Question

Two cells of emf E1 and E2 are to be compared in a potentiometer (E1>E2). when the cells are used in series correctly the balancing length obtained is 400 cm. when they are connected in series but E2 is connected with reverse polarities the balancing length obtained is 200cm. ratio of emf of cells is

Answer 1

Answer:

E₁:E₂ = 3:1

Explanation:

When the cells are connected initially, in series such that their voltages add up, corresponding length= 400cm

In Potentiometer, Voltage across the the wire is given by V=k(L)

  This is because

  V=i(R)

  V=i(ρ)$(\frac{L}{A})$ =K(L) [∵ wires of potentiometer has same cross-sectional area]

So when they are connected in first case, $\boxed{E1+E2 =K(400)} ---(1)$

And in second case,  $\boxed{E1-E2 =K(200)} ---(2)$

$\frac{(1)}{(2)}$ ⇒ $\frac{E1+E2}{E1-E2} = 2$

⇒ E₁ = 3E₂

∴  $\boxed{\frac{E1}{E2} =\frac{3}{1}}$

Hope this answer helped you :)

Answer 2

Physics

5 points

Answer:

Answer

4

4.0

E₁:E₂ = 3:1

Explanation:

When the cells are connected initially, in series such that their voltages add up, corresponding length= 400cm

In Potentiometer, Voltage across the the wire is given by V=k(L)

  This is because

  V=i(R)

=K(L) [∵ wires of potentiometer has same cross-sectional area]

  V=i(ρ)

$(\frac{L}{A})$

first case

So when they are connected in

$\boxed{E1+E2 =K(400)} ---(1)$

,

And in

second case

,  

$\boxed{E1-E2 =K(200)} ---(2)$

⇒

$\frac{(1)}{(2)}$

$\frac{E1+E2}{E1-E2} = 2$

⇒ E₁ = 3E₂

  V=i(R)

  V=i(ρ)

$(\frac{L}{A})$

Physics

5 points

,

⇒

$\frac{(1)}{(2)}$

$\frac{E1+E2}{E1-E2} = 2$

⇒ E₁ = 3E₂

$\boxed{\frac{E1}{E2} =\frac{3}{1}}$

∴  

Hope this answer helped you :)