MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Electric Bulbs

The balancing...

Question

The balancing lengths of potentiometer wire are $l_{1}$ and $l_{2}$ when two cells of emf $E_{1}$ and $E_{2}$ are connected in the secondary circuit in series. first to support each other and next to oppose each other. Then the ratio of $\frac{E_{1}}{E_{2}}$ is equal to [if $E_{1} > E_{2}$ ]

A

\frac{l_{1} - l_{2}}{l_{1} + l_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

\frac{l_{1} + l_{2}}{l_{1} - l_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

\frac{l_{1}}{l_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

\frac{l_{2}}{l_{1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{l_{1} + l_{2}}{l_{1} - l_{2}}$
For support each other :

$(E_{1} + E_{2}) \propto l_{1} \Rightarrow E_{1} + E_{2} = x l_{1} . . . . . . (1)$

For Oppose each other :

$(E_{1} - E_{2}) \propto l_{2} \Rightarrow E_{1} - E_{2} = x l_{2} . . . . . (2)$

On dividing (1) and (2),

$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{l_{1}}{l_{2}}$

By componendo and dividendo method.

$\frac{(E_{1} + E_{2}) + (E_{1} - E_{2})}{(E_{1} + E_{2}) - (E_{1} - E_{2})} = \frac{l_{1} + l_{2}}{l_{1} - l_{2}}$

$\frac{2 E_{1}}{2 E_{2}} = \frac{l_{1} + l_{2}}{l_{1} - l_{2}}$

$\frac{E_{1}}{E_{2}} = \frac{l_{1} + l_{2}}{l_{1} - l_{2}}$

 Hence, $(B)$ is the correct answer.

Suggest Corrections

thumbs-up

6

Similar questions

Q. Two cells of emf

E_{1}

E_{2} (E_{2} > E_{1})

are connected in series in a secondary circuit of a potentiometer experiment for determination of emf. The balancing length is found to be 825 cm. Now when the terminals of cell of emf

E_{1}

are reversed, then the balancing length is found to be 225 cm. The ratio of

E_{1}

E_{2}

Q. Two cells of emf

e 1

e 2

are joined in series and the balancing length of the potentiometer. wire is

625 c m

. if the terminals of e1 are reversed the balancing length obtained is

125 c m

e 2 > e 1

e 1 : e 2

Q. Two cells of e.m.f.

E_{1}

E_{2}

are joined in series and balancing length of the potentiometer wire is 625 cm. If the terminals of

E_{1}

are reversed, the balancing length obtained is 125 cm. Given

E_{2} > E_{1}

E_{1} : E_{2}

Q. Two cells of emf

E_{1}

E_{2}

(E_{1}, > E_{2})

are connected as shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300cm. When the same potentiometer is connected between A and C, the balancing length is 100cm. The ratio of

E_{1}

E_{2}

Q. Two cells of emfs

E_{1}

E_{2} (E_{1} > E_{2})

are connected as shown in fig.

When a potentiometer is connected between

A

B

, the balancing length of the potentiometer wire is

300 cm

. On connecting the same potentiometer between

A

C

, the balancing length is

100 cm

\frac{E_{1}}{E_{2}}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Electrical Instruments

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon