The Potentiometer wire of length 400 cm has a resistance of $20 \Omega$. It is connected in series with a resistance of $10 \Omega$ and a cell of emf 9 V having negligible resistance. A source of 2.4 V is balanced against a length L of the potentiometer wire. The value of L is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 160 cm
$I_{AB}$, Current in the potentiometer wire AB = $\frac{9}{20+10}=\frac{9}{30}=0.3A$
Voltage ($V_{AB}$) potential difference across AB
$=I_{AB}×R_{AB}$ [By Ohm's law]
$=0.3×20$
$=6V$
∴ Potential gradient (K) = $\frac{V}{I}=\frac{6}{400}=0.015cm^{-1}$
The emf 2.4 V is balanced against L of the potentiometer wire.
$2.4=KL$
$L=\frac{2.4}{0.015}=160cm$