Question

A potentiometer has a wire of 100 cm length and its resistance is 10 ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40 cm length of potentiometer wire, the value of E is:

• A. 0.2 V
• B. 0.4 V
• C. 0.08 V
• D. 0.16 V

Answer 1

The correct option is A 0.2 V

The potentiometer has $10\Omega$ resistance at $100cm$ length . at $40cm$ length , it'll have resistance $R_1=\frac{10}{100}\times 40=4\Omega$ Drives cell has emf $E_1$ connected to a secondary circuit having a cell of emf $E_2=2V$ and series resistance $R_2=40\Omega$ At balance point,current through both primary and secondary circuits would be same.

$\Rightarrow I=\frac{E_1}{R_1}=\frac{E_2}{R_2}$ (follows from ohm's law $V=1R$ )

$\Rightarrow \frac{E_1}{4}=\frac{E_2\times 2}{40}\Rightarrow E_2=\frac{2}{40}\times 4=0.2V$