AB is a potentiometer wire of length 100cm and its resistance is 10 ohm It is connected in series with a resistance R = 40 ohm and a battery of emf 2 volt and negligible internal resistance. If a source of unknown emf E is balanced by 40cm length of the potentiometer wire, the value of E is

Question

AB is a potentiometer wire of length 100cm and its resistance is 10 ohm It is connected in series with a resistance R = 40 ohm and a battery of emf 2 volt and negligible internal resistance. If a source of unknown emf E is balanced by 40cm length of the potentiometer wire, the value of E is (A) 0.8 V (B) 1.6 V (C) 0.08 (D) 0.16 V

Tardigrade Admin · Accepted Answer

As we known K =(( E / R + r )) ( r / L ) K=((2/10+40)) (10/1) K=(20/50) ∴ E=K ℓ [unknown emf] E=(20/50) × 40 × 10-2=0.16 volt

Q. AB is a potentiometer wire of length 100cm and its resistance is 10 ohm It is connected in series with a resistance $R = 40$ ohm and a battery of emf 2 volt and negligible internal resistance. If a source of unknown emf E is balanced by 40cm length of the potentiometer wire, the value of E is

$0.8 V$

$1.6 V$

$0.08$

$0.16 V$

As we known
$K = (\frac{E}{R + r}) \frac{r}{L}$
$K = (\frac{2}{10 + 40}) \frac{10}{1}$
$K = \frac{20}{50}$
$∴ E = K ℓ$ [unknown emf]
$E = \frac{20}{50} \times 40 \times 1 0^{- 2} = 0.16 v o lt$

Q. AB is a potentiometer wire of length 100cm and its resistance is 10 ohm It is connected in series with a resistance R=40 ohm and a battery of emf 2 volt and negligible internal resistance. If a source of unknown emf E is balanced by 40cm length of the potentiometer wire, the value of E is

0.8V

1.6V

0.08

0.16V