The length of potentiometer wire is $$200$$ cm and the emf of standard cell in primary circuit is E volts. It is employed to a battery of emf $$0.4$$ v the balance point is obtained at $$l=40$$ cm from positive end, the E of the battery is (cell$$ in primary is ideal and series resistance is $$zero)

Question

The length of potentiometer wire is $$200$$ cm     and the emf of standard cell in primary circuit is  E volts.     It is employed to a battery of emf $$0.4$$ v     the balance point is obtained at $$l=40$$ cm     from positive end, the E     of the battery is (cell$$ in primary is ideal and series resistance is  $$zero)

Infinity Learn · Accepted Answer

Length of potentiometer $$L=200$$ cm       Battery of emf $$E1=0.4$$ V     Balancing length $$l=40$$ cm     $$E1=ERPRP+RSlL$$     ⇒$$E=E1Ll$$               $$=0.4$$×$$20040$$             ∴  $$E=2v$$

The length of potentiometer wire is $200 c m$ and the emf of standard cell in primary circuit is $E volts .$ It is employed to a battery of emf $0.4 v$ the balance point is obtained at $l = 40 c m$ from positive end, the $E$ of the battery is (cell in primary is ideal and series resistance is zero)

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The length of potentiometer wire is 200 cm and the emf of standard cell in primary circuit is E volts. It is employed to a battery of emf 0.4 v the balance point is obtained at l=40 cm from positive end, the E of the battery is (cell in primary is ideal and series resistance is zero)

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The length of potentiometer wire is $200 c m$ and the emf of standard cell in primary circuit is $E volts .$ It is employed to a battery of emf $0.4 v$ the balance point is obtained at $l = 40 c m$ from positive end, the $E$ of the battery is (cell in primary is ideal and series resistance is zero)