#### Question

A potentiometer wire of length 1 m has a resistance of 5 Ω. It is connected to a 8 V battery in series with a resistance of 15 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.

#### Solution

From the figure:

Total resistance of the circuit, *R* = (*R _{AB}* + 15) Ω = 20 Ω

Current in the circuit ,

\[i = \frac{V}{R} = \frac{8}{20} A\]

∴ Voltage across *AB*, *V _{AB}* =

*i.R*= 2 V

_{AB}The emf of the cell connected as above is given by: \[e = \frac{l}{L} V_0\]

Here:*l* = 60 cm (balance point)

AB* = L* = 1 m = 100 cm (total length of the wire)

\[\therefore e = \frac{60}{100}\left( 2 \right) = 1 . 2 V\]

Is there an error in this question or solution?

Solution A Potentiometer Wire of Length 1 M Has a Resistance of 5 ω. It is Connected to a 8 V Battery in Series with a Resistance of 15 ω. Determine the Emf of the Primary Cell Which Gives a Bala Concept: Potentiometer.