Question

Two sinusoidal sources of emf have rms values E1 andE2 and a phase difference alpha. When connected in series,the resultant voltage 41.1 V. When one of the source isreversed, the resultant emfis 7.52 V. when phasedisplacement is made zero, the resultant emf is 42.5V.Find E., E2 and alpha.​

Answer 1

Answer:

3:1

Explanation:

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})(

A

L

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

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

E1+E2=K(400)

−−−(1)

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

E1−E2=K(200)

−−−(2)

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

(2)

(1)

⇒ \frac{E1+E2}{E1-E2} = 2

E1−E2

E1+E2

=2

⇒ E₁ = 3E₂

∴ \boxed{\frac{E1}{E2} =\frac{3}{1}}

E2

E1

=

1

3

Hope this answer helped you :)