A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5 Ω. When a resistance of 15 Ω is used for shunting null point moves to 300 cm. The internal resistance of the cell is Ω.

Question

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5 Ω. When a resistance of 15 Ω is used for shunting null point moves to 300 cm. The internal resistance of the cell is Ω. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(ε/ r +5) × 5=200 x .....(1) (ε × 15/ r +15)=300 x ....(2) ⇒ r =5 Ω

Q. A null point is found at $200 c m$ in potentiometer when cell in secondary circuit is shunted by $5 Ω$ . When a resistance of $15 Ω$ is used for shunting null point moves to $300 c m$ . The internal resistance of the cell is ___ $Ω$ .

$\frac{ε}{r + 5} \times 5 = 200 x$ .....(1)
$\frac{ε \times 15}{r + 15} = 300 x$ ....(2)
$\Rightarrow r = 5 Ω$

Q. A null point is found at 200cm in potentiometer when cell in secondary circuit is shunted by 5Ω. When a resistance of 15Ω is used for shunting null point moves to 300cm. The internal resistance of the cell is ___Ω.

r+5ε​×5=200x.....(1) r+15ε×15​=300x....(2) ⇒r=5Ω

Q. A null point is found at $200 c m$ in potentiometer when cell in secondary circuit is shunted by $5 Ω$ . When a resistance of $15 Ω$ is used for shunting null point moves to $300 c m$ . The internal resistance of the cell is ___ $Ω$ .

$\frac{ε}{r + 5} \times 5 = 200 x$ .....(1)
$\frac{ε \times 15}{r + 15} = 300 x$ ....(2)
$\Rightarrow r = 5 Ω$