Question

In a potentiometer circuit, two wires of same material of resistivity , $\rho$ one of radius of cross-section a and other of radius of cross-section 2a are joined in series. They are of length l and 2l respectively. This combination act as potentiometer wire of length 3l. The emf of the cell in primary circuit is E and internal resistance is $\frac{\rho l}{2\pi a^2}$. This cell is connected to the potentiometer wire by a conducting wire of negligible resistance with positive terminal of the cell connected to one end (call it A) of longer wire.

The maximum voltage which can be balanced on the potentiometer wire is

• A. $2E$
• B. $\frac{3E}{4}$
• C. $\frac{3E}{8}$
• D. $\frac{E}{4}$

Answer 1

The correct option is B $\frac{3E}{4}$


Total resistance of the wire (AD) $R=\frac{\rho \left(2l\right)}{4\pi a^2}+\frac{\rho \left(l\right)}{\pi a^2}=\frac{3\rho l}{2\pi a^2}$
Hence pd a cross the wire AD is $V=\frac{3E}{4}$ as R: r = 3:1