force of the primary cell:

$$E_\text{p}=\sigma\cdot x=3.75\cdot 0.6=2.25\text{ V}.$$